10.1
Before
Now
Why?
Key Vocabulary
• quadratic function
• parabola
• parent quadratic
function
• vertex
• axis of symmetry
Graph y 5 ax2 1 c
You graphed linear and exponential functions.
You will graph simple quadratic functions.
So you can solve a problem involving an antenna, as in Ex. 40.
A quadratic function is a nonlinear function that can be written in the
standard form y 5 ax2 1 bx 1 c where a Þ 0. Every quadratic function has a
U-shaped graph called a parabola. In this lesson, you will graph quadratic
functions where b 5 0.
For Your Notebook
KEY CONCEPT
Parent Quadratic Function
The most basic quadratic function in the family of quadratic functions,
called the parent quadratic function, is y 5 x2. The graph of y 5 x2 is
shown below.
The line that passes
through the vertex and
divides the parabola
into two symmetric
parts is called the axis
of symmetry. The axis of
symmetry for the graph of
y 5 x2 is the y-axis, x 5 0.
y
The lowest or
highest point
on a parabola is
the vertex. The
vertex of the
graph of y 5 x2
is (0, 0).
EXAMPLE 1
y 5 x2
1
1
(0, 0)
x
Graph y 5 ax 2 where ⏐a⏐ > 1
STEP 1 Make a table of values for y 5 3x2.
y
x
22
21
0
1
2
y
12
3
0
3
12
PLOT ADDITIONAL
POINTS
STEP 2 Plot the points from the table.
If you are having
difficulty seeing the
shape of the parabola,
plot additional points.
STEP 4 Compare the graphs of y 5 3x2 and y 5 x2 .
628
STEP 3 Draw a smooth curve through the points.
Both graphs open up and have the same
vertex, (0, 0), and axis of symmetry, x 5 0.
The graph of y 5 3x2 is narrower than the
graph of y 5 x2 because the graph of y 5 3x2
is a vertical stretch (by a factor of 3) of the
graph of y 5 x2.
Chapter 10 Quadratic Equations and Functions
y 5 3x2
4
y 5 x2
1
x
EXAMPLE 2
Graph y 5 ax 2 where ⏐a⏐ < 1
1 2
Graph y 5 2}
x . Compare the graph with the graph of y 5 x 2 .
4
y
STEP 1 Make a table of values for y 5 2}1x2.
4
MAKE A TABLE
To make the
calculations easier,
choose values of x that
are multiples of 2.
x
24
22
0
2
4
y
24
21
0
21
24
y 5 x2
1
3
STEP 2 Plot the points from the table.
x
y 5 2 14 x 2
STEP 3 Draw a smooth curve through the points.
STEP 4 Compare the graphs of y 5 2}1x2 and y 5 x2 . Both graphs have the
4
same vertex (0, 0), and the same axis of symmetry, x 5 0. However,
1
the graph of y 5 2}
x2 is wider than the graph of y 5 x2 and it opens
4
1
down. This is because the graph of y 5 2}
x2 is a vertical shrink
4
1 by a factor of }14 2 with a reflection in the x-axis of the graph of y 5 x .
2
GRAPHING QUADRATIC FUNCTIONS Examples 1 and 2 suggest the following
general result: a parabola opens up when the coefficient of x2 is positive and
opens down when the coefficient of x2 is negative.
EXAMPLE 3
Graph y 5 x 2 1 c
Graph y 5 x 2 1 5. Compare the graph with the graph of y 5 x 2 .
STEP 1 Make a table of values for y 5 x2 1 5.
x
22
21
0
1
2
y
9
6
5
6
9
y
STEP 2 Plot the points from the table.
y 5 x2 1 5
STEP 3 Draw a smooth curve through the points.
STEP 4 Compare the graphs of y 5 x2 1 5 and y 5 x2 .
Both graphs open up and have the same axis
of symmetry, x 5 0. However, the vertex of the
graph of y 5 x2 1 5, (0, 5), is different than the
vertex of the graph of y 5 x2, (0, 0), because
the graph of y 5 x2 1 5 is a vertical translation
(of 5 units up) of the graph of y 5 x2.
✓
GUIDED PRACTICE
2
y 5 x2
1
x
for Examples 1, 2, and 3
Graph the function. Compare the graph with the graph of y 5 x 2 .
1. y 5 24x2
1 2
2. y 5 }
x
3
3. y 5 x2 1 2
10.1 Graph y 5 ax 2 1 c
629
Graph y 5 ax 2 1 c
EXAMPLE 4
1 2
Graph y 5 }
x 2 4. Compare the graph with the graph of y 5 x 2 .
