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Graphs
1
A LOOK BACK
Appendix A reviews skills from
Intermediate Algebra.
30-year mortgage rates back under 6%
WASHINGTON (Reuters)—Interest rates on 30-year
mortgages fell under 6% this week, while rates on 15-year
and adjustable mortgages also eased, mortgage finance
company Freddie Mac said Thursday.
Thirty-year mortgage rates averaged 5.99% in the week
ended Aug. 5 compared with 6.08% a week earlier, Freddie Mac said.
Freddie Mac said 15-year mortgages averaged 5.40%,
down from 5.49% last week. One-year adjustable rate
mortgages dipped to an average of 4.08% from 4.17% last
week.
Freddie Mac is a mortgage finance company chartered
by Congress that buys mortgages from lenders and packages them into securities for investors or holds them in
its own portfolio.
USA Today, August 5, 2004.
—See Chapter Project 1.
A LOOK AHEAD In this chapter we make the
connection between algebra and geometry through the
rectangular coordinate system. The idea of using a system of
rectangular coordinates dates back to ancient times, when
such a system was used for surveying and city planning.
Apollonius of Perga, in 200 BC, used a form of rectangular
coordinates in his work on conics, although this use does
not stand out as clearly as it does in modern treatments.
Sporadic use of rectangular coordinates continued until the
1600s. By that time, algebra had developed sufficiently so
that René Descartes (1596–1650) and Pierre de Fermat
(1601–1665) could take the crucial step, which was the use
of rectangular coordinates to translate geometry problems
into algebra problems, and vice versa. This step was
important for two reasons. First, it allowed both geometers
and algebraists to gain new insights into their subjects, which
previously had been regarded as separate, but now were
seen to be connected in many important ways. Second,
these insights made the development of calculus possible,
which greatly enlarged the number of areas in which
mathematics could be applied and made possible a much
deeper understanding of these areas.
With the advent of modern technology, in particular
graphing utilities, now not only are we able to visualize the
dual roles of algebra and geometry, but we are also able to
solve many problems that required advanced methods
before this technology.
OUTLINE
1.1
1.2
1.3
Rectangular Coordinates; Graphing Utilities
1.4
1.5
Lines
Graphs of Equations in Two Variables
Solving Equations in One Variable Using a Graphing
Utility
Circles
Chapter Review Chapter Test Chapter Projects
1
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CHAPTER 1 Graphs
1.1
Rectangular Coordinates; Graphing Utilities
Before getting started, review the following:
PREPARING FOR THIS SECTION
• Algebra Review (Appendix, Section A.1, pp. 951–959)
• Geometry Review (Appendix, Section A.2, pp. 961–964)
Now work the ‘Are You Prepared?’ problems on page 8.
OBJECTIVES
1
2
Figure 1
y
4
2
–4
O
–2
2
4
x
–2
–4
Figure 2
y
4
3
(–3, 1)
1
–4
3
(–2, –3)
3
(3, 2)
2
O
3
x
4
2
(3, –2)
2
Figure 3
y
Quadrant II
x < 0, y > 0
Quadrant I
x > 0, y > 0
Quadrant III
x < 0, y < 0
Quadrant IV
x > 0, y < 0
x
Use the Distance Formula
Use the Midpoint Formula
We locate a point on the real number line by assigning it a single real number, called
the coordinate of the point. For work in a two-dimensional plane, we locate points by
using two numbers.
We begin with two real number lines located in the same plane: one horizontal
and the other vertical. We call the horizontal line the x-axis, the vertical line the
y-axis, and the point of intersection the origin O. See Figure 1. We assign coordinates to every point on these number lines using a convenient scale. In mathematics,
we usually use the same scale on each axis; in applications, a different scale is often
used on each axis.
The origin O has a value of 0 on both the x-axis and y-axis. Points on the x-axis
to the right of O are associated with positive real numbers, and those to the left of O
are associated with negative real numbers. Points on the y-axis above O are associated with positive real numbers, and those below O are associated with negative real
numbers. In Figure 1, the x-axis and y-axis are labeled as x and y, respectively, and
we have used an arrow at the end of each axis to denote the positive direction.
The coordinate system described here is called a rectangular or Cartesian*
coordinate system. The plane formed by the x-axis and y-axis is sometimes called
the xy-plane, and the x-axis and y-axis are referred to as the coordinate axes.
Any point P in the xy-plane can be located by using an ordered pair 1x, y2 of
real numbers. Let x denote the signed distance of P from the y-axis (signed means
that, if P is to the right of the y-axis, then x 7 0, and if P is to the left of the y-axis,
then x 6 0); and let y denote the signed distance of P from the x-axis. The ordered
pair 1x, y2, also called the coordinates of P, then gives us enough information to
locate the point P in the plane.
For example, to locate the point whose coordinates are 1-3, 12, go 3 units along
the x-axis to the left of O and then go straight up 1 unit. We plot this point by placing
a dot at this location. See Figure 2, in which the points with coordinates 1-3, 12,
1-2, - 32, 13, - 22, and 13, 22 are plotted.
The origin has coordinates 10, 02. Any point on the x-axis has coordinates of the
form 1x, 02, and any point on the y-axis has coordinates of the form 10, y2.
If 1x, y2 are the coordinates of a point P, then x is called the x-coordinate, or
abscissa, of P and y is the y-coordinate, or ordinate, of P. We identify the point P by
its coordinates 1x, y2 by writing P = 1x, y2. Usually, we will simply say “the point
1x, y2” rather than “the point whose coordinates are 1x, y2.”
The coordinate axes divide the xy-plane into four sections called quadrants, as
shown in Figure 3. In quadrant I, both the x-coordinate and the y-coordinate of all
points are positive; in quadrant II, x is negative and y is positive; in quadrant III,
both x and y are negative; and in quadrant IV, x is positive and y is negative. Points
on the coordinate axes belong to no quadrant.
NOW WORK PROBLEM
*
11.
Named after René Descartes (1596–1650), a French mathematician, philosopher, and theologian.
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SECTION 1.1 Rectangular Coordinates; Graphing Utilities
3
Graphing Utilities
All graphing utilities, that is, all graphing calculators and all computer software
graphing packages, graph equations by plotting points on a screen. The screen itself
actually consists of small rectangles, called pixels. The more pixels the screen has, the
better the resolution. Most graphing calculators have 48 pixels per square inch; most
computer screens have 32 to 108 pixels per square inch. When a point to be plotted
lies inside a pixel, the pixel is turned on (lights up). The graph of an equation is a collection of lighted pixels. Figure 4 shows how the graph of y = 2x looks on a TI-84
Plus graphing calculator.
The screen of a graphing utility will display the coordinate axes of a rectangular
coordinate system. However, you must set the scale on each axis. You must also include the smallest and largest values of x and y that you want included in the graph.
This is called setting the viewing rectangle or viewing window. Figure 5 illustrates a
typical viewing window.
To select the viewing window, we must give values to the following expressions:
Xmin:
the smallest value of x shown on the viewing window
Xmax:
the largest value of x shown on the viewing window
Xscl:
the number of units per tick mark on the x-axis
Ymin:
the smallest value of y shown on the viewing window
Ymax:
the largest value of y shown on the viewing window
Yscl:
the number of units per tick mark on the y-axis
Figure 6 illustrates these settings and their relation to the Cartesian coordinate
system.
Figure 4
y = 2x
Figure 5
Figure 6
y
Ymax
Yscl
x
Xmin
Xmax
Xscl
Ymin
If the scale used on each axis is known, we can determine the minimum and maximum values of x and y shown on the screen by counting the tick marks. Look again at
Figure 5. For a scale of 1 on each axis, the minimum and maximum values of x are - 10
and 10, respectively; the minimum and maximum values of y are also - 10 and 10. If the
scale is 2 on each axis, then the minimum and maximum values of x are - 20 and 20, respectively; and the minimum and maximum values of y are - 20 and 20, respectively.
Conversely, if we know the minimum and maximum values of x and y, we can
determine the scales being used by counting the tick marks displayed. We shall follow the practice of showing the minimum and maximum values of x and y in our illustrations so that you will know how the window was set. See Figure 7.
Figure 7
4
3
3
means
Xmin = - 3
Xmax = 3
Xscl = 1
4
Ymin = - 4
Ymax = 4
Yscl = 2
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CHAPTER 1 Graphs
EXAMPLE 1
Finding the Coordinates of a Point Shown on a Graphing
Utility Screen
Find the coordinates of the point shown in Figure 8. Assume the coordinates are
integers.
Figure 8
4
3
3
4
Solution
First we note that the viewing window used in Figure 8 is
Xmin = - 3
Xmax = 3
Xscl = 1
Ymin = - 4
Ymax = 4
Yscl = 2
The point shown is 2 tick units to the left on the horizontal axis 1scale = 12 and 1
tick up on the vertical scale 1scale = 22. The coordinates of the point shown are
1- 2, 22.
NOW WORK PROBLEMS
15
AND
25.
1 Use the Distance Formula
✓
If the same units of measurement, such as inches, centimeters, and so on, are used
for both the x-axis and y-axis, then all distances in the xy-plane can be measured
using this unit of measurement.
EXAMPLE 2
Finding the Distance between Two Points
Find the distance d between the points 11, 32 and 15, 62.
Solution
First we plot the points 11, 32 and 15, 62 and connect them with a straight line. See
Figure 9(a). We are looking for the length d. We begin by drawing a horizontal line
from 11, 32 to 15, 32 and a vertical line from 15, 32 to 15, 62, forming a right triangle,
as shown in Figure 9(b). One leg of the triangle is of length 4 (since ƒ 5 - 1 ƒ = 4) and
the other is of length 3 (since ƒ 6 - 3 ƒ = 3). By the Pythagorean Theorem, the square
of the distance d that we seek is
d2 = 4 2 + 32 = 16 + 9 = 25
d = 225 = 5
Figure 9
y
6
(5, 6)
y
6
d
d
3
(5, 6)
3
(1, 3)
6 x
3
(a)
3
(1, 3) 4 (5, 3)
3
(b)
6 x
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SECTION 1.1 Rectangular Coordinates; Graphing Utilities
5
The distance formula provides a straightforward method for computing the distance between two points.
Theorem
Distance Formula
The distance between two points P1 = 1x1 , y12 and P2 = 1x2 , y22, denoted by
d1P1 , P22, is
In Words
To compute the distance
between two points, find the
difference of the x-coordinates,
square it, and add this to the
square of the difference of the
y-coordinates. The square root
of this sum is the distance.
d1P1 , P22 = 41x2 - x122 + 1y2 - y122
(1)
Proof of the Distance Formula Let 1x1 , y12 denote the coordinates of point P1 ,
and let 1x2 , y22 denote the coordinates of point P2 . Assume that the line joining
P1 and P2 is neither horizontal nor vertical. Refer to Figure 10(a). The coordinates of
P3 are 1x2 , y12. The horizontal distance from P1 to P3 is the absolute value of the
difference of the x-coordinates, ƒ x2 - x1 ƒ . The vertical distance from P3 to P2 is the
absolute value of the difference of the y-coordinates, ƒ y2 - y1 ƒ . See Figure 10(b).
The distance d1P1 , P22 that we seek is the length of the hypotenuse of the right
triangle, so, by the Pythagorean Theorem, it follows that
3d1P1 , P2242 = ƒ x2 - x1 ƒ 2 + ƒ y2 - y1 ƒ 2
= 1x2 - x122 + 1y2 - y122
d1P1 , P22 = 41x2 - x122 + 1y2 - y122
Figure 10
y
y
y2
P2 (x2, y2)
y1
P3 (x2, y1)
P1 (x1, y1)
x1
x2
x
P2 (x2, y2)
y2
d(P1, P2)
y1
P1 (x1, y1)
⏐x2 x1⏐
x1
(a)
⏐y2 y1⏐
x2
P3 (x2, y1)
x
(b)
Now, if the line joining P1 and P2 is horizontal, then the y-coordinate of P1
equals the y-coordinate of P2 ; that is, y1 = y2 . Refer to Figure 11(a). In this case, the
distance formula (1) still works, because, for y1 = y2 , it reduces to
d1P1 , P22 = 41x2 - x122 + 02 = 41x2 - x122 = ƒ x2 - x1 ƒ
Figure 11
y
y
y2
P1 (x1, y1)
y1
d (P1, P2)
P2 (x2, y1)
y1
⏐x2 x1⏐
x1
x2
(a)
x
P2 (x1, y2)
⏐y2 y1⏐ d (P1, P2)
P1 (x1, y1)
x1
x
(b)
A similar argument holds if the line joining P1 and P2 is vertical. See Figure 11(b).
The distance formula is valid in all cases.
■
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CHAPTER 1 Graphs
EXAMPLE 3
Finding the Length of a Line Segment
Find the length of the line segment shown in Figure 12.
Figure 12
Solution
The length of the line segment is the distance between the points
P1 = 1x1 , y12 = 1-4, 52, and P2 = 1x2, y22 = 13, 22. Using the distance formula (1)
with x1 = - 4, y1 = 5, x2 = 3, and y2 = 2, the length d is
6
(4, 5)
d = 41x2 - x122 + 1y2 - y122 = 433 - 1- 4242 + 12 - 522
(3, 2)
5
5
= 472 + 1- 322 = 249 + 9 = 258 L 7.62
NOW WORK PROBLEM
0
31.
The distance between two points P1 = 1x1 , y12 and P2 = 1x2 , y22 is never a negative number. Furthermore, the distance between two points is 0 only when the points are
identical, that is, when x1 = x2 and y1 = y2 . Also, because 1x2 - x122 = 1x1 - x222
and 1y2 - y122 = 1y1 - y222, it makes no difference whether the distance is computed
from P1 to P2 or from P2 to P1 ; that is, d1P1 , P22 = d1P2 , P12.
The introduction to this chapter mentioned that rectangular coordinates enable us
to translate geometry problems into algebra problems, and vice versa. The next example shows how algebra (the distance formula) can be used to solve geometry problems.
EXAMPLE 4
Using Algebra to Solve Geometry Problems
Consider the three points A = 1-2, 12, B = 12, 32, and C = 13, 12.
(a) Plot each point and form the triangle ABC.
(b) Find the length of each side of the triangle.
(c) Verify that the triangle is a right triangle.
(d) Find the area of the triangle.
Solution
Figure 13
(b) d1A, B2 = 432 - 1- 2242 + 13 - 122 = 216 + 4 = 220 = 225
y
3
d1B, C2 = 413 - 222 + 11 - 322 = 21 + 4 = 25
B = (2, 3)
A = (–2, 1)
–3
(a) Points A, B, C and triangle ABC are plotted in Figure 13.
C = (3, 1)
3
x
d1A, C2 = 433 - 1- 2242 + 11 - 122 = 225 + 0 = 5
(c) To show that the triangle is a right triangle, we need to show that the sum of the
squares of the lengths of two of the sides equals the square of the length of the
third side. (Why is this sufficient?) Looking at Figure 13, it seems reasonable to
conjecture that the right angle is at vertex B. We shall check to see whether
3d1A, B242 + 3d1B, C242 = 3d1A, C242
We find that
3d1A, B242 + 3d1B, C242 = A 225 B +
2
A 25 B 2
= 20 + 5 = 25 = 3d1A, C242
so it follows from the converse of the Pythagorean Theorem that triangle ABC
is a right triangle.
(d) Because the right angle is at vertex B, the sides AB and BC form the base and
height of the triangle. Its area is
Area =
1
1
1Base21Height2 = A 225 B A 25 B = 5 square units
2
2
NOW WORK PROBLEM
49.