2
y
STEP 1 Make a table of values for y 5 }1x2 2 4.
2
y 5 x2
x
24
22
0
2
4
y
4
22
24
22
4
1
1
x
STEP 2 Plot the points from the table.
y 5 12 x 2 2 4
STEP 3 Draw a smooth curve through the points.
STEP 4 Compare the graphs of y 5 }1x2 2 4 and y 5 x2 . Both graphs open up
2
and have the same axis of symmetry, x 5 0. However, the graph of
1 2
y5}
x 2 4 is wider and has a lower vertex than the graph of y 5 x2
2
1 2
because the graph of y 5 }
x 2 4 is a vertical shrink and a vertical
2
translation of the graph of y 5 x2.
✓
GUIDED PRACTICE
for Example 4
Graph the function. Compare the graph with the graph of y 5 x 2 .
4. y 5 3x2 2 6
3
4
5. y 5 25x2 1 1
6. y 5 }x2 2 2
For Your Notebook
KEY CONCEPT
y 5 ax2 , a > 0
y 5 ax2 , a < 0
y
y 5 x2 1 c
y
y
a < 21
a 5 21
21 < a < 0
a>1
a51
0<a<1
x
x
x
Compared with the
graph of y 5 x2, the
graph of y 5 ax2 is:
Compared with the
graph of y 5 x2, the
graph of y 5 ax2 is:
Compared with the
graph of y 5 x2, the
graph of y 5 x2 1 c is:
• a vertical stretch if
• a vertical stretch with
• an upward vertical
a > 1,
• a vertical shrink if
0 < a < 1.
a reflection in the
x-axis if a < 21,
• a vertical shrink with
a reflection in the
x-axis if 21 < a < 0.
630
c>0
c50
c<0
Chapter 10 Quadratic Equations and Functions
translation if c > 0,
• a downward vertical
translation if c < 0.
★
EXAMPLE 5
Standardized Test Practice
How would the graph of the function y 5 x 2 1 6 be affected if the
function were changed to y 5 x 2 1 2?
A The graph would shift 2 units up.
B The graph would shift 4 units up.
C The graph would shift 4 units down.
ELIMINATE CHOICES
You can eliminate
choice D because
changing the value
of c in a function of
the form y 5 x2 1 c
translates the graph up
or down.
D The graph would shift 4 units to the left.
Solution
The vertex of the graph of y 5 x2 1 6 is 6 units above the origin, or (0, 6). The
vertex of the graph of y 5 x2 1 2 is 2 units above the origin, or (0, 2). Moving
the vertex from (0, 6) to (0, 2) translates the graph 4 units down.
c The correct answer is C.
EXAMPLE 6
A B C D
Use a graph
SOLAR ENERGY A solar trough has a reflective parabolic surface that is used
to collect solar energy. The sun’s rays are reflected from the surface toward
a pipe that carries water. The heated water produces steam that is used to
produce electricity.
y
The graph of the function
y 5 0.09x2 models the cross
section of the reflective
surface where x and y are
measured in meters. Use
the graph to find the
domain and range of the
function in this situation.
3
3
x
Solution
STEP 1 Find the domain. In the graph, the reflective surface extends
5 meters on either side of the origin. So, the domain is 25 ≤ x ≤ 5.
STEP 2 Find the range using the fact that the lowest point on the reflective
surface is (0, 0) and the highest point, 5, occurs at each end.
y 5 0.09(5)2 5 2.25
Substitute 5 for x. Then simplify.
The range is 0 ≤ y ≤ 2.25.
✓
GUIDED PRACTICE
for Examples 5 and 6
7. Describe how the graph of the function y 5 x2 1 2 would be affected if
the function were changed to y 5 x2 2 2.
8. WHAT IF? In Example 6, suppose the reflective surface extends just
4 meters on either side of the origin. Find the domain and range of the
function in this situation.
10.1 Graph y 5 ax 2 1 c
631
10.1
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 7 and 41
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 22, 33, 43, and 44
SKILL PRACTICE
1. VOCABULARY Copy and complete: Every quadratic function has a
U-shaped graph called a(n) ? .
2.
★ WRITING Explain how you can tell whether the graph of a quadratic
function opens up or down.
MATCHING Match the quadratic function with its graph.
1 2
1 2
3. y 5 }
x 24
4. y 5 }
x 22
2
2
A.
B.
y
1
2
5. y 5 2}x2 1 2
C.
y
y
1
1
1
1
EXAMPLES
1, 2, and 3
on pp. 628–629
for Exs. 6–23
1
1
x
GRAPHING QUADRATIC FUNCTIONS Graph the function. Compare the graph
with the graph of y 5 x 2 .