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SECTION 1.1 Rectangular Coordinates; Graphing Utilities
2 Use the Midpoint Formula
✓
Figure 14
y
P2 = (x 2, y2)
y2
M = (x, y )
y
y1
7
y – y1
x – x1
P1 = (x1, y1)
y2 – y
x2 – x
B = (x 2, y )
We now derive a formula for the coordinates of the midpoint of a line segment. Let
P1 = 1x1 , y12 and P2 = 1x2 , y22 be the endpoints of a line segment, and let M = 1x, y2
be the point on the line segment that is the same distance from P1 as it is from P2 . See
Figure 14. The triangles P1 AM and MBP2 are congruent.* [Do you see why? Angle
AP1M = angle BMP2 ,† angle P1 MA = angle MP2B, and d1P1 , M2 = d1M, P22 is
given. So, we have angle–side–angle.] Hence, corresponding sides are equal in length.
That is,
x - x1 = x2 - x
2x = x1 + x2
x1 + x2
x =
2
A = (x, y1)
x1
x
x2
x
Theorem
and
y - y1 = y2 - y
2y = y1 + y2
y1 + y2
y =
2
Midpoint Formula
The midpoint M = 1x, y2 of the line segment from P1 = 1x1 , y12 to
P2 = 1x2 , y22 is
In Words
To find the midpoint of a line
segment, average the
x-coordinates and average the
y-coordinates of the endpoints.
M = 1x, y2 = ¢
EXAMPLE 5
x1 + x2 y1 + y2
,
≤
2
2
(2)
Finding the Midpoint of a Line Segment
Find the midpoint of a line segment from P1 = 1- 5, 52 to P2 = 13, 12. Plot the
points P1 and P2 and their midpoint. Check your answer.
Solution
We apply the midpoint formula (2) using x1 = - 5, y1 = 5, x2 = 3, and y2 = 1.
Then the coordinates 1x, y2 of the midpoint M are
x =
y
M (–1, 3)
–5
and
y =
y1 + y2
5 + 1
=
= 3
2
2
That is, M = 1- 1, 32. See Figure 15.
Figure 15
P1 (–5, 5)
x1 + x2
-5 + 3
=
= -1
2
2
✔ CHECK: Because M is the midpoint, we check the answer by verifying that
d1P1 , M2 = d1M, P22:
5
d1P1 , M2 = 43- 1 - 1-5242 + 13 - 522 = 216 + 4 = 220
P2 (3, 1)
5
d1M, P22 = 433 - 1-1242 + 11 - 322 = 216 + 4 = 220
x
NOW WORK PROBLEM
*
55.
The following statement is a postulate from geometry. Two triangles are congruent if their sides are the
same length (SSS), or if two sides and the included angle are the same (SAS), or if two angles and the included sides are the same (ASA).
†
Another postulate from geometry states that the transversal P1 P2 forms congruent corresponding angles
with the parallel line segments P1A and MB.
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CHAPTER 1 Graphs
c1.1 Assess Your Understanding
»
‘Are You Prepared?’
Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. On the real number line the origin is assigned the number
3. If 3 and 4 are the legs of a right triangle, the hypotenuse is
_____. (p. 954)
_____. (pp. 961–962)
2. If - 3 and 5 are the coordinates of two points on the real
number line, the distance between these points is _____.
(p. 956)
4. Use the converse of the Pythagorean Theorem to show that
a triangle whose sides are of lengths 11, 60, and 61 is a right
triangle. (p. 962)
Concepts and Vocabulary
5. If 1x, y2 are the coordinates of a point P in the xy-plane,
then x is called the _____ of P and y is the _____ of P.
8. True or False: The distance between two points is sometimes
a negative number.
6. The coordinate axes divide the xy-plane into four sections
called _____.
9. True or False: The point 1-1, 42 lies in quadrant IV of the
Cartesian plane.
7. If three distinct points P, Q, and R all lie on a line and if
d1P, Q2 = d1Q, R2, then Q is called the _____ of the line
segment from P to R.
10. True or False: The midpoint of a line segment is found by averaging the x-coordinates and averaging the y-coordinates
of the endpoints.
Skill Building
In Problems 11 and 12, plot each point in the xy-plane. Tell in which quadrant or on what coordinate axis each point lies.
11. (a)
(b)
(c)
(d)
(e)
(f)
A
B
C
D
E
F
=
=
=
=
=
=
1- 3, 22
16, 02
1 - 2, - 22
16, 52
10, - 32
16, -32
12. (a)
(b)
(c)
(d)
(e)
(f)
A
B
C
D
E
F
=
=
=
=
=
=
11, 42
1- 3, -42
1-3, 42
14, 12
10, 12
1-3, 02
13. Plot the points 12, 02, 12, -32, 12, 42, 12, 12, and 12, - 12. Describe the set of all points of the form 12, y2, where y is a real number.
14. Plot the points 10, 32, 11, 32, 1- 2, 32, 15, 32 and 1 - 4, 32. Describe the set of all points of the form 1x, 32, where x is a real number.
In Problems 15–18, determine the coordinates of the points shown. Tell in which quadrant each point lies. Assume the coordinates are integers.
15.
16.
10
5
5
17.
10
5
10
18.
5
5
5
10
10
5
10
10
5
10
In Problems 19–24, select a setting so that each given point will lie within the viewing window.
19. 1 -10, 52, 13, -22, 14, -12
22. 1 -80, 602, 120, -302, 1- 20, -402
20. 15, 02, 16, 82, 1-2, -32
23. 10, 02, 1100, 52, 15, 1502
In Problems 25–30, determine the viewing window used.
25.
26.
4
6
6
24. 10, - 12, 1100, 502, 1 -10, 302
27.
2
3
21. 140, 202, 1- 20, -802, 110, 402
3
3
6
4
2
6
1
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SECTION 1.1 Rectangular Coordinates; Graphing Utilities
28.
29.
4
9
30.
10
9
3
12
P2 = (–2, 1) 2 P = (0, 0)
1
2 P = (2, 1)
2
P1 = (0, 0)
–1
2
22
9
10
4
2
In Problems 31–44, find the distance d1P1 , P22 between the points P1 and P2 .
31.
32.
33.
y
y
–2
8
x
–2
2
–1
P2 (–2, 2)
x
34.
y
2
P1 = (–1, 1) 2
P1 (1, 1)
2 x
–2
–2
35. P1 = 13, - 42; P2 = 15, 42
36. P1 = 1-1, 02; P2 = 12, 42
39. P1 = 14, - 32; P2 = 16, 42
40. P1 = 1-4, -32; P2 = 16, 22
37. P1 = 1 - 3, 22; P2 = 16, 02
y
P2 = (2, 2)
–1
2
x
38. P1 = 12, -32; P2 = 14, 22
41. P1 = 1 - 0.2, 0.32; P2 = 12.3, 1.12
42. P1 = 11.2, 2.32; P2 = 1-0.3, 1.12
43. P1 = 1a, b2; P2 = 10, 02
44. P1 = 1a, a2; P2 = 10, 02
In Problems 45–48, find the length of the line segment. Assume that the endpoints of each line segment have integer coordinates.
45.
46.
47.
48.
12
12
6
18
6
6
12
18
12
12
6
12
12
12
6
6
In Problems 49–54, plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area.
49. A = 1-2, 52; B = 11, 32; C = 1-1, 02
50. A = 1- 2, 52; B = 112, 32; C = 110, -112
51. A = 1 -5, 32; B = 16, 02; C = 15, 52
53. A = 14, -32; B = 10, - 32; C = 14, 22
52. A = 1- 6, 32; B = 13, -52; C = 1 -1, 52
54. A = 14, -32; B = 14, 12; C = 12, 12
In Problems 55–64, find the midpoint of the line segment joining the points P1 and P2 .
55. P1 = 13, - 42; P2 = 15, 42
56. P1 = 1-2, 02; P2 = 12, 42
59. P1 = 14, - 32; P2 = 16, 12
60. P1 = 1-4, -32; P2 = 12, 22
57. P1 = 1 - 3, 22; P2 = 16, 02
61. P1 = 1 - 0.2, 0.32; P2 = 12.3, 1.12
63. P1 = 1a, b2; P2 = 10, 02
58. P1 = 12, -32; P2 = 14, 22
62. P1 = 11.2, 2.32; P2 = 1-0.3, 1.12
64. P1 = 1a, a2; P2 = 10, 02
Applications and Extensions
65. Find all points having an x-coordinate of 2 whose distance
from the point 1 - 2, - 12 is 5.
66. Find all points having a y-coordinate of -3 whose distance
from the point 11, 22 is 13.
69. The medians of a triangle are the line segments from each
vertex to the midpoint of the opposite side (see the figure).
Find the lengths of the medians of the triangle with vertices
at A = 10, 02, B = 16, 02, and C = 14, 42.
C
Median
67. Find all points on the x-axis that are 5 units from the point
14, - 32.
68. Find all points on the y-axis that are 5 units from the point
14, 42.
Midpoint
A
B
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70. An equilateral triangle is one in which all three sides are of
equal length. If two vertices of an equilateral triangle are
10, 42 and 10, 02, find the third vertex. How many of these
triangles are possible?
s
s
s
71. Geometry Find the midpoint of each diagonal of a square
with side of length s. Draw the conclusion that the diagonals
of a square intersect at their midpoints.
[Hint: Use (0, 0), (0, s), (s, 0), and (s, s) as the vertices of the
square.]
a 23 a
72. Geometry Verify that the points (0, 0), (a, 0) a ,
b
2
2
are the vertices of an equilateral triangle. Then show that
the midpoints of the three sides are the vertices of a second
equilateral triangle (refer to Problem 70).
In Problems 73–76, find the length of each side of the triangle determined by the three points P1 , P2 , and P3 . State whether the triangle is
an isosceles triangle, a right triangle, neither of these, or both. (An
isosceles triangle is one in which at least two of the sides are of
equal length.)
73. P1 = 12, 12; P2 = 1- 4, 12; P3 = 1-4, -32
78. Little League Baseball The layout of a Little League playing field is a square, 60 feet on a side.* How far is it directly
from home plate to second base (the diagonal of the square)?
*SOURCE: Little League Baseball, Official Regulations and
Playing Rules, 2005.
79. Baseball Refer to Problem 77. Overlay a rectangular coordinate system on a major league baseball diamond so that
the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis
lies in the direction from home plate to third base.
(a) What are the coordinates of first base, second base, and
third base? Use feet as the unit of measurement.
(b) If the right fielder is located at 1310, 152, how far is it
from there to second base?
(c) If the center fielder is located at 1300, 3002, how far is it
from there to third base?
80. Little League Baseball Refer to Problem 78. Overlay a
rectangular coordinate system on a Little League baseball
diamond so that the origin is at home plate, the positive
x-axis lies in the direction from home plate to first base, and
the positive y-axis lies in the direction from home plate to
third base.
(a) What are the coordinates of first base, second base, and
third base? Use feet as the unit of measurement.
(b) If the right fielder is located at 1180, 202, how far is it
from there to second base?
(c) If the center fielder is located at 1220, 2202, how far is it
from there to third base?
81. A Dodge Neon and a Mack truck leave an intersection at
the same time. The Neon heads east at an average speed of
30 miles per hour, while the truck heads south at an average
speed of 40 miles per hour. Find an expression for their distance apart d (in miles) at the end of t hours.
74. P1 = 1 - 1, 42; P2 = 16, 22; P3 = 14, -52
75. P1 = 1 - 2, -12; P2 = 10, 72; P3 = 13, 22
76. P1 = 17, 22; P2 = 1- 4, 02; P3 = 14, 62
77. Baseball A major league baseball “diamond” is actually a
square, 90 feet on a side (see the figure). What is the distance directly from home plate to second base (the diagonal
of the square)?
82. A hot-air balloon, headed due east at an average speed of 15
miles per hour and at a constant altitude of 100 feet, passes
over an intersection (see the figure). Find an expression for
the distance d (measured in feet) from the balloon to the intersection t seconds later.
15 mph
2nd base
100 ft
90 ft
Pitching
rubber
3rd base
1st base
90 ft
Home plate
‘Are You Prepared?’ Answers
1. 0
2. 8
3. 5
4. 112 + 602 = 612
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SECTION 1.2
1.2
Graphs of Equations in Two Variables
11
Graphs of Equations in Two Variables
Before getting started, review the following:
PREPARING FOR THIS SECTION
• Linear Equations (Appendix, Section A.5, pp. 986–988)
• Quadratic Equations (Appendix, Section A.5, pp. 988–995)
Now work the ‘Are You Prepared?’ problems on page 21.
OBJECTIVES
1
2
3
4
5
6
7
8
Graph Equations by Hand by Plotting Points
Graph Equations Using a Graphing Utility
Use a Graphing Utility to Create Tables
Find Intercepts from a Graph
Find Intercepts from an Equation
Use a Graphing Utility to Approximate Intercepts
Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin
Know How to Graph Key Equations
1 Graph Equations by Hand by Plotting Points
✓
An equation in two variables, say x and y, is a statement in which two expressions involving x and y are equal.The expressions are called the sides of the equation. Since an
equation is a statement, it may be true or false, depending on the value of the variables.
Any values of x and y that result in a true statement are said to satisfy the equation.
For example, the following are all equations in two variables x and y:
x2 + y2 = 5
2x - y = 6
y = 2x + 5
x2 = y
The first of these, x2 + y2 = 5, is satisfied for x = 1, y = 2, since
12 + 2 2 = 1 + 4 = 5. Other choices of x and y also satisfy this equation. It is not
satisfied for x = 2 and y = 3, since 2 2 + 32 = 4 + 9 = 13 Z 5.
The graph of an equation in two variables x and y consists of the set of points in
the xy-plane whose coordinates 1x, y2 satisfy the equation.
EXAMPLE 1
Determining Whether a Point Is on the Graph of an Equation
Determine if the following points are on the graph of the equation 2x - y = 6.
(a) 12, 32
Solution
(b) 12, -22
(a) For the point 12, 32, we check to see if x = 2, y = 3 satisfies the equation
2x - y = 6.
2x - y = 2122 - 3 = 4 - 3 = 1 Z 6
The equation is not satisfied, so the point 12, 32 is not on the graph.
(b) For the point 12, -22, we have
2x - y = 2122 - 1- 22 = 4 + 2 = 6
The equation is satisfied, so the point 12, -22 is on the graph.
NOW WORK PROBLEM
21.
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EXAMPLE 2
Graphing an Equation by Hand by Plotting Points
Graph the equation:
Solution
Figure 16
y
25
(0, 5)
– 25 (–5, – 5)
(10, 25)
(1, 7)
25 x
y = 2x + 5
We want to find all points 1x, y2 that satisfy the equation. To locate some of these
points (and thus get an idea of the pattern of the graph), we assign some numbers to
x and find corresponding values for y.
If
x = 0
Then
y = 2102 + 5 = 5
x = 1
y = 2112 + 5 = 7
x = -5
y = 21-52 + 5 = - 5
x = 10
y = 21102 + 5 = 25
Point on Graph
10, 52
11, 72
1- 5, -52
110, 252
– 25
By plotting these points and then connecting them, we obtain the graph of the
equation (a line), as shown in Figure 16.
EXAMPLE 3
Graphing an Equation by Hand by Plotting Points
Graph the equation:
Solution
Table 1
y = x2
Table 1 provides several points on the graph. In Figure 17 we plot these points and
connect them with a smooth curve to obtain the graph (a parabola).
x
y x2
(x, y)
-4
16
(- 4, 16)
-3
9
( - 3, 9)
-2
4
(- 2, 4)
-1
1
( - 1, 1)
0
0
(0, 0)
1
1
(1, 1)
2
4
(2, 4)
3
9
(3, 9)
4
16
(4, 16)
Figure 17
y
20
(– 4, 16)
(–3, 9)
(4, 16)
15
10
(3, 9)
5
(–2, 4)
(2, 4)
(1, 1)
(–1, 1)
(0, 0)
–4
4
x
The graphs of the equations shown in Figures 16 and 17 do not show all points.
For example, in Figure 16, the point 120, 452 is a part of the graph of y = 2x + 5,
but it is not shown. Since the graph of y = 2x + 5 could be extended out as far as
we please, we use arrows to indicate that the pattern shown continues. It is important when illustrating a graph to present enough of the graph so that any viewer of
the illustration will “see” the rest of it as an obvious continuation of what is actually
there. This is referred to as a complete graph.