6. y 5 8x2
7. y 5 22x2
11 2
10. y 5 }
x
8. y 5 23x2
9. y 5 5x2
3
4
2
2 2
11. y 5 }
x
12. y 5 2}x2
1 2
13. y 5 2}
x
14. y 5 }x2
3
8
1 2
15. y 5 2}
x
16. y 5 x2 2 7
17. y 5 x2 1 9
18. y 5 x2 1 6
19. y 5 x2 2 4
20. y 5 x2 2 1
21. y 5 x2 1 }
22.
3
5
9
7
4
★ MULTIPLE CHOICE What is the vertex of the graph of the function
3
y 5 2}
x2 1 7?
4
A (27, 0)
B (0, 27)
C (0, 7)
D (7, 0)
23. ERROR ANALYSIS Describe and correct the error in drawing and
comparing the graphs of y 5 x2 and y 5 x2 2 2.
Both graphs open up and have the
same axis of symmetry. However,
the vertex of the graph of y 5 x2 2 2,
(0, 2), is 2 units above the vertex of
the graph of y 5 x2 , (0, 0).
y
4
y = x2 – 2
y = x2
1
632
x
x
x
Chapter 10 Quadratic Equations and Functions
EXAMPLE 4
GRAPHING QUADRATIC FUNCTIONS Graph the function. Compare the graph
on p. 630
for Exs. 24–32
with the graph of y 5 x 2 .
24. y 5 7x2 1 7
25. y 5 2x2 1 5
26. y 5 2x2 2 12
27. y 5 22x2 2 1
28. y 5 23x2 2 2
29. y 5 }x2 2 3
1 2
30. y 5 }
x 1 10
1 2
31. y 5 }
x 25
32. y 5 2}x2 1 9
5
EXAMPLE 5
33.
on p. 631
for Exs. 33–36
2
3
4
2
3
★ MULTIPLE CHOICE How would the graph of the function y 5 x2 1 3 be
affected if the function were changed to y 5 x2 1 9?
A The graph would shift 9 units to the right.
B The graph would shift 6 units up.
C The graph would shift 9 units up.
D The graph would shift 6 units down.
COMPARING GRAPHS Tell how you can obtain the graph of g from the graph
of f using transformations.
34. f(x) 5 x2 2 5
35. f(x) 5 3x2 2 11
2
2
g(x) 5 x 1 8
g(x) 5 3x 2 16
36. f(x) 5 4x2
g(x) 5 2x2
CHALLENGE Write a function of the form y 5 ax 2 1 c whose graph passes
through the two given points.
38. (2, 1), (5, 220)
37. (21, 9), (0, 3)
39. (22, 216.5), (1, 4.5)
PROBLEM SOLVING
GRAPHING CALCULATOR You may wish to use a graphing calculator to
complete the following Problem Solving exercises.
EXAMPLE 6
on p. 631
for Exs. 40–41
40. ASTRONOMY A cross section of the parabolic
surface of the antenna shown can be modeled
by the graph of the function y 5 0.012x2 where
x and y are measured in meters.
a. Find the domain of the function in this situation.
b. Find the range of the function in this situation.
y
8
24
x
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
41. SAILING Sailors need to consider the speed of the wind when adjusting
the sails on their boat. The force F (in pounds per square foot) on a sail
when the wind is blowing perpendicular to the sail can be modeled by
the function F 5 0.004v 2 where v is the wind speed (in knots).
a. Graph the function for wind speeds from 0 knots to 50 knots.
b. Use the graph to estimate the wind speed that will produce a force of
1 pound per square foot on a sail.
c. Estimate the wind speed that will produce a force of 5 pounds per
square foot on a sail.
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
10.1 Graph y 5 ax 2 1 c
633
42. FALLING OBJECTS Two acorns drop from an oak tree. One falls 45 feet,
REVIEW
VERTICAL
MOTION
while the other falls 32 feet.
a. For each acorn, write an equation that gives the height h (in feet) of
For help with
the vertical
motion model,
see p. 575.
the acorn as a function of the time t (in seconds) it has fallen.
b. Describe how the graphs of the two equations are related.
43.
★ SHORT RESPONSE The breaking strength w (in pounds)
of a manila rope can be modeled by the function
w = 8900d2 where d is the diameter (in inches)
of the rope.
d
a. Graph the function.
b. If a manila rope has 4 times the breaking strength
of another manila rope, does the rope have
4 times the diameter of the other rope? Explain.