One way to obtain a complete graph of an equation is to plot a sufficient number of points on the graph until a pattern becomes evident. Then these points are
connected with a smooth curve following the suggested pattern. But how many
points are sufficient? Sometimes knowledge about the equation tells us. For exam-
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SECTION 1.2
Graphs of Equations in Two Variables
13
ple, we will learn in Section 1.4 that, if an equation is of the form y = mx + b, then
its graph is a line. In this case, only two points are needed to obtain the graph.
One purpose of this book is to investigate the properties of equations in order
to decide whether a graph is complete. Sometimes we shall graph equations by plotting points. Shortly, we shall investigate various techniques that will enable us to
graph an equation without plotting so many points. Other times we shall graph
equations using a graphing utility.
2 Graph Equations Using a Graphing Utility
✓
From Examples 2 and 3, we see that a graph can be obtained by plotting points in a
rectangular coordinate system and connecting them. Graphing utilities perform
these same steps when graphing an equation. For example, the TI-84 Plus determines 95 evenly spaced input values,* uses the equation to determine the output values, plots these points on the screen, and finally (if in the connected mode) draws a
line between consecutive points.
To graph an equation in two variables x and y using a graphing utility requires
that the equation be written in the form y = 5expression in x6. If the original equation is not in this form, replace it by equivalent equations until the form
y = 5expression in x6 is obtained. Most graphing utilities require the following
steps.
Steps for Graphing an Equation Using a Graphing Utility
STEP 1: Solve the equation for y in terms of x.
STEP 2: Get into the graphing mode of your graphing utility. The screen will
, prompting you to enter the expression inusually display Y =
volving x that you found in Step 1. (Consult your manual for the correct way to enter the expression; for example, y = x 2 might be
entered as x^2 or as x*x or as xxY2).
STEP 3: Select the viewing window. Without prior knowledge about the behavior of the graph of the equation, it is common to select the
standard viewing window† initially. The viewing window is then adjusted based on the graph that appears. In this text, the standard viewing window will be
Xmin = - 10
Ymin = - 10
Xmax = 10
Ymax = 10
Xscl = 1
Yscl = 1
STEP 4: Graph.
STEP 5: Adjust the viewing window until a complete graph is obtained.
*
These input values depend on the values of Xmin and Xmax. For example, if Xmin = -10 and Xmax = 10,
then the first input value will be -10 and the next input value will be - 10 + 110 - 1 -1022>94 = - 9.7872,
and so on.
†
Some graphing utilities have a ZOOM-STANDARD feature that automatically sets the viewing
window to the standard viewing window and graphs the equation.
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CHAPTER 1 Graphs
EXAMPLE 4
Graphing an Equation on a Graphing Utility
Graph the equation:
6x2 + 3y = 36
STEP 1: We solve for y in terms of x.
6x2 + 3y = 36
3y = - 6x2 + 36
y = - 2x2 + 12
Subtract 6x2 from both sides of the equation.
Divide both sides of the equation by 3 and simplify.
STEP 2: From the graphing mode, enter the expression - 2x2 + 12 after the prompt
Y =
. Figure 18 shows the expression entered on a TI-84 Plus.
STEP 3: Set the viewing window to the standard viewing window.
STEP 4: Graph. The screen should look like Figure 19.
STEP 5: The graph of y = - 2x 2 + 12 is not complete. The value of Ymax must be
increased so that the top portion of the graph is visible. After increasing the
value of Ymax to 12, we obtain the graph in Figure 20.* The graph is now
complete.
Figure 18
Figure 19
Figure 20
12
10
10
10
10
10
10
10
3 Use a Graphing Utility to Create Tables
✓
In addition to graphing equations, graphing utilities can also be used to create a
table of values that satisfy the equation. This feature is especially useful in determining an appropriate viewing window when graphing an equation. Many graphing
utilities require the following steps to create a table.
Steps for Creating a Table of Values Using a Graphing Utility
STEP 1: Solve the equation for y in terms of x.
STEP 2: Enter the expression in x following the Y = prompt of the graphing
utility.
STEP 3: Set up the table. Graphing utilities typically have two modes for
creating tables. In the AUTO mode, the user determines a starting
point for the table (TblStart) and ¢Tbl (pronounced “delta-table”).
The ¢Tbl feature determines the increment for x. The ASK mode
requires the user to enter values of x and then the utility determines
the corresponding value of y.
STEP 4: Create the table. The user can scroll within the table if the table was
created in AUTO mode.
*
Some graphing utilities have a ZOOM-FIT feature that determines the appropriate Ymin and Ymax for
a given Xmin and Xmax. Consult your owner’s manual for the appropriate keystrokes.
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SECTION 1.2
EXAMPLE 5
Graphs of Equations in Two Variables
15
Creating a Table Using a Graphing Utility
Create a table that displays the points on the graph of 6x2 + 3y = 36 for
x = - 3, - 2, - 1, 0, 1, 2, and 3.
Solution
Table 2
STEP 1: We solved the equation for y in Example 4 and obtained y = - 2x2 + 12.
STEP 2: Enter the expression in x following the Y = prompt.
STEP 3: We set up the table in the AUTO mode with TblStart = - 3 and
¢Tbl = 1.
STEP 4: Create the table. The screen should look like Table 2.
In looking at Table 2, we notice that y = 12 when x = 0. This information
could have been used to help to create the initial viewing window by letting us know
that Ymax needs to be at least 12 in order to get a complete graph.
4 Find Intercepts from a Graph
✓
The points, if any, at which a graph crosses or touches the coordinate axes are called
the intercepts. See Figure 21. The x-coordinate of a point at which the graph crosses
or touches the x-axis is an x-intercept, and the y-coordinate of a point at which the
graph crosses or touches the y-axis is a y-intercept. For a graph to be complete, all its
intercepts must be displayed.
Figure 21
Graph
crosses
y -axis
y
Graph
crosses
x-axis
x
Graph
touches
x-axis
EXAMPLE 6
Intercepts
Finding Intercepts from a Graph
Find the intercepts of the graph in Figure 22. What are its x-intercepts? What are its
y-intercepts?
Solution
Figure 22
y
4
1-3, 02, 10, 32,
(0, 3)
( 3–2 , 0)
4 (3, 0)
The x-intercepts are - 3,
(4.5, 0)
3
a , 0b,
2
4
a0, - b, 10, -3.52, 14.5, 02
3
3
4
, and 4.5; the y-intercepts are - 3.5, - , and 3.
2
3
5 x
(0,
(0, 3.5)
The intercepts of the graph are the points
4
–3
)
In Example 6, you should notice the following usage: If we do not specify the
type of intercept (x- versus y-), then we report the intercept as an ordered pair.
However, if we specify the type of intercept, then we only report the coordinate of
the intercept. For x-intercepts, we report the x-coordinate of the intercept; for
y-intercepts, we report the y-coordinate of the intercept.
NOW WORK PROBLEM
27(a).
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CHAPTER 1 Graphs
5 Find Intercepts from an Equation
✓
The intercepts of a graph can be found from its equation by using the fact that
points on the x-axis have y-coordinates equal to 0 and points on the y-axis have
x-coordinates equal to 0.
Procedure for Finding Intercepts
1. To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in
the equation and solve for x.
2. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in
the equation and solve for y.
Because the x-intercepts of the graph of an equation are those x-values for
which y = 0, they are also called the zeros (or roots) of the equation.
EXAMPLE 7
Finding Intercepts from an Equation
Find the x-intercept(s) and the y-intercept(s) of the graph of y = x2 - 4.
Solution
To find the x-intercept(s), we let y = 0 and obtain the equation
x2 - 4 = 0
1x + 221x - 22 = 0
x - 2 = 0
x + 2 = 0
or
x = - 2 or
x = 2
Factor.
Zero-Product Property
The equation has two solutions, -2 and 2. The x-intercepts (or zeros) are -2
and 2.
To find the y-intercept(s), we let x = 0 in the equation.
y = x2 - 4
= 02 - 4 = - 4
The y-intercept is - 4.
Sometimes the TABLE feature of a graphing utility will reveal the intercepts of
an equation. Table 3 shows a table for the equation y = x2 - 4. Can you find the intercepts in the table?
Table 3
NOW WORK PROBLEM
43.
6 Use a Graphing Utility to Approximate Intercepts
✓
We can use a graphing utility to approximate the intercepts of the graph of an
equation, as illustrated in the next example.
EXAMPLE 8
Finding Intercepts Using a Graphing Utility
Use a graphing utility to approximate the intercepts of the equation y = x3 - 16.
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SECTION 1.2
Graphs of Equations in Two Variables
17
Figure 23(a) shows the graph of y = x3 - 16.
The eVALUEate feature of a TI-84 Plus graphing calculator accepts as input a
value of x and determines the value of y. If we let x = 0, we find that the y-intercept
is - 16. See Figure 23(b).
The ZERO feature of a TI-84 Plus is used to find the x-intercept(s). See Figure
23(c). Rounded to two decimal places, the x-intercept is 2.52.
Solution
Figure 23
10
–5
5
–25
(a)
(b)
NOW WORK PROBLEM
(c)
53.
7 Test an Equation for Symmetry
✓
We just saw the role that intercepts play in obtaining key points on the graph of an
equation. Another helpful tool for graphing equations by hand involves symmetry,
particularly symmetry with respect to the x-axis, the y-axis, and the origin.
A graph is said to be symmetric with respect to the x-axis if, for every point
1x, y2 on the graph, the point 1x, -y2 is also on the graph.
A graph is said to be symmetric with respect to the y-axis if, for every point
1x, y2 on the graph, the point 1-x, y2 is also on the graph.
A graph is said to be symmetric with respect to the origin if, for every point
1x, y2 on the graph, the point 1-x, -y2 is also on the graph.
Figure 24 illustrates the definition. Notice that, when a graph is symmetric with
respect to the x-axis, the part of the graph above the x-axis is a reflection or mirror
image of the part below it, and vice versa. And when a graph is symmetric with respect to the y-axis, the part of the graph to the right of the y-axis is a reflection of the
part to the left of it, and vice versa. Symmetry with respect to the origin may be
viewed in two ways:
1. As a reflection about the y-axis, followed by a reflection about the x-axis
2. As a projection along a line through the origin so that the distances from the
origin are equal
Figure 24
y
y
y
(–x, y )
(x, y )
(x, y )
(x, y )
(x, y )
(x, –y )
(x, y )
(x, y )
x
(x, –y )
(–x, y )
(x, y )
x
x
(– x, – y )
(x, –y )
(– x, – y )
Symmetry with respect
to the x-axis
Symmetry with respect
to the y-axis
Symmetry with respect
to the origin
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CHAPTER 1 Graphs
EXAMPLE 9
Symmetric Points
(a) If a graph is symmetric with respect to the x-axis and the point 14, 22 is on the
graph, then the point 14, -22 is also on the graph.
(b) If a graph is symmetric with respect to the y-axis and the point 14, 22 is on the
graph, then the point 1- 4, 22 is also on the graph.
(c) If a graph is symmetric with respect to the origin and the point 14, 22 is on the
graph, then the point 1-4, - 22 is also on the graph.
NOW WORK PROBLEM
11.
When the graph of an equation is symmetric with respect to the x-axis, the
y-axis, or the origin, the number of points that you need to plot in order to see the
pattern is reduced. For example, if the graph of an equation is symmetric with respect to the y-axis, then, once points to the right of the y-axis are plotted, an equal
number of points on the graph can be obtained by reflecting them about the y-axis.
Because of this, before we graph an equation, we first want to determine whether it
has any symmetry. The following tests are used for this purpose.
Tests for Symmetry
To test the graph of an equation for symmetry with respect to the
Replace y by - y in the equation. If an equivalent equation results,
the graph of the equation is symmetric with respect to the x-axis.
Replace x by - x in the equation. If an equivalent equation results,
the graph of the equation is symmetric with respect to the y-axis.
Replace x by -x and y by -y in the equation. If an equivalent
equation results, the graph of the equation is symmetric with
respect to the origin.
x-Axis
y-Axis
Origin
EXAMPLE 10
Testing an Equation for Symmetry
4x2
for symmetry.
x + 1
x-Axis: To test for symmetry with respect to the x-axis, replace y by -y. Since
4x2
4x2
is not equivalent to y = 2
-y = 2
, the graph of the
x + 1
x + 1
equation is not symmetric with respect to the x-axis.
y-Axis: To test for symmetry with respect to the y-axis, replace x by - x. Since
41-x22
4x2
4x2
is equivalent to y = 2
y =
= 2
, the graph of
2
1-x2 + 1
x + 1
x + 1
the equation is symmetric with respect to the y-axis.
Origin: To test for symmetry with respect to the origin, replace x by -x and y
by -y.
41-x22
-y =
Replace x by - x and y by - y.
1-x22 + 1
Test y =
Solution
2
4x2
x2 + 1
4x2
y = - 2
x + 1
-y =
Simplify.
Multiply both sides by - 1.
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SECTION 1.2
Graphs of Equations in Two Variables
19
Since the result is not equivalent to the original equation, the graph of
4x2
the equation y = 2
is not symmetric with respect to the origin.
x +1
Figure 25
Seeing the Concept
4x2
Figure 25 shows the graph of y = 2
using a graphing utility. Do you see the symmetry with
x + 1
respect to the y-axis?
5
⫺5
5
NOW WORK PROBLEMS
27(b)
AND
63.
8 Know How to Graph Key Equations
✓
⫺5
The next three examples use intercepts, symmetry, and point plotting to obtain the
graphs of key equations. It is important to know the graphs of these key equations
because we use them later. The first of these is y = x3.
EXAMPLE 11
Graphing the Equation y x3 by Finding Intercepts and
Checking for Symmetry
Graph the equation y = x3 by hand by plotting points. Find any intercepts and
check for symmetry first.
Solution
First, we seek the intercepts. When x = 0, then y = 0; and when y = 0, then x = 0.
The origin 10, 02 is the only intercept. Now we test for symmetry.
x-Axis:
y-Axis:
Origin:
Figure 26
Replace y by -y. Since - y = x3 is not equivalent to y = x3, the graph
is not symmetric with respect to the x-axis.
Replace x by -x. Since y = 1-x23 = - x3 is not equivalent to y = x3,
the graph is not symmetric with respect to the y-axis.
Replace x by - x and y by -y. Since -y = 1-x23 = - x3 is equivalent
to y = x3 (multiply both sides by - 1), the graph is symmetric with
respect to the origin.
y
8
(0, 0)
6
(1, 1)
(2, 8)
8
To graph by hand, we use the equation to obtain several points on the graph.
Because of the symmetry, we only need to locate points on the graph for which
x Ú 0. See Table 4. Points on the graph could also be obtained using the TABLE
feature on a graphing utility. See Table 5. Do you see the symmetry with respect to
the origin from the table? Figure 26 shows the graph.
(2, 8)
Table 4
(1, 1)
6
x
Table 5
x
y x3
(x, y)
0
0
(0, 0)
1
1
(1, 1)
2
8
(2, 8)
3
27
(3, 27)
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CHAPTER 1 Graphs
EXAMPLE 12
Graphing the Equation x y2
Graph the equation x = y2. Find any intercepts and check for symmetry first.
Solution
Table 6
y
x y2
(x, y)
0
0
(0, 0)
1
1
(1, 1)
2
4
(4, 2)
3
9
(9, 3)
The lone intercept is 10, 02. The graph is symmetric with respect to the x-axis since
x = 1-y22 is equivalent to x = y2. The graph is not symmetric with respect to the
y-axis or the origin.
To graph x = y2 by hand, we use the equation to obtain several points on the
graph. Because the equation is solved for x, it is easier to assign values to y and use
the equation to determine the corresponding values of x. See Table 6. Because of the
symmetry, we can restrict ourselves to points whose y-coordinates are positive. We
then use the symmetry to find additional points on the graph. For example, since
11, 12 is on the graph, so is 11, - 12. Since 14, 22 is on the graph, so is 14, - 22, and so
on. We plot these points and connect them with a smooth curve to obtain Figure 27.
To graph the equation x = y2 using a graphing utility, we must write the equation in the form y = 5expression in x6. We proceed to solve for y.
Figure 27
x = y2
y
6
y2 = x
y = ; 1x
(9, 3)
(1, 1)
(4, 2)
(0, 0)
2
(1, 1)
5
(4, 2)
10 x
(9, 3)
Square Root Method
To graph x = y2, we need to graph both Y1 = 1x and Y2 = - 1x on the same
screen. Figure 28 shows the result. Table 7 shows various values of y for a given
value of x when Y1 = 1x and Y2 = - 1x. Notice that when x 6 0 we get an error.
Can you explain why?
Figure 28
6
Table 7
Y1 x
2
10
6
Y2 x
Look again at either Figure 27 or 28. If we restrict y so that y Ú 0, the equation
x = y2, y Ú 0, may be written as y = 1x. The portion of the graph of x = y2 in
quadrant I is the graph of y = 1x. See Figure 29.
Figure 29
6
y
6
Y1 x
(1, 1)
(4, 2)
(9, 3)
(0, 0)
2
10
6
2
5
10 x
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SECTION 1.2
EXAMPLE 13
Graphing the Equation y Graph the equation y =
Solution
Table 8
1
x
x
y 1
10
10
1
3
3
1
a , 3b
3
1
2
2
1
a , 2b
2
1
1
(1, 1)
2
1
2
a 2,
1
b
2
3
1
3
a 3,
1
b
3
10
1
10
a 10,
(x, y)
a
1
, 10 b
10
1
b
10
Graphs of Equations in Two Variables
21
1
x
1
. Find any intercepts and check for symmetry first.
x
We check for intercepts first. If we let x = 0, we obtain a 0 in the denominator,
which is not defined. We conclude that there is no y-intercept. If we let y = 0, we get
1
the equation = 0, which has no solution. We conclude that there is no x-intercept.
x
1
The graph of y = does not cross or touch the coordinate axes.
x
Next we check for symmetry.
1
1
x-Axis: Replacing y by - y yields -y = , which is not equivalent to y = .
x
x
1
1
y-Axis: Replacing x by - x yields y =
= - , which is not equivalent to
-x
x
1
y = .
x
1
Origin: Replacing x by - x and y by - y yields -y = - , which is equivalent to
x
1
y = . The graph is symmetric only with respect to the origin.
x
We can use the equation to form Table 8 and obtain some points on the graph.
Because of symmetry, we only find points 1x, y2 for which x is positive. From Table 8
1
we infer that, if x is a large and positive number, then y = is a positive number
x
1
close to 0. We also infer that if x is a positive number close to 0, then y = is a large
x
and positive number. Armed with this information, we can graph the equation.
1
Figure 30 illustrates some of these points and the graph of y = . Observe how
x
the absence of intercepts and the existence of symmetry with respect to the origin
were utilized. Figure 31 confirms our algebraic analysis using a TI-84 Plus.
Figure 30
Figure 31
y
3
Y1 1x
(––12 , 2)
(1, 1)
3
(2, ––12 )
3
(2,
1
––
2
4
x
3
3
)
(1, 1)
( ––12 , 2)
4
3
1.2 Assess Your Understanding
‘Are You Prepared?’
Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Solve the equation 21x + 32 - 1 = - 7. (p. 986)
2. Solve the equation 2x2 - 9x - 5 = 0. (pp. 988–995)
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CHAPTER 1 Graphs
Concepts and Vocabulary
3. The points, if any, at which a graph crosses or touches the coordinate axes are called _____.
4. Because the x-intercepts of the graph of an equation are
those x-values for which y = 0, they are also called _____ or
_____.
5. If for every point 1x, y2 on the graph of an equation the
point 1 -x, y2 is also on the graph, then the graph is symmetric with respect to the _____.
6. If the graph of an equation is symmetric with respect to the
y-axis and -4 is an x-intercept of this graph, then _____ is
also an x-intercept.
7. If the graph of an equation is symmetric with respect to the
origin and 13, -42 is a point on the graph, then _____ is also
a point on the graph.
8. True or False: To find the y-intercepts of the graph of an
equation, let x = 0 and solve for y.
9. True or False: To graph an equation using a graphing utility,
we must first solve for x in terms of y.
10. True or False: If a graph is symmetric with respect to the
x-axis, then it cannot be symmetric with respect to the
y-axis.
Skill Building
In Problems 11–20, plot each point. Then plot the point that is symmetric to it with respect to (a) the x-axis; (b) the y-axis; (c) the origin.
11. 13, 42
12. 15, 32
16. 1 -1, -12
13. 1 - 2, 12
17. 1-3, -42
14. 14, -22
18. 14, 02
15. 15, -22
19. 10, -32
20. 1 -3, 02
In Problems 21–26, tell whether the given points are on the graph of the equation.
21. Equation: y = x 4 - 1x
Points: 10, 02; 11, 12; 1- 1, 02
22. Equation: y = x 3 - 21x
Points: 10, 02; 11, 12; 11, - 12
24. Equation: y 3 = x + 1
Points: 11, 22; 10, 12; 1-1, 02
23. Equation: y 2 = x2 + 9
Points: 10, 32; 13, 02; 1- 3, 02
25. Equation: x2 + y2 = 4
Points: 10, 22; 1- 2, 22; A 22, 22 B
26. Equation: x2 + 4y2 = 4
Points:
1
10, 12; 12, 02; a2, b
2
In Problems 27–34, the graph of an equation is given. (a) Find the intercepts. (b) Indicate whether the graph is symmetric with respect
to the x-axis, the y-axis, or the origin.
27.
28.
29.
30.
y
y
y
y
3
3
3
3x
–3
––
3x
2
––
2
1
x
3
32.
y
3
3x
–3
–3
3 x
4
3
–3
31.
4
1
33.
y
3
y
40
x
3x
3
3x
–3
34.
y
6
6
6
6
–3
40
In Problems 35–38, draw a complete graph so that it has the type of symmetry indicated.
35. y-axis
36. x-axis
y
9
37. Origin
y
y
5
4
(0, 2)
(5, 3)
(0, 0)
(–4, 0)
5x
–5
9x
(2, –5)
–9
–9
(0, –9)
38. y-axis
–5
y
(0, 4)
4
, 2)
(––
2
(2, 2)
(, 0)
x
(0, 0)
3x
–3
–4
–2
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SECTION 1.2
Graphs of Equations in Two Variables
23
In Problems 39–50, determine the intercepts algebraically and graph each equation by plotting points. Be sure to label the intercepts. Verify
your results using a graphing utility.
39. y = x + 2
40. y = x - 6
41. y = 2x + 8
42. y = 3x - 9
43. y = x2 - 1
44. y = x2 - 9
45. y = - x2 + 4
46. y = - x2 + 1
47. 2x + 3y = 6
48. 5x + 2y = 10
49. 9x2 + 4y = 36
50. 4x2 + y = 4
In Problems 51–58, graph each equation using a graphing utility. Use a graphing utility to approximate the intercepts rounded to two decimal places. Use the TABLE feature to help to establish the viewing window.
51. y = 2x - 13
52. y = - 3x + 14
53. y = 2x2 - 15
54. y = - 3x2 + 19
55. 3x - 2y = 43
56. 4x + 5y = 82
57. 5x2 + 3y = 37
58. 2x2 - 3y = 35
In Problems 59–74, list the intercepts and test for symmetry. Verify your results using a graphing utility.
59. y2 = x + 4
60. y2 = x + 9
61. y = 1
3x
62. y = 1
5x
63. x2 + y - 9 = 0
64. y2 - x - 4 = 0
65. 9x2 + 4y2 = 36
66. 4x2 + y2 = 4
67. y = x3 - 27
68. y = x4 - 1
69. y = x2 - 3x - 4
70. y = x2 + 4
71. y =
3x
72. y =
2
x + 9
x2 - 4
2x
73. y =
-x3
2
x - 9
74. y =
x4 + 1
2x5
In Problems 75–78, draw a quick sketch of each equation.
75. y = x3
76. x = y2
79. If 13, b2 is a point on the graph of y = 4x + 1, what is b?
81. If 1a, 42 is a point on the graph of y = x2 + 3x, what is a?
77. y = 1x
78. y =
1
x
80. If 1- 2, b2 is a point on the graph of 2x + 3y = 2, what is b?
82. If 1a, - 52 is a point on the graph of y = x2 + 6x, what is a?
Discussion and Writing
In Problem 83, use a graphing utility.
83. (a) Graph y = 3x2 , y = x, y = ƒ x ƒ , and y = 11x22, noting which graphs are the same.
(b) Explain why the graphs of y = 3x2 and y = ƒ x ƒ are
the same.
(c) Explain why the graphs of y = x and y = 11x22 are
not the same.
(d) Explain why the graphs of y = 3x2 and y = x are not
the same.
84. Explain what is meant by a complete graph.
85. What is the standard viewing window?
86. Draw a graph of an equation that contains two x-intercepts;
at one the graph crosses the x-axis, and at the other the
graph touches the x-axis.
‘Are You Prepared?’ Answers
1. 5 - 66
1
2. e - , 5 f
2
87. Make up an equation with the intercepts 12, 02, 14, 02, and
10, 12. Compare your equation with a friend’s equation.
Comment on any similarities.
88. Draw a graph that contains the points 1 - 2, - 12, 10, 12,
11, 32, and 13, 52. Compare your graph with those of other
students. Are most of the graphs almost straight lines? How
many are “curved”? Discuss the various ways that these
points might be connected.
89. An equation is being tested for symmetry with respect to
the x-axis, the y-axis, and the origin. Explain why, if two of
these symmetries are present, the remaining one must also
be present.
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CHAPTER 1 Graphs
1.3
Solving Equations in One Variable Using a Graphing Utility
Before getting started, review the following:
PREPARING FOR THIS SECTION
• Linear, Rational, and Quadratic Equations (Appendix, Section A.5, pp. 984–997)
Now work the ‘Are You Prepared?’ problems on page 26.
OBJECTIVE
1
Solve Equations in One Variable Using a Graphing Utility
1 Solve Equations in One Variable Using a Graphing Utility
✓
In this text, we present two methods for solving equations: algebraic and graphical.We
shall see as we proceed through this book that some equations can be solved using algebraic techniques that result in exact solutions. For other equations, however, there
are no algebraic techniques that lead to an exact solution. For such equations, a graphing utility can often be used to investigate possible solutions. When a graphing utility
is used to solve an equation, usually approximate solutions are obtained.
One goal of this text is to determine when equations can be solved algebraically. If
an algebraic method for solving an equation exists, we shall use it to obtain an exact solution. A graphing utility can then be used to support the algebraic result. However, if
an equation must be solved for which no algebraic techniques are available, a graphing
utility will be used to obtain approximate solutions. Unless otherwise stated, we shall
follow the practice of giving approximate solutions rounded to two decimal places.
The ZERO (or ROOT) feature of a graphing utility can be used to find the solutions of an equation when one side of the equation is 0. In using this feature to
solve equations, we make use of the fact that the x-intercepts (or zeros) of the graph
of an equation are found by letting y = 0 and solving the equation for x. Solving an
equation for x when one side of the equation is 0 is equivalent to finding where the
graph of the corresponding equation crosses or touches the x-axis.
EXAMPLE 1
Using ZERO (or ROOT) to Approximate Solutions of an Equation
Find the solution(s) of the equation x3 - x + 1 = 0. Round answers to two decimal places.
Solution
The solutions of the equation x3 - x + 1 = 0 are the same as the x-intercepts of
the graph of Y1 = x3 - x + 1. We begin by graphing the equation.
Figure 32 shows the graph. From the graph there appears to be one x-intercept
(solution to the equation) between - 2 and -1.
Using the ZERO (or ROOT) feature of our graphing utility, we determine that
the x-intercept, and thus the solution to the equation, is x = - 1.32, rounded to two
decimal places. See Figure 33.
Figure 32
Figure 33
10
–3
10
–3
3
–5
NOW WORK PROBLEM
3
–5
5.
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SECTION 1.3
Solving Equations in One Variable Using a Graphing Utility
25
A second method for solving equations using a graphing utility involves the
INTERSECT feature of the graphing utility. This feature is used most effectively
when one side of the equation is not 0.
EXAMPLE 2
Using INTERSECT to Approximate Solutions of an Equation
Find the solution(s) to the equation 4x4 - 3 = 2x + 1. Round answers to two decimal places.
Solution
Figure 34
10
–4
4
–10
We begin by graphing each side of the equation as follows: graph Y1 = 4x4 - 3 and
Y2 = 2x + 1. See Figure 34. At the point of intersection of the graphs, the value of
the y-coordinate is the same. Thus, the x-coordinate of the point of intersection represents the solution to the equation. Do you see why?
The INTERSECT feature on a graphing utility determines the point of intersection of the graphs. Using this feature, we find that the graphs intersect at
1-0.87, - 0.732 and 11.12, 3.232, rounded to two decimal places. See Figures 35(a)
and (b). The solutions of the equation are x = - 0.87 and x = 1.12, rounded to two
decimal places.
Figure 35
10
–4
10
4 –4
–10
(a)
4
–10
(b)
NOW WORK PROBLEM
7.
The steps to follow for solving equations graphically are given next.
Steps for Solving Equations Graphically Using ZERO (or ROOT)
STEP 1: Write the equation in the form 5expression in x6 = 0.
STEP 2: Graph Y1 = 5expression in x6.
STEP 3: Use ZERO (or ROOT) to determine each x-intercept of the graph.
Steps for Solving Equations Graphically Using INTERSECT
STEP 1: Graph Y1 = 5expression in x on the left side of the equation6;
Y2 = 5expression in x on the right side of the equation6.
STEP 2: Use INTERSECT to determine the x-coordinate of each point of
intersection.
In the next example, we solve an equation both algebraically and graphically.
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CHAPTER 1 Graphs
EXAMPLE 3
Solving an Equation Algebraically and Graphically
Solve the equation:
31x - 22 = 51x - 12
Algebraic Solution
31x - 22
3x - 6
3x - 6 - 5x
-2x - 6
-2x - 6 + 6
-2x
-2 x
-2
51x - 12
5x - 5
5x - 5 - 5x
-5
-5 + 6
1
1
=
-2
1
x = 2
=
=
=
=
=
=
Graphing Solution
Graph Y1 = 31x - 22 and Y2 = 51x - 12.
See Figure 36. Using INTERSECT, we find
the point of intersection to be 1 -0.5, -7.52.
The solution of the equation is x = - 0.5.
Remove the parentheses.
Subtract 5x from each side.
Simplify.
Add 6 to each side.
Simplify.
Figure 36
Divide each side by -2.
5
–5
Simplify.
1
in the expression in x on the left side of the
2
1
equation and simplify. Let x = - in the expression in x on the right
2
side of the equation and simplify. If the two expressions are equal, the solution checks.
5
✔ CHECK: Let x = -
–15
1
5
15
- 2b = 3a - b = 2
2
2
1
3
15
51x - 12 = 5a - - 1b = 5a - b = 2
2
2
31x - 22 = 3a -
Since the two expressions are equal, the solution x = -
1
checks.
2
NOW WORK PROBLEM
19.
1.3 Assess Your Understanding
‘Are You Prepared?’
Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
2. Solve the equation 2x + 3 = 41x - 12 + 1. (p. 986)
1. Solve the equation 2x2 + 5x + 2 = 0. (pp. 984–997)
Concepts and Vocabulary
3. To solve an equation of the form 5expression in x6 = 0
using a graphing utility, we graph Y1 = 5expression in x6
and use _____ to determine each x-intercept of the graph.
4. True or False: In using a graphing utility to solve an equation, exact solutions are always obtained.
Skill Building
In Problems 5–16, use a graphing utility to approximate the real solutions, if any, of each equation rounded to two decimal places. All solutions lie between - 10 and 10.