44.
★ EXTENDED RESPONSE For an engineering contest, you have to create
a container for an egg so that the container can be dropped from a height
of 30 feet without breaking the egg.
a. The distance y (in feet) that the container falls is given by the function
y 5 16t 2 where t is the time (in seconds) the container has fallen.
Graph the function.
b. The height y (in feet) of the dropped container is given by the function
y 5 216t 2 1 30 where t is the time (in seconds) since the container is
dropped. Graph the function.
c. How are the graphs from part (a) and part (b) related? Explain how
you can use each graph to find the number of seconds after which the
container has fallen 10 feet.
"MHFCSB
at classzone.com
45. CHALLENGE The kinetic energy E (in joules) of an object in motion is
1
given by E 5 }
mv 2 where m is the object’s mass (in kilograms) and v
2
is the object’s velocity (in meters per second). Suppose a baseball has
918.75 joules of energy when traveling 35 meters per second. Use this
information to write and graph an equation that gives the energy E of the
baseball as a function of its velocity v.
MIXED REVIEW
PREVIEW
Evaluate the expression. (p. 8)
Prepare for
Lesson 10.2 in
Exs. 46–51.
46. x2 1 5 when x 5 2
47. 4y 2 1 1 when y 5 0
48. 16 1 3m2 when m 5 5
49. 5b2 1 11 when b 5 10
50. 20 2 8w 2 when w 5 1
51. 7z2 2 22 when z 5 3
Solve the linear system using substitution. (p. 435)
52. x 5 20 2 3y
2x 2 y 5 24
53. y 5 2x 2 10
54. x 5 11y 1 4
11 5 3x 2 2y
2x 2 17y 5 13
Use the FOIL pattern to find the product. (p. 562)
55. (2p 1 3)(p 1 2)
634
56. (7x 1 5)(3x 2 1)
EXTRA PRACTICE for Lesson 10.1, p. 947
57. (5n 2 10)(5n 2 9)
ONLINE QUIZ at classzone.com
10.2
Before
Now
Why?
Key Vocabulary
• minimum value
• maximum value
Graph
y 5 ax2 1 bx 1 c
You graphed simple quadratic functions.
You will graph general quadratic functions.
So you can investigate a cable’s height, as in Example 4.
You can use the properties below to graph any quadratic function. You will
justify the formula for the axis of symmetry in Exercise 38 on page 639.
For Your Notebook
KEY CONCEPT
Properties of the Graph of a Quadratic Function
The graph of y 5 ax2 1 bx 1 c is a parabola that:
• opens up if a . 0 and opens down if a , 0.
y 5 ax 2 1 bx 1 c, a > 0
• is narrower than the graph of y 5 x2
if ⏐a⏐ . 1 and wider if ⏐a⏐ , 1.
y
(0, c)
b
• has an axis of symmetry of x 5 2}
.
2a
x
b
• has a vertex with an x-coordinate of 2}
.
2a
b
x 5 2 2a
• has a y-intercept of c. So, the point (0, c) is
on the parabola.
EXAMPLE 1
Find the axis of symmetry and the vertex
Consider the function y 5 22x 2 1 12x 2 7.
a. Find the axis of symmetry of the graph of the function.
b. Find the vertex of the graph of the function.
Solution
a. For the function y 5 22x2 1 12x 2 7, a 5 22 and b 5 12.
b
IDENTIFY THE
VERTEX
Because the vertex
lies on the axis of
symmetry, x 5 3, the
x-coordinate of the
vertex is 3.
12
x 5 2}
5 2}
53
2a
2(22)
Substitute 22 for a and 12 for b. Then simplify.
b
2a
b. The x-coordinate of the vertex is 2} , or 3.
To find the y-coordinate, substitute 3 for x in the function and find y.
y 5 –2(3)2 1 12(3) 2 7 5 11
Substitute 3 for x. Then simplify.
c The vertex is (3, 11).
10.2 Graph y 5 ax2 1 bx 1 c
635
EXAMPLE 2
Graph y 5 ax 2 1 bx 1 c
Graph y 5 3x 2 2 6x 1 2.
STEP 1 Determine whether the parabola opens up or down. Because a . 0,
the parabola opens up.
AVOID ERRORS
26
b
Be sure to include the
negative sign before
the fraction when
calculating the axis of
symmetry.
STEP 2 Find and draw the axis of symmetry: x 5 2}
5 2}
5 1.
2a
2(3)
STEP 3 Find and plot the vertex.
y
b
(21, 11)
The x-coordinate of the vertex is 2}
,
2a
or 1.
To find the y-coordinate, substitute
1 for x in the function and simplify.
(3, 11)
x51
axis of
symmetry
y 5 3(1)2 2 6(1) 1 2 5 21
So, the vertex is (1, 21).
3
STEP 4 Plot two points. Choose two x-values
(0, 2)
less than the x-coordinate of the vertex.
Then find the corresponding y-values.
(2, 2)
3
REVIEW
REFLECTIONS
For help with
reflections, see p. 922.
x
0
21
y
2
11
STEP 5 Reflect the points plotted in Step 4 in the axis of symmetry.
STEP 6 Draw a parabola through the plotted points.
"MHFCSB
✓
vertex (1, 21)
at classzone.com
GUIDED PRACTICE
for Examples 1 and 2
1. Find the axis of symmetry and the vertex of the graph of the function
y 5 x2 2 2x 2 3.
2. Graph the function y 5 3x2 1 12x 2 1. Label the vertex and axis
of symmetry.
For Your Notebook
KEY CONCEPT
Minimum and Maximum Values
For y 5 ax2 1 bx 1 c, the y-coordinate of the vertex is the minimum value
of the function if a . 0 or the maximum value of the function if a , 0.
y 5 ax2 1 bx 1 c, a > 0
y 5 ax2 1 bx 1 c, a < 0
y
y
maximum
x
minimum
636
Chapter 10 Quadratic Equations and Functions
x
x
EXAMPLE 3
Find the minimum or maximum value
Tell whether the function f(x) 5 23x 2 2 12x 1 10 has a minimum value or a
maximum value. Then find the minimum or maximum value.
Solution
Because a 5 23 and 23 , 0, the parabola opens down and the function has a
maximum value. To find the maximum value, find the vertex.
b
212
b
x 5 2}
5 2}
5 22
2a
2(23)
The x-coordinate is 2}
2a .
f(22) 5 23(22)2 2 12(22) 1 10 5 22
Substitute 22 for x. Then simplify.
c The maximum value of the function is f(22) 5 22.
EXAMPLE 4
Find the minimum value of a function
SUSPENSION BRIDGES The suspension cables between the two towers of the
Mackinac Bridge in Michigan form a parabola that can be modeled by the
graph of y 5 0.000097x2 2 0.37x 1 549 where x and y are measured in feet.
What is the height of the cable above the water at its lowest point?
500
x
500
Solution
The lowest point of the cable is at the vertex of the parabola. Find the
x-coordinate of the vertex. Use a 5 0.000097 and b 5 20.37.
b
20.37
x 5 2}
5 2}
≈ 1910
2a
2(0.000097)
Use a calculator.
Substitute 1910 for x in the equation to find the y-coordinate of the vertex.
y ≈ 0.000097(1910)2 2 0.37(1910) 1 549 ≈ 196
c The cable is about 196 feet above the water at its lowest point.
✓
GUIDED PRACTICE
for Examples 3 and 4
3. Tell whether the function f(x) 5 6x2 1 18x 1 13 has a minimum value or a
maximum value. Then find the minimum or maximum value.
4. SUSPENSION BRIDGES The cables between the two towers of the Takoma
Narrows Bridge form a parabola that can be modeled by the graph of the
equation y 5 0.00014x2 2 0.4x 1 507 where x and y are measured in feet.
What is the height of the cable above the water at its lowest point? Round
your answer to the nearest foot.
10.2 Graph y 5 ax2 1 bx 1 c
637
10.2
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 9 and 41
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 12, 27, 37, 42, and 44
SKILL PRACTICE
1. VOCABULARY Explain how you can tell whether a quadratic function has
a maximum value or minimum value without graphing the function.
2.
★ WRITING Describe the steps you would take to graph a quadratic
function in standard form.
EXAMPLE 1
FINDING AXIS OF SYMMETRY AND VERTEX Find the axis of symmetry and
on p. 635
for Exs. 3–14
the vertex of the graph of the function.
3. y 5 2x2 2 8x 1 6
4. y 5 x2 2 6x 1 11
5. y 5 23x2 1 24x 2 22
6. y 5 2x2 2 10x
7. y 5 6x2 1 6x
8. y 5 4x2 1 7
2
3
12.
1
1 2
10. y 5 }
x 1 8x 2 9
9. y 5 2} x2 2 1
11. y 5 2}
x2 1 3x 2 2
4
2
★ MULTIPLE CHOICE What is the vertex of the graph of the function
y 5 23x2 1 18x 2 13?