5. x3 - 4x + 2 = 0
4
2
8. - x + 1 = 2x - 3
11. - x3 -
5 2
7
x + x + 2 = 0
3
2
6. x3 - 8x + 1 = 0
4
3
9. x - 2x + 3x - 1 = 0
12. -x 4 + 3x3 +
7 2
15
x x + 2 = 0
3
2
7. -2x4 + 5 = 3x - 2
10. 3x4 - x3 + 4x2 - 5 = 0
2
5
2
1
13. - x4 - 2x3 + x = - x2 +
3
2
3
2
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SECTION 1.4 Lines
14.
1 3
1
x - 5x = x2 - 4
4
5
15. x4 - 5x2 + 2x + 11 = 0
27
16. - 3x4 + 8x2 - 2x - 9 = 0
In Problems 17–36, solve each equation algebraically. Verify your solution using a graphing utility.
17. 213 + 2x2 = 31x - 42
x + 1
x + 2
+
= 5
21.
3
7
25. 1x + 721x - 12 = 1x + 122
29. 3x2 = 4x + 4
18. 312 - x2 = 2x - 1
2x + 1
+ 16 = 3x
22.
3
26. 1x + 221x - 32 = 1x - 322
30. 5x2 = 13x + 6
33. 2x + 1 = 4
34. 2x - 2 = 3
19. 8x - 12x + 12 = 3x - 13
5
4
+
= 3
23.
y
y
27. x2 - 3x - 28 = 0
31. x3 + x2 - 4x - 4 = 0
2
3
-8
35.
+
=
x + 2
x - 1
5
20. 5 - 12x - 12 = 10 - x
18
4
- 5 =
24.
y
2y
28. x2 - 7x - 18 = 0
32. x3 + 2x2 - 9x - 18 = 0
1
5
21
36.
=
x + 1
x - 4
4
‘Are You Prepared?’ Answers
1
1. e -2, - f
2
1.4
2. 536
Lines
OBJECTIVES
1
2
3
4
5
6
7
8
9
10
Calculate and Interpret the Slope of a Line
Graph Lines Given a Point and the Slope
Find the Equation of a Vertical Line
Use the Point–Slope Form of a Line; Identify Horizontal Lines
Find the Equation of a Line Given Two Points
Write the Equation of a Line in Slope–Intercept Form
Identify the Slope and y-Intercept of a Line from Its Equation
Graph Lines Written in General Form Using Intercepts
Find Equations of Parallel Lines
Find Equations of Perpendicular Lines
In this section we study a certain type of equation that contains two variables, called
a linear equation, and its graph, a line.
1 Calculate and Interpret the Slope of a Line
✓
Figure 37
Line
Rise
Run
Consider the staircase illustrated in Figure 37. Each step contains exactly the same
horizontal run and the same vertical rise. The ratio of the rise to the run, called the
slope, is a numerical measure of the steepness of the staircase. For example, if the
run is increased and the rise remains the same, the staircase becomes less steep. If
the run is kept the same, but the rise is increased, the staircase becomes more steep.
This important characteristic of a line is best defined using rectangular coordinates.
Let P = 1x1 , y12 and Q = 1x2 , y22 be two distinct points. If x1 Z x2 , the slope m
of the nonvertical line L containing P and Q is defined by the formula
m =
y2 - y1
,
x2 - x1
x1 Z x2
(1)
If x1 = x2 , L is a vertical line and the slope m of L is undefined (since this results in division by 0).
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CHAPTER 1 Graphs
Figure 38(a) provides an illustration of the slope of a nonvertical line; Figure 38(b)
illustrates a vertical line.
Figure 38
L
L
Q = (x 2, y2)
y2
Q = (x 1, y2)
y1
P = (x 1, y1)
Rise = y2 – y1
P = (x 1, y1)
y1
y2
Run = x2 – x1
x1
x2
(a) Slope of L is m =
x1
y_______
2 – y1
x2 – x1
(b) Slope is undefined; L is vertical
As Figure 38(a) illustrates, the slope m of a nonvertical line may be viewed as
m =
y2 - y1
Rise
=
x2 - x1
Run
We can also express the slope m of a nonvertical line as
m =
Change in y
y2 - y1
¢y
=
=
x2 - x1
Change in x
¢x
That is, the slope m of a nonvertical line L measures the amount that y changes as x
changes from x1 to x2 . This is called the average rate of change of y with respect to x.
Two comments about computing the slope of a nonvertical line may prove helpful:
1. Any two distinct points on the line can be used to compute the slope of the line.
(See Figure 39 for justification.)
Figure 39
Triangles ABC and PQR are similar
(equal angles). Hence, ratios of corresponding sides are proportional so that
y2 - y1
Slope using P and Q =
x2 - x1
d(B, C)
=
= Slope using A and B
d(A, C)
y
Q = (x 2, y2)
y2 – y1
P = (x 1, y1)
B
x2 – x1
R
x
A
C
2. The slope of a line may be computed from P = 1x1 , y12 to Q = 1x2 , y22 or from
Q to P because
y1 - y2
y2 - y1
=
x2 - x1
x1 - x2
EXAMPLE 1
Finding and Interpreting the Slope of a
Line Containing Two Points
The slope m of the line containing the points 11, 22 and 15, -32 may be computed as
m =
-3 - 2
-5
5
=
= 5 - 1
4
4
or as m =
2 - 1-32
5
5
=
= 1 - 5
-4
4
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SECTION 1.4 Lines
29
For every 4-unit change in x, y will change by -5 units. That is, if x increases by 4
units, then y will decrease by 5 units. The average rate of change of y with respect to
5
x is - .
4
Figure 40
7
NOW WORK PROBLEMS
(4, 4)
4
AND
13.
Square Screens
4
4
(4, 4)
To get an undistorted view of slope, the same scale must be used on each axis.
However, most graphing utilities have a rectangular screen. Because of this, using
the same interval for both x and y will result in a distorted view. For example,
Figure 40 shows the graph of the line y = x connecting the points 1- 4, -42 and
14, 42. We expect the line to bisect the first and third quadrants, but it doesn’t. We
need to adjust the selections for Xmin, Xmax, Ymin, and Ymax so that a square
screen results. On most graphing utilities, this is accomplished by setting the ratio
of x to y at 3 : 2.*
Figure 41 shows the graph of the line y = x on a square screen using a TI-84
Plus. Notice that the line now bisects the first and third quadrants. Compare this illustration to Figure 40.
To get a better idea of the meaning of the slope m of a line, consider the following:
4
Figure 41
(4, 4)
4
6
6
(4, 4)
Seeing the Concept
4
On the same square screen, graph the following equations:
Figure 42
Y6 6x
Y5 2x
Y4 x
Y3 12 x
3
3
Y1 0
2
1
Slope of line is .
4
1
x
2
1
Slope of line is .
2
Y4 = x
Slope of line is 1.
Y5 = 2x
Slope of line is 2.
Y6 = 6x
Slope of line is 6.
See Figure 42.
Seeing the Concept
On the same square screen, graph the following equations:
Figure 43
Y6 6x
Y5 2x
Y4 x
Y3 12 x
Y2 14 x
Slope of line is 0.
1
Y2 = x
4
Y3 =
Y2 14 x
2
Y1 = 0
2
3
3
Y1 = 0
Slope of line is 0.
1
Y2 = - x
4
1
Slope of line is - .
4
1
Y3 = - x
2
Y4 = - x
1
Slope of line is - .
2
Slope of line is -1.
Y5 = - 2x
Slope of line is -2.
Y6 = - 6x
Slope of line is - 6.
See Figure 43.
Y1 0
2
*
Most graphing utilities have a feature that automatically squares the viewing window. Consult your
owner’s manual for the appropriate keystrokes.
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Figures 42 and 43 illustrate the following facts:
1. When the slope of a line is positive, the line slants upward from left to right.
2. When the slope of a line is negative, the line slants downward from left to right.
3. When the slope is 0, the line is horizontal.
Figures 42 and 43 also illustrate that the closer the line is to the vertical position,
the greater the magnitude of the slope.
2 Graph Lines Given a Point and the Slope
✓
The next example illustrates how the slope of a line can be used to graph the line.
EXAMPLE 2
Graphing a Line Given a Point and a Slope
Draw a graph of the line that contains the point 13, 22 and has a slope of:
(a)
Solution
Figure 44
3
Slope =
4
y
6
(7, 5)
3
4
(b) -
4
5
Rise
3
. The fact that the slope is
means that for every horizontal
Run
4
movement (run) of 4 units to the right there will be a vertical movement (rise)
of 3 units. If we start at the given point 13, 22 and move 4 units to the right and 3
units up, we reach the point 17, 52. By drawing the line through this point and
the point 13, 22, we have the graph. See Figure 44.
(b) The fact that the slope is
(a) Slope =
Rise = 3
(3, 2)
-
Run = 4
–2
5
4
-4
Rise
=
=
5
5
Run
10 x
means that for every horizontal movement of 5 units to the right there will be a
corresponding vertical movement of - 4 units (a downward movement). If we
start at the given point 13, 22 and move 5 units to the right and then 4 units
down, we arrive at the point 18, - 22. By drawing the line through these points,
we have the graph. See Figure 45.
Figure 45
4
Slope = 5
y
(–2, 6)
6
Rise = 4
(3, 2) Run = 5
Run = –5
Rise = – 4
10 x
–2
–2
(8, –2)
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SECTION 1.4 Lines
31
Alternatively, we can set
-
4
4
Rise
=
=
5
-5
Run
so that for every horizontal movement of - 5 units (a movement to the left) there
will be a corresponding vertical movement of 4 units (upward). This approach
brings us to the point 1-2, 62, which is also on the graph shown in Figure 45. NOW WORK PROBLEM
19.
3 Find the Equation of a Vertical Line
✓
Now that we have discussed the slope of a line, we are ready to derive equations of
lines. As we shall see, there are several forms of the equation of a line. Let’s start
with an example.
EXAMPLE 3
Graphing a Line
Graph the equation:
Solution
x = 3
To graph x = 3 by hand, recall that we are looking for all points 1x, y2 in the plane
for which x = 3. No matter what y-coordinate is used, the corresponding
x-coordinate always equals 3. Consequently, the graph of the equation x = 3 is a
vertical line with x-intercept 3 and undefined slope. See Figure 46(a).
To use a graphing utility, we need to express the equation in the form
y = 5expression in x6. But x = 3 cannot be put into this form, so an alternative
method must be used. Consult your manual to determine the methodology required
to draw vertical lines. Figure 46(b) shows the graph that you should obtain.
Figure 46
x = 3
y
4
(3, 3)
x3
5
(3, 2)
(3, 1)
1
1
(3, 0)
(3, 1)
5 x
2
6
5
(b)
(a)
As suggested by Example 3, we have the following result:
Theorem
Equation of a Vertical Line
A vertical line is given by an equation of the form
x = a
where a is the x-intercept.
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Figure 47
y
L
4 Use the Point–Slope Form of a Line;
✓
(x, y )
y – y1
(x 1, y1)
x – x1
Identify Horizontal Lines
Now let L be a nonvertical line with slope m and containing the point 1x1 , y12. See
Figure 47. For any other point 1x, y2 on L, we have
m =
x
Theorem
or y - y1 = m1x - x12
y - y1
x - x1
Point–Slope Form of an Equation of a Line
An equation of a nonvertical line of slope m that contains the point 1x1 , y12 is
y - y1 = m1x - x12
EXAMPLE 4
(2)
Using the Point–Slope Form of a Line
An equation of the line with slope 4 and containing the point 11, 22 can be found by
using the point–slope form with m = 4, x1 = 1, and y1 = 2.
y - y1 = m1x - x12
y - 2 = 41x - 12
y = 4x - 2
m = 4, x1 = 1, y1 = 2
See Figure 48 for the graph.
Figure 48
y
6
(2, 6)
8
Y1 4x 2
Rise 4
(1, 2)
Run 1
2
8
10 x
5
10
4
EXAMPLE 5
Finding the Equation of a Horizontal Line
Find an equation of the horizontal line containing the point 13, 22.
Solution
Because all the y-values are equal on a horizontal line, the slope of a horizontal line
is 0. To get an equation, we use the point–slope form with m = 0, x1 = 3, and y1 = 2.
y - y1
y - 2
y - 2
y
See Figure 49 for the graph.
=
=
=
=
m1x - x12
0 # 1x - 32
0
2
m = 0, x1 = 3, and y1 = 2
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SECTION 1.4 Lines
Figure 49
33
y
4
y2
(3, 2)
5
Y1 2
1
5
x
5
5
1
5
As suggested by Example 5, we have the following result:
Theorem
Equation of a Horizontal Line
A horizontal line is given by an equation of the form
y = b
where b is the y-intercept.
5 Find the Equation of a Line Given Two Points
✓
We use the slope formula and the point–slope form of a line to find the equation
given two points.
EXAMPLE 6
Finding an Equation of a Line Given Two Points
Find an equation of the line L containing the points 12, 32 and 1-4, 52. Graph the
line L.
Solution
Figure 50
y
(– 4, 5)
We first compute the slope of the line.
5 - 3
1
2
m =
=
= -4 - 2
-6
3
1
We use the point 12, 32 and the slope m = - to get the point–slope form of the
3
equation of the line.
(2, 3)
1
y - 3 = - 1x - 22
3
2
–4
–2
10
x
See Figure 50 for the graph.
In the solution to Example 6, we could have used the other point, 1-4, 52, instead of the point 12, 32. The equation that results, although it looks different, is
equivalent to the equation that we obtained in the example. (Try it for yourself.)
NOW WORK PROBLEM
33.
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6 Write the Equation of a Line in Slope–Intercept Form
✓
Another useful equation of a line is obtained when the slope m and y-intercept b are
known. In this event, we know both the slope m of the line and a point 10, b2 on the
line; then we may use the point–slope form, equation (2), to obtain the following
equation:
y - b = m1x - 02 or y = mx + b
Theorem
Slope–Intercept Form of an Equation of a Line
An equation of a line L with slope m and y-intercept b is
y = mx + b
(3)
Figure 51 y = mx + 2
Y5 3x 2
Y3 x 2
4
Y4 3x 2
Y2 x 2
Seeing the Concept
To see the role that the slope m plays, graph the following lines on the same square screen.
Y1 2
6
Y1 = 2
6
Y2 = x + 2
Y3 = - x + 2
Y4 = 3x + 2
4
Figure 52
Y5 = - 3x + 2
See Figure 51. What do you conclude about the lines y = mx + 2?
y = 2x + b
Y4 2x 4
4
Y2 2x 1
Y1 2x
Y3 2x 1
Y5 2x 4
Seeing the Concept
To see the role of the y-intercept b, graph the following lines on the same square screen.
Y1 = 2x
Y2 = 2x + 1
6
Y3 = 2x - 1
6
Y4 = 2x + 4
Y5 = 2x - 4
4
See Figure 52. What do you conclude about the lines y = 2x + b?
7 Identify the Slope and y-Intercept of a Line from Its Equation
✓
When the equation of a line is written in slope–intercept form, it is easy to find the
slope m and y-intercept b of the line. For example, suppose that the equation of a
line is
y = - 2x + 3
Compare it to y = mx + b.
y = -2x + 3
y =
q
q
mx + b
The slope of this line is -2 and its y-intercept is 3.
NOW WORK PROBLEM
65.
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EXAMPLE 7
35
Finding the Slope and y-Intercept
Find the slope m and y-intercept b of the equation 2x + 4y = 8. Graph the equation.
Solution
To obtain the slope and y-intercept, we transform the equation into its slope–intercept form by solving for y.
2x + 4y = 8
4y = - 2x + 8
1
y = - x + 2
2
Figure 53
2x + 4y = 8
y
4
(0, 2)
–3
1
The coefficient of x, - , is the slope, and the y-intercept is 2. We can graph the
2
1
line using the fact that the y-intercept is 2 and the slope is - . Then, starting at
2
the point 10, 22, go to the right 2 units and then down 1 unit to the point 12, 12.