A (23, 294)
B (23, 214)
C (3, 213)
D (3, 14)
ERROR ANALYSIS Describe and correct the error in finding the axis of
symmetry of the graph of the given function.
3
2
13. y 5 2x2 1 16x 2 1
14. y 5 2}x2 1 18x 2 5
18
16
b
5}
54
x5}
b
x 5 2}
5 2}
5 –6
3
The axis of symmetry is x 5 4.
The axis of symmetry is x 5 26.
2a
21 } 2
2a
2(2)
2
EXAMPLE 2
GRAPHING QUADRATIC FUNCTIONS Graph the function. Label the vertex
on p. 636
for Exs. 15–27
and axis of symmetry.
15. y 5 x2 1 6x 1 2
16. y 5 x2 1 4x 1 8
17. y 5 2x2 1 7x 1 21
18. y 5 5x2 1 10x 2 3
19. y 5 4x2 1 x 2 32
20. y 5 24x2 1 4x 1 8
21. y 5 23x2 2 2x 2 5
22. y 5 28x2 2 12x 1 1
1
1
23. y 5 2x2 1 }
x1}
1 2
24. y 5 }
x 1 6x 2 9
3
1
25. y 5 2} x2 1 6x 1 3
2
27.
4
2
1 2
26. y 5 2}
x 2x11
4
★ MULTIPLE CHOICE Which function has the
y
graph shown?
A y 5 22x2 1 8x 1 3
(2, 5)
5
(0, 3)
1
B y 5 2}
x2 1 2x 1 3
2
1 2
C y5}
x 1 2x 1 3
2
2
D y 5 2x 1 8x 1 3
638
Chapter 10 Quadratic Equations and Functions
1
x
EXAMPLE 3
on p. 637
for Exs. 28–36
MAXIMUM AND MINIMUM VALUES Tell whether the function has a minimum
value or a maximum value. Then find the minimum or maximum value.
28. f(x) 5 x2 2 6
29. f(x) 5 25x2 1 7
30. f(x) 5 4x2 1 32x
31. f(x) 5 23x2 1 12x 2 20
32. f(x) 5 x2 1 7x 1 8
33. f(x) 5 22x2 2 x 1 10
1 2
34. f(x) 5 }
x 2 2x 1 5
35. f(x) 5 2} x2 1 9x
2
37.
3
8
1 2
36. f(x) 5 }
x 1 7x 1 11
4
★ WRITING Compare the graph of y 5 x 1 4x 1 1 with the graph of
y 5 x2 2 4x 1 1.
2
38. REASONING Follow the steps below to justify the equation for the axis
of symmetry for the graph of y 5 ax2 1 bx 1 c. Because the graph of
y 5 ax2 1 bx 1 c is a vertical translation of the graph of y 5 ax2 1 bx,
the two graphs have the same axis of symmetry. Use the function
y 5 ax2 1 bx in place of y 5 ax2 1 bx 1 c.
a. Find the x-intercepts of the graph of y 5 ax2 1 bx. (You can do this
by finding the zeros of the function y 5 ax2 1 bx using factoring.)
b. Because a parabola is symmetric about its axis of symmetry, the
axis of symmetry passes through a point halfway between the
x-intercepts of the parabola. Find the x-coordinate of this point.
What is an equation of the vertical line through this point?
39. CHALLENGE Write a function of the form y 5 ax2 1 bx whose graph
contains the points (1, 6) and (3, 6).
PROBLEM SOLVING
GRAPHING CALCULATOR You may wish to use a graphing calculator to
complete the following Problem Solving exercises.
EXAMPLE 4
on p. 637
for Exs. 40–42
40. SPIDERS Fishing spiders can propel themselves across water and leap
vertically from the surface of the water. During a vertical jump, the
height of the body of the spider can be modeled by the function
y 5 24500x2 1 820x 1 43 where x is the duration (in seconds) of the
jump and y is the height (in millimeters) of the spider above the surface
of the water. After how many seconds does the spider’s body reach its
maximum height? What is the maximum height?
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
41. ARCHITECTURE The parabolic arches that support the roof of the
Dallas Convention Center can be modeled by the graph of the equation
y 5 20.0019x2 1 0.71x where x and y are measured in feet. What is the
height h at the highest point of the arch as shown in the diagram?
y
h
20
30
x
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
10.2 Graph y 5 ax2 1 bx 1 c
639
42.
★ EXTENDED RESPONSE Students are selling packages of flower bulbs
to raise money for a class trip. Last year, when the students charged
$5 per package, they sold 150 packages. The students want to increase
the cost per package. They estimate that they will lose 10 sales for each
$1 increase in the cost per package. The sales revenue R (in dollars)
generated by selling the packages is given by the function
R 5 (5 1 n)(150 2 10n) where n is the number of $1 increases.
a. Write the function in standard form.
b. Find the maximum value of the function.
c. At what price should the packages be sold to generate the most sales
revenue? Explain your reasoning.
y
43. AIRCRAFT An aircraft hangar is a large
building where planes are stored. The
opening of one airport hangar is a
parabolic arch that can be modeled by the
graph of the equation y 5 20.007x2 1 1.7x
where x and y are measured in feet. Graph
the function. Use the graph to determine
how wide the hangar is at its base.
44.
50
x
50
★ SHORT RESPONSE The casts of some Broadway shows go on tour,
performing their shows in cities across the United States. For the period
1990–2001, the number of tickets sold S (in millions) for Broadway road
tours can be modeled by the function S 5 332 1 132t 2 10.4t 2 where t is
the number of years since 1990. Was the greatest number of tickets for
Broadway road tours sold in 1995? Explain.
45. CHALLENGE During an archery competition,
an archer shoots an arrow from 1.5 meters
off of the ground. The arrow follows the
parabolic path shown and hits the ground in
front of the target 90 meters away. Use the
y-intercept and the points on the graph to
write an equation for the graph that models
the path of the arrow.
y
2
vertex (18, 1.6)
(0, 1.5)
1
10
(90, 0) x
MIXED REVIEW
Graph the equation. (pp. 215, 225, 244)
47. x 2 5y 5 15
2
48. y 5 2}
x26
49. 23(4 2 2x) 2 9 (p. 96)
50. 2(1 2 a) 2 5a (p. 96)
51. } (p. 103)
52. (22mn)4 (p. 489)
53. 5 p (7w 7)2 (p. 489)
46. y 5 3
3
Simplify.
12y 2 4
24
6u3 uv2
54. } p } (p. 495)
v
36
PREVIEW
Find the zeros of the polynomial function.
Prepare for
Lesson 10.3 in
Exs. 55–58.
55. f(x) 5 x2 2 4x 2 21 (p. 583)
56. f(x) 5 x2 1 10x 1 24 (p. 583)
57. f(x) 5 5x2 1 18x 1 9 (p. 593)
58. f(x) 5 2x2 1 4x 2 6 (p. 593)
640
EXTRA PRACTICE for Lesson 10.2, p. 947
ONLINE QUIZ at classzone.com
Extension
Use after Lesson 10.2
Graph Quadratic Functions in
Intercept Form
GOAL Graph quadratic functions in intercept form.
Key Vocabulary
• intercept form
In Lesson 10.2 you graphed quadratic functions written in standard
form. Quadratic functions can also be written in intercept form,
y 5 a(x 2 p)(x 2 q) where a Þ 0. In this form, the x-intercepts of the graph
can easily be determined.
For Your Notebook
KEY CONCEPT
Graph of Intercept Form y 5 a(x 2 p)(x 2 q)
Characteristics of the graph of y 5 a(x 2 p)(x 2 q):
y
• The x-intercepts are p and q.
x5
p1q
2
• The axis of symmetry is halfway between
(p, 0) and (q, 0). So, the axis of symmetry
p1q
2
is x 5 }.
(q, 0)
(p, 0)
x
• The parabola opens up if a . 0 and opens
down if a , 0.
EXAMPLE 1
Graph a quadratic function in intercept form
Graph y 5 2(x 1 1)(x 2 5).
FIND ZEROS OF A
FUNCTION
Notice that the x-intercepts
of the graph are also the
zeros of the function:
0 5 2(x 1 1)( x 2 5)
x 1 1 5 0 or x 2 5 5 0
x 5 21 or x 5 5
Solution
STEP 1 Identify and plot the x-intercepts. Because p 5 21 and q 5 5,
the x-intercepts occur at the points (21, 0) and (5, 0).
STEP 2 Find and draw the axis of symmetry.
y
p1q
15
x 5 } 5 21
}52
2
2
(2, 9)
STEP 3 Find and plot the vertex.
The x-coordinate of the vertex is 2.
5
To find the y-coordinate of the vertex,
substitute 2 for x and simplify.
y 5 2(2 1 1)(2 2 5) 5 9
So, the vertex is (2, 9).
STEP 4 Draw a parabola through the vertex
(21, 0)
(5, 0)
1
x
and the points where the x-intercepts
occur.
Extension: Graph Quadratic Functions in Intercept Form
641
EXAMPLE 2
Graph a quadratic function
Graph y 5 2x 2 2 8.