See Figure 53.
2
1
(2, 1)
3
y = mx + b
x
NOW WORK PROBLEM
71.
8 Graph Lines Written in General Form Using Intercepts
✓
The form of the equation of the line in Example 7, 2x + 4y = 8, is called the
general form.
The equation of a line L is in general form when it is written as
Ax + By = C
(4)
where A, B, and C are real numbers and A and B are not both 0.
When we want to graph an equation that is written in general form, we can solve
the equation for y and write the equation in slope–intercept form as we did in Example 7. Another approach to graphing the equation would be to find its intercepts.
Remember, the intercepts of the graph of an equation are the points where the
graph crosses or touches a coordinate axis.
EXAMPLE 8
Graphing an Equation in General Form Using Its Intercepts
Graph the equation 2x + 4y = 8 by finding its intercepts.
Solution
To obtain the x-intercept, let y = 0 in the equation and solve for x.
2x + 4y
2x + 4102
2x
x
=
=
=
=
8
8
8
4
Let y = 0.
Divide both sides by 2.
The x-intercept is 4 and the point 14, 02 is on the graph of the equation.
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To obtain the y-intercept, let x = 0 in the equation and solve for y.
Figure 54
y
4
2x + 4y
2102 + 4y
4y
y
(0, 2)
(4, 0)
–3
3
x
=
=
=
=
8
8
8
2
Let x = 0.
Divide both sides by 4.
The y-intercept is 2 and the point 10, 22 is on the graph of the equation.
We plot the points 14, 02 and 10, 22 in a Cartesian plane and draw a line through
the points. See Figure 54.
NOW WORK PROBLEM
85.
Every line has an equation that is equivalent to an equation written in general
form. For example, a vertical line whose equation is
x = a
can be written in the general form
1#x + 0#y = a
A = 1, B = 0, C = a
A horizontal line whose equation is
y = b
can be written in the general form
0#x + 1#y = b
A = 0, B = 1, C = b
Lines that are neither vertical nor horizontal have general equations of the form
Ax + By = C
A Z 0 and B Z 0
Because the equation of every line can be written in general form, any equation
equivalent to equation (4) is called a linear equation.
9 Find Equations of Parallel Lines
✓
When two lines (in the plane) do not intersect (that is, they have no points in common), they are said to be parallel. Look at Figure 55. There we have drawn two lines
and have constructed two right triangles by drawing sides parallel to the coordinate
axes. These lines are parallel if and only if the right triangles are similar. (Do you see
why? Two angles are equal.) And the triangles are similar if and only if the ratios of
corresponding sides are equal.
Figure 55
The distinct lines are parallel if and
only if their slopes are equal.
y
Rise
Run
Rise
Run
x
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SECTION 1.4 Lines
37
This suggests the following result:
Theorem
Criterion for Parallel Lines
Two nonvertical lines are parallel if and only if their slopes are equal and they
have different y-intercepts.
The use of the words “if and only if” in the preceding theorem means that actually two statements are being made, one the converse of the other.
If two nonvertical lines are parallel, then their slopes are equal and they have
different y-intercepts.
If two nonvertical lines have equal slopes and they have different y-intercepts,
then they are parallel.
EXAMPLE 9
Showing That Two Lines Are Parallel
Show that the lines given by the following equations are parallel:
L1 :
Solution
2x + 3y = 6,
L2 :
4x + 6y = 0
To determine whether these lines have equal slopes and different y-intercepts, we
write each equation in slope–intercept form:
L1 :
L2 :
2x + 3y = 6
4x + 6y = 0
6y = - 4x
3y = - 2x + 6
2
y = - x + 2
3
2
Slope = - ; y-intercept = 2
3
2
y = - x
3
2
Slope = - ; y-intercept = 0
3
2
Because these lines have the same slope, - , but different y-intercepts, the lines are
3
parallel. See Figure 56.
Figure 56
Parallel lines
y
5
4
6
6
L1
4
EXAMPLE 10
5
L2
5 x
L1
L2
Finding a Line That Is Parallel to a Given Line
Find an equation for the line that contains the point 12, -32 and is parallel to the
line 2x + y = 6.
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Solution
Since the two lines are to be parallel, the slope of the line that we seek equals the
slope of the line 2x + y = 6. We begin by writing the equation of the line
2x + y = 6 in slope–intercept form.
2x + y = 6
y = - 2x + 6
The slope is -2. Since the line that we seek contains the point 12, -32, we use
the point–slope form to obtain
y - y1 = m1x - x12
Point–slope form
y - 1- 32 = - 21x - 22
m = - 2, x1 = 2, y1 = - 3
y + 3 = - 2x + 4
Simplify
y = - 2x + 1
Slope–intercept form
2x + y = 1
General form
This line is parallel to the line 2x + y = 6 and contains the point 12, -32. See
Figure 57.
y
Figure 57
Y1 2x 1
8
6
6
Y2 2x 6
12
6
6 x
2x y 6
4
Figure 58
Perpendicular lines
(2, 3)
5
2x y 1
y
NOW WORK PROBLEM
10
✓
90°
x
Theorem
53.
Find Equations of Perpendicular Lines
When two lines intersect at a right angle (90°), they are said to be perpendicular. See
Figure 58.
The following result gives a condition, in terms of their slopes, for two lines to
be perpendicular.
Criterion for Perpendicular Lines
Two nonvertical lines are perpendicular if and only if the product of their
slopes is -1.
Here we shall prove the “only if” part of the statement:
If two nonvertical lines are perpendicular, then the product of their slopes is -1.
You are asked to prove the “if” part of the theorem; that is:
If two nonvertical lines have slopes whose product is - 1, then the lines are perpendicular.
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SECTION 1.4 Lines
Figure 59
y
Slope m2
A = (1, m2)
Slope m1
Rise = m 2
x
Run = 1
O
1
Rise = m1
B = (1, m1)
39
Proof Let m1 and m2 denote the slopes of the two lines. There is no loss in generality (that is, neither the angle nor the slopes are affected) if we situate the lines so
that they meet at the origin. See Figure 59. The point A = 11, m22 is on the line having slope m2 , and the point B = 11, m12 is on the line having slope m1 . (Do you see
why this must be true?)
Suppose that the lines are perpendicular. Then triangle OAB is a right triangle.
As a result of the Pythagorean Theorem, it follows that
3d1O, A242 + 3d1O, B242 = 3d1A, B242
(5)
By the distance formula, we can write the squares of these distances as
3d1O, A242 = 11 - 022 + 1m2 - 022 = 1 + m22
3d1O, B242 = 11 - 022 + 1m1 - 022 = 1 + m21
3d1A, B242 = 11 - 122 + 1m2 - m122 = m22 - 2m1m2 + m21
Using these facts in equation (5), we get
11 + m222 + 11 + m212 = m22 - 2m1m2 + m21
which, upon simplification, can be written as
m1 m2 = - 1
If the lines are perpendicular, the product of their slopes is -1.
■
You may find it easier to remember the condition for two nonvertical lines to be
perpendicular by observing that the equality m1 m2 = - 1 means that m1 and m2 are
1
1
.
negative reciprocals of each other; that is, either m1 = or m2 = m2
m1
EXAMPLE 11
Finding the Slope of a Line Perpendicular to Another Line
3
2
If a line has slope , any line having slope - is perpendicular to it.
2
3
EXAMPLE 12
Finding the Equation of a Line Perpendicular to a Given Line
Find an equation of the line that contains the point 11, -22 and is perpendicular to
the line x + 3y = 6. Graph the two lines.
Solution
We first write the equation of the given line in slope–intercept form to find its slope.
x + 3y = 6
Proceed to solve for y.
3y = - x + 6
1
Place in the form y = mx + b.
y = - x + 2
3
1
The given line has slope - . Any line perpendicular to this line will have slope 3.
3
Because we require the point 11, -22 to be on this line with slope 3, we use the
point–slope form of the equation of a line.
y - y1 = m1x - x12
y - 1- 22 = 31x - 12
Point–slope form
m = 3, x1 = 1, y1 = - 2.
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To obtain other forms of the equation, we proceed as follows:
y + 2
y + 2
y
3x - y
=
=
=
=
31x - 12
3x - 3
3x - 5
5
Simplify
Slope–intercept form
General form
Figure 60 shows the graphs.
Figure 60
y
y 3x 5
6
Y2 13 x 2
6
Y1 3x 5
x 3y 6
4
WARNING
2
Be sure to use a square screen
when you graph perpendicular lines.
Otherwise, the angle between the
two lines will appear distorted. ■
6
x
12
2
2
2
4
6
(1, 2)
4
6
NOW WORK PROBLEM
59.
1.4 Assess Your Understanding
Concepts and Vocabulary
1. The slope of a vertical line is _____; the slope of a horizontal
line is _____.
2. Two nonvertical lines have slopes m1 and m2 , respectively.
The lines are parallel if _____ and the _____ are unequal; the
lines are perpendicular if _____.
3. A horizontal line is given by an equation of the form _____,
where b is the _____.
4. True or False: Vertical lines have an undefined slope.
5. True or False: The slope of the line 2y = 3x + 5 is 3.
6. True or False: Perpendicular lines have slopes that are reciprocals of one another.
Skill Building
In Problems 7–10, (a) find the slope of the line and (b) interpret the slope.
7.
8.
y
2
(–2, 1) 2
(2, 1)
–1
10.
y
(–2, 2)
2
(1, 1)
y
(–1, 1)
2
(2, 2)
(0, 0)
(0, 0)
–2
9.
y
2
x
–2
–1
2
x
–2
–1
2
x
–2
–1
2
x
In Problems 11–18, plot each pair of points and determine the slope of the line containing them. Graph the line.
11. 12, 32; 14, 02
15. 1 -3, -12; 12, -12
12. 14, 22; 13, 42
16. 14, 22; 1- 5, 22
13. 1 -2, 32; 12, 12
14. 1- 1, 12; 12, 32
17. 1- 1, 22; 1-1, -22
18. 12, 02; 12, 22
In Problems 19–26, graph, by hand, the line containing the point P and having slope m.
19. P = 11, 22; m = 3
20. P = 12, 12; m = 4
21. P = 12, 42; m = -
3
4
22. P = 11, 32; m = -
2
5
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SECTION 1.4 Lines
23. P = 1 -1, 32; m = 0
24. P = 12, - 42; m = 0
25. P = 10, 32; slope undefined
41
26. P = 1- 2, 02; slope undefined
In Problems 27–32, the slope and a point on a line are given. Use this information to locate three additional points on the line. Answers
may vary.
[Hint: It is not necessary to find the equation of the line. See Example 2.]
27. Slope 4; point 11, 22
28. Slope 2; point 1- 2, 32
4
30. Slope ; point 1-3, 22
3
3
29. Slope - ; point 12, -42
2
31. Slope - 2; point 1 -2, -32
32. Slope - 1; point 14, 12
In Problems 33–40, find an equation of the line L.
33.
34.
y
2
(2, 1)
(–2, 1) 2
L
37.
2
–1
x
–2
(–1, 3)
3
38.
y
3
(3, 3)
–2
39.
–3
L
3 x
–1
x
2
L is parallel to y = 2 x
2
–1
40.
y
3
(1, 2)
–3
L
(2, 2)
(1, 1)
x
y
3
(1, 2)
L
L
y = 2x
2
(–1, 1)
–2
L
y
3
y
x
2
–1
36.
y
L
(0, 0)
(0, 0)
–2
35.
y
3 x
y = –x
L is parallel to y = –x
–3
y = 2x
3 x
(–1, 1)
–3
L
L is perpendicular
to y = 2x
3 x
y = –x
L is perpendicular
to y = –x
In Problems 41–64, find an equation for the line with the given properties. Express your answer using either the general form or the
slope–intercept form of the equation of a line, whichever you prefer.
41. Slope = 3; containing the point 1- 2, 32
42. Slope = 2; containing the point 14, - 32
2
43. Slope = - ; containing the point 11, - 12
3
44. Slope =
45. Containing the points 11, 32 and 1 -1, 22
46. Containing the points 1 -3, 42 and 12, 52
47. Slope = - 3; y-intercept = 3
48. Slope = - 2; y-intercept = - 2
49. x-intercept = 2; y-intercept = - 1
50. x-intercept = - 4; y-intercept = 4
51. Slope undefined; containing the point 12, 42
52. Slope undefined; containing the point 13, 82
55. Parallel to the line 2x - y = - 2; containing the point 10, 02
56. Parallel to the line x - 2y = - 5; containing the point 10, 02
53. Parallel to the line y = 2x; containing the point 1- 1, 22
57. Parallel to the line x = 5; containing the point 14, 22
59. Perpendicular to the line y =
containing the point 11, - 22
1
x + 4;
2
61. Perpendicular to the line 2x + y = 2;
containing the point 1 - 3, 02
63. Perpendicular to the line x = 8; containing the point 13, 42
1
; containing the point 13, 12
2
54. Parallel to the line y = - 3x; containing the point 1- 1, 22
58. Parallel to the line y = 5; containing the point 14, 22
60. Perpendicular to the line y = 2x - 3;
containing the point 11, - 22
62. Perpendicular to the line x - 2y = - 5;
containing the point 10, 42
64. Perpendicular to the line y = 8; containing the point 13, 42
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In Problems 65–84, find the slope and y-intercept of each line. Graph the line by hand. Check your graph using a graphing utility.
65. y = 2x + 3
1
y = x - 1
2
1
x + y = 2
3
69. y =
1
x + 2
2
66. y = - 3x + 4
67.
71. x + 2y = 4
72. - x + 3y = 6
73. 2x - 3y = 6
74. 3x + 2y = 6
75. x + y = 1
76. x - y = 2
77. x = - 4
78. y = - 1
79. y = 5
80. x = 2
81. y - x = 0
82. x + y = 0
83. 2y - 3x = 0
84. 3x + 2y = 0
70. y = 2x +
1
2
68.
In Problems 85–94, (a) find the intercepts of the graph of each equation and (b) graph the equation by hand.
85. 2x + 3y = 6
86. 3x - 2y = 6
87. -4x + 5y = 40
89. 7x + 2y = 21
90. 5x + 3y = 18
91.
93. 0.2x - 0.5y = 1
94. - 0.3x + 0.4y = 1.2
95. Find an equation of the x-axis.
88. 6x - 4y = 24
1
1
x + y = 1
2
3
92. x -
2
y = 4
3
96. Find an equation of the y-axis.
In Problems 97–100, match each graph with the correct equation:
x
(a) y = x
(b) y = 2x
(c) y =
(d) y = 4x
2
97.
98.
8
6
99.
4
12
6
3
12
4
8
100.
8
2
3
3
3
2
8
In Problems 101–104, write an equation of each line. Express your answer using either the general form or the slope–intercept form of the
equation of a line, whichever you prefer.
101.
102.
4
6
6
103.
2
3
4
3
2
104.
2
3
3
2
2
3
3
2
Applications and Extensions
105. Truck Rentals A truck rental company rents a moving
truck for one day by charging $29 plus $0.07 per mile. Write
a linear equation that relates the cost C, in dollars, of renting
the truck to the number x of miles driven. What is the cost of
renting the truck if the truck is driven 110 miles? 230 miles?
106. Cost Equation The fixed costs of operating a business are
the costs incurred regardless of the level of production.
Fixed costs include rent, fixed salaries, and costs of buying
machinery. The variable costs of operating a business are the
costs that change with the level of output. Variable costs include raw materials, hourly wages, and electricity. Suppose
that a manufacturer of jeans has fixed costs of $500 and
variable costs of $8 for each pair of jeans manufactured.
Write a linear equation that relates the cost C, in dollars, of
manufacturing the jeans to the number x pairs of jeans manufactured. What is the cost of manufacturing 400 pairs of
jeans? 740 pairs?
107. Cost of Sunday Home Delivery The cost to the Chicago Tribune for Sunday home delivery is approximately $0.53 per
newspaper with fixed costs of $1,070,000. Write an equation
that relates the cost C and the number x of copies delivered.
SOURCE: Chicago Tribune, 2002.
108. Wages of a Car Salesperson Dan receives $375 per week
for selling new and used cars at a car dealership in Oak
Lawn, Illinois. In addition, he receives 5% of the profit on
any sales that he generates. Write an equation that relates
Dan’s weekly salary S when he has sales that generate a
profit of x dollars.
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SECTION 1.4 Lines
109. Electricity Rates in Illinois Commonwealth Edison Company supplies electricity to residential customers for a
monthly customer charge of $7.58 plus 8.275 cents per kilowatt-hour for up to 400 kilowatt-hours.
43
(d) What is the monthly charge for using 500 kilowatthours?
(e) Interpret the slope of the line.
SOURCE: Florida Power & Light Company, January 2003.
111. Measuring Temperature The relationship between Celsius
(°C) and Fahrenheit (°F) degrees of measuring temperature
is linear. Find an equation relating °C and °F if 0°C corresponds to 32°F and 100°C corresponds to 212°F. Use the
equation to find the Celsius measure of 70°F.
112. Measuring Temperature The Kelvin (K) scale for measuring temperature is obtained by adding 273 to the Celsius
temperature.
(a) Write an equation relating K and °C.
(b) Write an equation relating K and °F (see Problem 111).
(a) Write an equation that relates the monthly charge C, in
dollars, to the number x of kilowatt-hours used in a
month, 0 … x … 400.
(b) Graph this equation.
(c) What is the monthly charge for using 100 kilowatthours?
(d) What is the monthly charge for using 300 kilowatthours?
(e) Interpret the slope of the line.
SOURCE: Commonwealth Edison Company, December 2002.
110. Electricity Rates in Florida Florida Power & Light Company supplies electricity to residential customers for a
monthly customer charge of $5.25 plus 6.787 cents per
kilowatt-hour for up to 750 kilowatt-hours.
(a) Write an equation that relates the monthly charge C, in
dollars, to the number x of kilowatt-hours used in a
month, 0 … x … 750.
(b) Graph this equation.
(c) What is the monthly charge for using 200 kilowatthours?
113. Product Promotion A cereal company finds that the number of people who will buy one of its products in the first
month that it is introduced is linearly related to the amount
of money it spends on advertising. If it spends $40,000 on
advertising, then 100,000 boxes of cereal will be sold, and if
it spends $60,000, then 200,000 boxes will be sold.
(a) Write an equation describing the relation between the
amount A spent on advertising and the number x of
boxes sold.
(b) How much advertising is needed to sell 300,000 boxes of
cereal?
(c) Interpret the slope.
114. Show that the line containing the points 1a, b2 and 1b, a2,
a Z b, is perpendicular to the line y = x. Also show that the
midpoint of 1a, b2 and 1b, a2 lies on the line y = x.
115. The equation 2x - y = C defines a family of lines, one line
for each value of C. On one set of coordinate axes, graph the
members of the family when C = - 4, C = 0, and C = 2.
Can you draw a conclusion from the graph about each
member of the family?
116. Rework Problem 115 for the family of lines Cx + y = - 4.
Discussion and Writing
117. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(a) 2x + 3y = 6
(e) x - y = - 1
(b) - 2x + 3y = 6
(f) y = 3x - 5
(c) 3x - 4y = - 12
(g) y = 2x + 3
(d) x - y = 1
(h) y = - 3x + 3
118. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(e) x - y = - 1
(a) 2x + 3y = 6
(f) y = - 2x - 1
(b) 2x - 3y = 6
1
(c) 3x + 4y = 12
(g) y = - x + 10
(d) x - y = 1
2
(h) y = x + 4
y
y
x
x
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CHAPTER 1 Graphs
119. The figure shows the graph of two parallel lines. Which of
the following pairs of equations might have such a graph?
(d) x - y = - 2
(a) x - 2y = 3
2x - 2y = - 4
x + 2y = 7
(e) x + 2y = 2
x + 2y = - 1
(b) x + y = 2
x + y = -1
(c) x - y = - 2
x - y = 1
121. The accepted symbol used to denote the slope of a line is
the letter m. Investigate the origin of this symbolism. Begin
by consulting a French dictionary and looking up the French
word monter. Write a brief essay on your findings.
122. The term grade is used to describe the inclination of a road.
How does this term relate to the notion of slope of a line? Is
a 4% grade very steep? Investigate the grades of some
mountainous roads and determine their slopes. Write a brief
essay on your findings.
y
Steep
x
7% Grade
120. The figure shows the graph of two perpendicular lines. Which
of the following pairs of equations might have such a graph?
(a) y - 2x = 2
(d) y - 2x = 2
y + 2x = - 1
x + 2y = - 1
(b) y - 2x = 0
2y + x = 0
(e) 2x + y = - 2
2y + x = - 2
(c) 2y - x = 2
2y + x = - 2
y
x
1.5
123. Carpentry Carpenters use the term pitch to describe the
steepness of staircases and roofs. How does pitch relate to
slope? Investigate typical pitches used for stairs and for
roofs. Write a brief essay on your findings.
124. Can the equation of every line be written in slope–intercept
form? Why?
125. Does every line have exactly one x-intercept and one
y-intercept? Are there any lines that have no intercepts?
126. What can you say about two lines that have equal slopes and
equal y-intercepts?
127. What can you say about two lines with the same x-intercept
and the same y-intercept? Assume that the x-intercept is not 0.
128. If two lines have the same slope, but different x-intercepts,
can they have the same y-intercept?
129. If two lines have the same y-intercept, but different slopes,
can they have the same x-intercept?
Circles
OBJECTIVES
1
2
3
Write the Standard Form of the Equation of a Circle
Graph a Circle by Hand and by Using a Graphing Utility
Work with the General Form of the Equation of a Circle
1 Write the Standard Form of the Equation of a Circle
✓
One advantage of a coordinate system is that it enables us to translate a geometric
statement into an algebraic statement, and vice versa. Consider, for example, the following geometric statement that defines a circle.
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SECTION 1.5 Circles
45
A circle is a set of points in the xy-plane that are a fixed distance r from a fixed
point 1h, k2. The fixed distance r is called the radius, and the fixed point 1h, k2
is called the center of the circle.
Figure 61 shows the graph of a circle. To find the equation, we let 1x, y2 represent the coordinates of any point on a circle with radius r and center 1h, k2. Then the
distance between the points 1x, y2 and 1h, k2 must always equal r. That is, by the distance formula
Figure 61
y
(x, y )
r
2
2
41x - h2 + 1y - k2 = r
(h, k )
x
or, equivalently,
1x - h22 + 1y - k22 = r2
The standard form of an equation of a circle with radius r and center 1h, k2 is
1x - h22 + 1y - k22 = r2
(1)
The standard form of an equation of a circle of radius r with center at the origin 10, 02 is
x2 + y2 = r2
If the radius r = 1, the circle whose center is at the origin is called the unit
circle and has the equation
x2 + y2 = 1
See Figure 62.
Figure 62
Unit circle x2 + y2 = 1
y
2
2 Y1 1 x
3
1
3
2 Y 1 x 2
2
1
(0,0)
1
1
x
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CHAPTER 1 Graphs
EXAMPLE 1
Writing the Standard Form of the Equation of a Circle
Write the standard form of the equation of the circle with radius 5 and center
1-3, 62.
Solution
Using the form of equation (1) and substituting the values r = 5, h = - 3, and
k = 6, we have
1x - h22 + 1y - k22 = r2
1x + 322 + 1y - 622 = 25
NOW WORK PROBLEM
5.
2 Graph a Circle by Hand and by Using a Graphing Utility
✓
The graph of any equation of the form (1) is that of a circle with radius r and center
1h, k2.
EXAMPLE 2
Graphing a Circle by Hand and by Using a Graphing Utility
Graph the equation:
Solution
1x + 322 + 1y - 222 = 16
The graph of the equation is a circle. To graph the equation by hand, we first compare the given equation to the standard form of the equation of a circle. The comparison yields information about the circle.
1x + 322 + 1y - 222 = 16
1x - 1- 3222 + 1y - 222 = 4 2
q
q
q
1x - h22 + 1y - k22 = r2
We see that h = - 3, k = 2, and r = 4. The circle has center 1 -3, 22 and a radius of
4 units. To graph this circle, we first plot the center 1- 3, 22. Since the radius is 4, we
can locate four points on the circle by plotting points 4 units to the left, to the right,
up, and down from the center. These four points can then be used as guides to obtain
the graph. See Figure 63.
To graph a circle on a graphing utility, we must write the equation in the form
y = 5expression involving x6.* We must solve for y in the equation
Figure 63
(–3, 6)
y
6
1x + 322 + 1y - 222 = 16
4
(–7, 2)
–10
(–3, 2)
(1, 2)
2 x
–5
(–3, –2)
1y - 222 = 16 - 1x + 322
y - 2 = ; 416 - 1x + 322
y = 2 ; 416 - 1x + 322
Subtract (x + 3)2 from both sides.
Use the Square Root Method.
Add 2 to both sides.
*
Some graphing utilities (e.g., TI-83, TI-84, and TI-86) have a CIRCLE function that allows the user to
enter only the coordinates of the center of the circle and its radius to graph the circle.
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SECTION 1.5 Circles
47
To graph the circle, we graph the top half
Figure 64
Y1 = 2 + 416 - 1x + 322
6
Y1
and the bottom half
–9
Y2 = 2 - 416 - 1x + 322
Also, be sure to use a square screen. Otherwise, the circle will appear distorted.
Figure 64 shows the graph on a TI-84 Plus.
3
Y2
–2
NOW WORK PROBLEMS
EXAMPLE 3
21(a)
AND
(b).
Finding the Intercepts of a Circle
For the circle 1x + 322 + 1y - 222 = 16, find the intercepts, if any, of its graph.
Solution
This is the equation discussed and graphed in Example 2. To find the x-intercepts, if
any, let y = 0 and solve for x. Then
1x + 322 + 1y - 222 = 16
1x + 322 + 10 - 222 = 16
y = 0
1x + 32 + 4 = 16
2
Simplify.
1x + 32 = 12
2
Simplify.
x + 3 = ; 212
x = - 3 ; 223
Apply the Square Root Method.
Solve for x.
The x-intercepts are -3 - 223 L - 6.46 and -3 + 223 L 0.46.
To find the y-intercepts, if any, we let x = 0 and solve for y. Then
1x + 322 + 1y - 222 = 16
10 + 322 + 1y - 222 = 16
x = 0
9 + 1y - 22 = 16
2
Simplify.
1y - 22 = 7
2
Simplify.
y - 2 = ; 27
y = 2 ; 27
Apply the Square Root Method.
Solve for y.
The y-intercepts are 2 - 27 L - 0.65 and 2 + 27 L 4.65.
Look back at Figure 63 to verify the approximate locations of the intercepts. NOW WORK PROBLEM
21(C).
3 Work with the General Form of the Equation of a Circle
✓
If we eliminate the parentheses from the standard form of the equation of the circle
given in Example 3, we get
1x + 322 + 1y - 222 = 16
x2 + 6x + 9 + y2 - 4y + 4 = 16
which we find, upon simplifying, is equivalent to
x2 + y2 + 6x - 4y - 3 = 0
It can be shown that any equation of the form
x2 + y2 + ax + by + c = 0
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CHAPTER 1 Graphs
has a graph that is a circle, or a point, or has no graph at all. For example, the graph
of the equation x2 + y2 = 0 is the single point 10, 02. The equation
x 2 + y2 + 5 = 0, or x2 + y2 = - 5, has no graph, because sums of squares of real
numbers are never negative. When its graph is a circle, the equation
x2 + y2 + ax + by + c = 0
is referred to as the general form of the equation of a circle.
If an equation of a circle is in the general form, we use the method of completing the square to put the equation in standard form so that we can identify its center
and radius.
EXAMPLE 4
Graphing a Circle Whose Equation Is in General Form
Graph the equation x2 + y2 + 4x - 6y + 12 = 0
Solution
We complete the square in both x and y to put the equation in standard form. Group
the expression involving x, group the expression involving y, and put the constant on
the right side of the equation. The result is
1x2 + 4x2 + 1y2 - 6y2 = - 12
Next, complete the square of each expression in parentheses. Remember that any
number added on the left side of the equation must be added on the right.
(x2 4x 4) (y2 6y 9) 12 4 9
( 42 )
2
(6
)
2
2
4
9
(x 2) (y 3)2 1
2
Factor.
We recognize this equation as the standard form of the equation of a circle with radius 1 and center 1 -2, 32. To graph the equation by hand, use the center 1- 2, 32 and
the radius 1. See Figure 65 (a).
To graph the equation using a graphing utility, we need to solve for y.
1y - 322 = 1 - 1x + 222
y - 3 = ; 41 - 1x + 222
Use the Square Root Method.
y = 3 ; 41 - 1x + 222
Add 3 to both sides.
Figure 65(b) illustrates the graph.
Figure 65
y
(2, 4)
4
1
(3, 3)
(1, 3)
Y1 3 1 (x 2)2 5
(2, 3)
(2, 2)
3
1 x
4.5
1.5
Y2 3 1 (x 2)2
(b)
(a)
NOW WORK PROBLEM
25.
1
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SECTION 1.5 Circles
EXAMPLE 5
49
Finding the General Equation of a Circle
Find the general equation of the circle whose center is 11, -22 and whose graph
contains the point 14, -22.
Solution
To find the equation of a circle, we need to know its center and its radius. Here, we
know that the center is 11, -22. Since the point 14, - 22 is on the graph, the radius r
will equal the distance from 14, - 22 to the center 11, -22. See Figure 66. Thus,
Figure 66
r = 414 - 122 + 3- 2 - 1- 2242
= 29 = 3
The standard form of the equation of the circle is
y
3
x
5
r
1x - 122 + 1y + 222 = 9
(4, 2)
(1, 2)
Eliminating the parentheses and rearranging terms, we get the general equation
5
x2 + y2 - 2x + 4y - 4 = 0
Overview
The discussion in Sections 1.4 and 1.5 about circles and lines dealt with two main
types of problems that can be generalized as follows:
1. Given an equation, classify it and graph it.
2. Given a graph, or information about a graph, find its equation.
This text deals with both types of problems. We shall study various equations,
classify them, and graph them. Although the second type of problem is usually more
difficult to solve than the first, in many instances a graphing utility can be used to
solve such problems.
1.5 Assess Your Understanding
Concepts and Vocabulary
3. True or False: The radius of the circle x2 + y2 = 9 is 3.
1. True or False: Every equation of the form
2
2
x + y + ax + by + c = 0
4. True or False: The center of the circle
has a circle as its graph.
2. For a circle, the _____ is the distance from the center to any
point on the circle.
is 13, -22.
1x + 322 + 1y - 222 = 13
Skill Building
In Problems 5–8, find the center and radius of each circle. Write the standard form of the equation.
5. y
6.
7. y
y
(4, 2)
8. y
(2, 3)
(1, 2)
(0, 1)
(2, 1)
(0, 1)
(1, 2)
x
(1, 0)
x
x
x
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CHAPTER 1 Graphs
In Problems 9–18, write the standard form of the equation and the general form of the equation of each circle of radius r and center 1h, k2.
By hand, graph each circle.
9. r = 2; 1h, k2 = 10, 02
13. r = 5; 1h, k2 = 14, -32
1
17. r = ;
2
1
1h, k2 = a , 0 b
2
10. r = 3; 1h, k2 = 10, 02
11. r = 2; 1h, k2 = 10, 22
14. r = 4; 1h, k2 = 12, - 32
1
18. r = ;
2
15. r = 4; 1h, k2 = 1- 2, 12
1
1h, k2 = a0, - b
2
12. r = 3; 1h, k2 = 11, 02
16. r = 7;
1h, k2 = 1- 5, - 22
In Problems 19–30, (a) find the center 1h, k2 and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any, of the graphs.
19. x2 + y2 = 4
20. x2 + 1y - 122 = 1
21. 21x - 322 + 2y2 = 8
22. 31x + 122 + 31y - 122 = 6
23. x2 + y2 - 2x - 4y - 4 = 0
24. x2 + y2 + 4x + 2y - 20 = 0
25. x2 + y2 + 4x - 4y - 1 = 0
26. x2 + y2 - 6x + 2y + 9 = 0
27. x2 + y2 - x + 2y + 1 = 0
29. 2x2 + 2y2 - 12x + 8y - 24 = 0
30. 2x2 + 2y2 + 8x + 7 = 0
28. x2 + y2 + x + y -
1
= 0
2
In Problems 31–36, find the general form of the equation of each circle.
31. Center at the origin and containing the point 1- 2, 32
33. Center 12, 32 and tangent to the x-axis
35. With endpoints of a diameter at 11, 42 and 1-3, 22
In Problems 37–40, match each graph with the correct equation.
(a) 1x - 322 + 1y + 322 = 9
(b) 1x + 122 + 1y - 222 = 4
37.
38.
4
6
6
6
9
9
4
32. Center 11, 02 and containing the point 1 -3, 22
34. Center 1-3, 12 and tangent to the y-axis
36. With endpoints of a diameter at 14, 32 and 10, 12
(c) 1x - 122 + 1y + 222 = 4
39.
(d) 1x + 322 + 1y - 322 = 9
40.
4
6
6
9
6
4
6
9
6
In Problems 41–44, find the standard form of the equation of each circle. Assume that the center has integer coordinates and that the radius
is an integer.
41.
42.
43.
44.
5
3
6
6
9
6
3
9
4
5
5
8
4
2
5
0
Applications and Extensions
45. Weather Satellites Earth is represented on a map of a portion of the solar system so that its surface is the circle with
equation x2 + y2 + 2x + 4y - 4091 = 0. A weather satellite circles 0.6 unit above Earth with the center of its circular
orbit at the center of Earth. Find the equation for the orbit
of the satellite on this map.
46. The tangent line to a circle may be defined as the line that
intersects the circle in a single point, called the point of tangency (see the figure).
y
r
x
r
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Chapter Review
If the equation of the circle is x2 + y2 = r2 and the equation of the tangent line is y = mx + b, show that:
(a) r211 + m22 = b2
[Hint: The quadratic equation x 2 + 1mx + b22 = r2
has exactly one solution.]
2
2
-r m r
, ≤.
b
b
(c) The tangent line is perpendicular to the line containing
the center of the circle and the point of tangency.
47. The Greek method for finding the equation of the tangent
line to a circle used the fact that at any point on a circle the
lines containing the center and the tangent line are perpendicular (see Problem 46). Use this method to find an equation of the tangent line to the circle x2 + y2 = 9 at the point
11, 2 122.
(b) The point of tangency is ¢
51
48. Use the Greek method described in Problem 47 to find an
equation of the tangent line to the circle
x2 + y2 - 4x + 6y + 4 = 0 at the point 13, 2 12 - 32.
49. Refer to Problem 46. The line x - 2y + 4 = 0 is tangent to
a circle at 10, 22. The line y = 2x - 7 is tangent to the same
circle at 13, - 12. Find the center of the circle.
50. Find an equation of the line containing the centers of the
two circles
x2 + y2 - 4x + 6y + 4 = 0
2
and
2
x + y + 6x + 4y + 9 = 0
51. If a circle of radius 2 is made to roll along the x-axis, what is
an equation for the path of the center of the circle?
Discussion and Writing
52. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(a) 1x - 222 + 1y + 322 = 13
(b) 1x - 222 + 1y - 222 = 8
(c) 1x - 222 + 1y - 322 = 13
(d) 1x + 222 + 1y - 222 = 8
(e) x2 + y2 - 4x - 9y = 0
(f) x2 + y2 + 4x - 2y = 0
(g) x2 + y2 - 9x - 4y = 0
(h) x2 + y2 - 4x - 4y = 4
53. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(a) 1x - 222 + y2 = 3
(e) x2 + y2 + 10x + 16 = 0
2
2
(b) 1x + 22 + y = 3
(f) x2 + y2 + 10x - 2y = 1
2
2
(c) x + 1y - 22 = 3
(g) x2 + y2 + 9x + 10 = 0
2
2
(d) 1x + 22 + y = 4
(h) x2 + y2 - 9x - 10 = 0
y
x
y
x
54. Explain how the center and radius of a circle can be used to
graph a circle by hand.
55. Explain how the center and radius of a circle can be used to
establish an initial viewing window.
56. If the circumference of a circle is 6p, what is its radius?
Chapter Review
Things to Know
Formulas
Distance formula (p. 5)
Midpoint formula (p. 7)
Slope (p. 27)
Parallel lines (p. 37)
Perpendicular lines (p. 38)
d = 41x2 - x122 + 1y2 - y122
1x, y2 = ¢
x1 + x2 y1 + y2
,
≤
2
2
y2 - y1
m =
, if x1 Z x2 ; undefined if x1 = x2
x2 - x1
Equal slopes 1m1 = m22 with different y-intercepts
Product of slopes is -1 1m1 # m2 = - 12
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CHAPTER 1 Graphs
Equations
x = a
Vertical line (p. 31)
y - y1 = m1x - x12; m is the slope of the line, 1x1 , y12 is a
point on the line
Point–slope form of the
equation of a line (p. 32)
y = b
y = mx + b; m is the slope of the line, b is the y-intercept
Horizontal line (p. 33)
Slope–intercept form of the
equation of a line (p. 34)
General form of the
equation of a line (p. 35)
Standard form of the
equation of a circle (p. 45)
Equation of the unit circle (p. 45)
1x - h22 + 1y - k22 = r2;
r is the radius of the circle, 1h, k2 is the center of the circle
x2 + y2 = 1
General form of the
equation of a circle (p. 48)
x2 + y2 + ax + by + c = 0
Ax + By = C; A, B not both 0
Objectives
Section
1.1
1.2
1.3
1.4
1.5
1
✓
2
✓
1
✓
2
✓
3
✓
4
✓
5
✓
6
✓
7
✓
8
✓
1
✓
1
✓
2
✓
3
✓
4
✓
5
✓
6
✓
7
✓
8
✓
9
✓
10
✓
1
✓
2
✓
3
✓
You should be able to
Review Exercises
Á
Use the distance formula (p. 4)
Use the midpoint formula (p. 7)
Graph equations by hand by plotting points (p. 11)
Graph equations using a graphing utility (p. 13)
Use a graphing utility to create tables (p. 14)
Find intercepts from a graph (p. 15)
Find intercepts from an equation (p. 16)
Use a graphing utility to approximate intercepts (p. 16)
Test an equation for symmetry with respect to the x-axis, the y-axis,
and the origin (p. 17)
Know how to graph key equations (p. 19)
Solve equations in one variable using a graphing utility (p. 24)
Calculate and interpret the slope of a line (p. 27)
Graph lines given a point and the slope (p. 30)
Find the equation of a vertical line (p. 31)
Use the point–slope form of a line; identify horizontal lines (p. 32)
Find the equation of a line given two points (p. 33)
Write the equation of a line in slope–intercept form (p. 34)
Identify the slope and y-intercept of a line from its equation (p. 34)
Graph lines written in general form using intercepts (p. 35)
Find equations of parallel lines (p. 36)
Find equations of perpendicular lines (p. 38)
Write the standard form of the equation of a circle (p. 44)
Graph a circle by hand and by using a graphing utility (p. 46)
Work with the general form of the equation of a circle (p. 47)
1(a)–6(a), 51, 52(a), 54–56
1(b)–6(b), 55
9–14
8
8
7
9–14, 47–50
8
15–22
23, 24
25–28
1(c)–6(c); 1(d)–6(d), 53
57
31
29, 30
32–34
29–38
39–42
9, 10
35, 36
37, 38
43–46
47–50
47–50, 55
Review Exercises
In Problems 1–6, find the following for each pair of points:
(a) The distance between the points.
(b) The midpoint of the line segment connecting the points.
(c) The slope of the line containing the points.
(d) Then interpret the slope found in part (c).
1. 10, 02; 14, 22
2. 10, 02; 1-4, 62
3. 11, - 12; 1-2, 32
4. 1- 2, 22; 11, 42
5. 14, - 42; 14, 82
6. 1- 3, 42; 12, 42
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Chapter Review
53
8. Graph y = - x2 + 15 using a graphing utility. Create a table
of values to determine a good initial viewing window. Use a
graphing utility to approximate the intercepts.
7. List the intercepts of the following graph.
y
2
4
4
x
2
In Problems 9–14, determine the intercepts and graph each equation by hand by plotting points. Verify your results using a graphing
utility. Label the intercepts on the graph.
9. 2x - 3y = 6
10. x + 2y = 4
11. y = x2 - 9
12. y = x2 + 4
13. x2 + 2y = 16
14. 2x2 - 4y = 24
In Problems 15–22, test each equation for symmetry with respect to the x-axis, the y-axis, and the origin.
15. 2x = 3y2
16. y = 5x
17. x2 + 4y2 = 16
19. y = x4 + 2x2 + 1
20. y = x3 - x
21. x2 + x + y2 + 2y = 0
18. 9x2 - y2 = 9
22. x2 + 4x + y2 - 2y = 0
24. Sketch a graph of y = 1x.
23. Sketch a graph of y = x3.
In Problems 25–28, use a graphing utility to approximate the solutions of each equation rounded to two decimal places. All solutions lie
between - 10 and 10.
25. x3 - 5x + 3 = 0
26. - x3 + 3x + 1 = 0
27. x4 - 3 = 2x + 1
28. - x4 + 7 = x2 - 2
In Problems 29–38, find an equation of the line having the given characteristics. Express your answer using either the general form or the
slope–intercept form of the equation of a line, whichever you prefer. Graph the line.
29. Slope = - 2; containing the point 13, -12
30. Slope = 0; containing the point 1- 5, 42
31. Slope undefined; containing the point 1-3, 42
32. x-intercept = 2; containing the point 14, - 52
33. y-intercept = - 2; containing the point 15, -32
34. Containing the points 13, -42 and 12, 12
35. Parallel to the line 2x - 3y = - 4; containing the point 1-5, 32 36. Parallel to the line x + y = 2; containing the point 11, -32
37. Perpendicular to the line x + y = 2; containing the point 14, -32
38. Perpendicular to the line 3x - y = - 4; containing the point 1- 2, 42
In Problems 39–42, find the slope and y-intercept of each line.
1
5
1
2
42. - x + y = 2
x + y = 10
2
2
3
2
In Problems 43–46, find the standard form of the equation of the circle whose center and radius are given.
43. 1h, k2 = 1-2, 32; r = 4
44. 1h, k2 = 13, 42; r = 4
45. 1h, k2 = 1 -1, -22; r = 1
46. 1h, k2 = 12, - 42; r = 3
39. 4x + 6y = 36
40. 7x - 3y = 42
41.
In Problems 47–50, find the center and radius of each circle. Graph each circle by hand. Determine the intercepts of the graph of each circle.
47. x2 + y 2 - 2x + 4y - 4 = 0 48. x2 + y2 + 4x - 4y - 1 = 0
49. 3x2 + 3y2 - 6x + 12y = 0 50. 2x 2 + 2y2 - 4x = 0
51. Show that the points A = 13, 42, B = 11, 12,
C = 1 - 2, 32 are the vertices of an isosceles triangle.
56. Find two numbers y such that the distance from 1- 3, 22 to
15, y2 is 10.
2
57. Graph the line with slope containing the point 11, 22.
3
58. Create four problems that you might be asked to do given
the two points 1- 3, 42 and 16, 12. Each problem should involve a different concept. Be sure that your directions are
clearly stated.
59. Describe each of the following graphs in the xy-plane. Give
justification.
(a) x = 0
(b) y = 0
(c) x + y = 0
(d) xy = 0
(e) x2 + y2 = 0
and
52. Show that the points A = 1 -2, 02, B = 1-4, 42, and
C = 18, 52 are the vertices of a right triangle in two ways:
(a) By using the converse of the Pythagorean Theorem
(b) By using the slopes of the lines joining the vertices
53. Show that the points A = 12, 52, B = 16, 12,
C = 18, - 12 lie on a straight line by using slopes.
and
54. Show that the points A = 11, 52, B = 12, 42, and
C = 1 - 3, 52 lie on a circle with center 1- 1, 22. What is the
radius of this circle?
55. The endpoints of the diameter of a circle are 1- 3, 22 and
15, - 62. Find the center and radius of the circle. Write the
general equation of this circle.
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CHAPTER 1 Graphs
Chapter Test
1. Suppose the points 1- 2, -32 and 14, 52 are the endpoints of
a diameter of a circle.
(a) Find the distance between the two points.
(b) Find the midpoint of the line segment connecting the
two points.
(c) Find the standard equation of the circle containing the
two points.
(d) Graph the circle by hand.
In Problems 2 and 3, graph each equation by hand by plotting points. Determine the intercepts and label them on the graph.
2. y = x2 - x - 2
3. 2x - 7y = 21
4. Given the linear equation 10x - 6y = 20, do the following:
(a) Find the slope of the line.
(b) Find the intercepts of the graph of the line.
(c) Graph the equation using the intercepts.
(d) Find the equation of the line that is perpendicular to the
given line that contains the point 1 -1, 52.
(e) Find the equation of the line that is parallel to the given
line that contains the point 1 -1, 52.
In Problems 5–7, use a graphing utility to approximate the real solutions of each equation rounded to two decimal places.
All solutions lie between - 10 and 10.
5. 2x3 - x2 - 2x + 1 = 0
6. x4 - 5x2 - 8 = 0
7. - x3 + 7x - 2 = x2 + 3x - 3
2
8. Test the equation 5x - 2y = 7 for symmetry with respect to the x-axis, the y-axis, and the origin.
Chapter Projects
(b) 4.85% on 15-year mortgages
1. For each rate, calculate the monthly payment for a
loan of $200,000.
2. Then compute the total amount paid over the term of
each loan.
3. Calculate the interest paid on each loan.
On August 5, 2004, average interest rates were:
(a) 5.99% on 30-year mortgages
(b) 5.40% on 15-year mortgages
4. For each rate, calculate the monthly payment for a
loan of $200,000.
5. Then compute the total amount paid over the term of
each loan.
6. Calculate the interest paid on each loan.
7. Solve equation (1) for L.
1. Home Mortgages While you may not be in the
8. If you can afford to pay $1000 per month for a mortmarket for a home right now, it is probably an event that will
gage payment, calculate the amount you can borrow
occur for you within the next few years. Formula (1) gives
on August 5, 2004.
the monthly payment P required to pay off a loan L at an
(a) On a 30-year mortgage.
annual rate of interest r, expressed as a decimal, but usually
(b) On a 15-year mortgage.
given as a percent. The time t, measured in months, is the du9.
Check
with your local lending institution for current
ration of the loan, so a 15-year mortgage requires
rates for 30-year and 15-year mortages. Calculate how
t = 12 * 15 = 180 monthly payments.
much you can borrow with a $1000 per month payment.
P = monthly payment
r
10. Repeat Problem 9 if you can afford a payment of
L = loan amount
12
$1300 per month.
P = LD
r
=
annual
rate
of
interest,
(1)
T
r -t
11.
Do
you think that the interest rate plays an important
expressed as a decimal
b
1 - a1 +
role in determining how much you can afford to pay
12
t = length of loan, in months
for a house?
On July 8, 2003, average interest rates were:
12. Comment on the two types of mortgages: 30-year and
(a) 5.52% on 30-year mortgages
15-year. Which would you take? Why?
The following projects are available on the Instructor’s Resource Center (IRC):
2.
3.
Project at Motorola Mobile Phone Usage
Economics Isocost Lines