Solution
STEP 1 Rewrite the quadratic function in intercept form.
y 5 2x2 2 8
2
Write original function.
5 2(x 2 4)
Factor out common factor.
5 2(x 1 2)(x 2 2)
Difference of two squares pattern
STEP 2 Identify and plot the x-intercepts. Because p 5 22 and q 5 2,
the x-intercepts occur at the points (22, 0) and (2, 0).
STEP 3 Find and draw the axis of symmetry.
p1q
12
x 5 } 5 22
}50
2
2
y
(22, 0)
21
1
(2, 0)
STEP 4 Find and plot the vertex.
The x-coordinate of the vertex is 0.
The y-coordinate of the vertex is:
y 5 2(0)2 2 8 5 28
So, the vertex is (0, 28).
(0, 28)
STEP 5 Draw a parabola through the vertex and
the points where the x-intercepts occur.
"MHFCSB
at classzone.com
PRACTICE
EXAMPLE 1
on p. 641 for
Exs. 1–9
Graph the quadratic function. Label the vertex, axis of symmetry, and
x-intercepts.
1. y 5 (x 1 2)(x 2 3)
2. y 5 (x 1 5)(x 1 2)
3. y 5 (x 1 9)2
4. y 5 22(x 2 5)(x 1 1)
5. y 5 25(x 1 7)(x 1 2)
6. y 5 3(x 2 6)(x 2 3)
1
7. y 5 2}
(x 1 4)(x 2 2)
8. y 5 (x 2 7)(2x 2 3)
9. y 5 2(x 1 10)(x 2 3)
2
EXAMPLE 2
10. y 5 2x2 1 8x 2 16
11. y 5 2x2 2 9x 2 18
12. y 5 12x2 2 48
on p. 642 for
Exs. 10–15
13. y 5 26x2 1 294
14. y 5 3x2 2 24x 1 36
15. y 5 20x2 2 6x 2 2
16. Follow the steps below to write an equation of the
y
parabola shown.
a. Find the x-intercepts.
b. Use the values of p and q and the coordinates of
the vertex to find the value of a in the equation
y 5 a(x 2 p)(x 2 q).
c. Write a quadratic equation in intercept form.
642
Chapter 10 Quadratic Equations and Functions
1
1
x
x
10.4
Before
Now
Why?
Key Vocabulary
• square root,
p. 110
• perfect square,
p. 111
Use Square Roots to
Solve Quadratic Equations
FPO
You solved a quadratic equation by graphing.
You will solve a quadratic equation by finding square roots.
So you can solve a problem about a falling object, as in Example 5.
To use square roots to solve a quadratic equation of the form ax2 1 c 5 0, first
isolate x2 on one side to obtain x2 5 d. Then use the following information
about the solutions of x2 5 d to solve the equation.
For Your Notebook
KEY CONCEPT
Solving x 2 5 d by Taking Square Roots
d>0
• If d 5 0, then x2 5 d has one solution: x 5 0.
d50
• If d , 0, then x2 5 d has no solution.
d<0
x 5 6Ïd .
READING
Recall that in this
course, solutions
refers to real-number
solutions.
y
• If d . 0,}then x2 5 d has two solutions:
EXAMPLE 1
x
Solve quadratic equations
Solve the equation.
a. 2x2 5 8
b. m2 2 18 5 218
c. b2 1 12 5 5
Solution
ANOTHER WAY
You can also use
factoring to solve
2x2 2 8 5 0:
2x2 2 8 5 0
2(x2 2 4) 5 0
2(x 2 2)(x 1 2) 5 0
x 5 2 or x 5 22
a. 2x2 5 8
Write original equation.
x2 5 4
Divide each side by 2.
}
x 5 6Ï4 5 62
Take square roots of each side. Simplify.
c The solutions are 22 and 2.
b. m2 2 18 5 218
m2 5 0
m50
Write original equation.
Add 18 to each side.
The square root of 0 is 0.
c The solution is 0.
c. b2 1 12 5 5
2
b 5 27
Write original equation.
Subtract 12 from each side.
c Negative real numbers do not have real square roots. So, there is
no solution.
652
Chapter 10 Quadratic Equations and Functions
SIMPLIFYING SQUARE ROOTS In cases where you need to take the square
root of a fraction whose numerator and denominator are perfect squares,
Î25
}
16
the radical can be written as a fraction. For example, }
can be written
2
16
4
4
as }
because 1 }
.
2 5}
5
5
25
EXAMPLE 2
Take square roots of a fraction
Solve 4z2 5 9.
Solution
4z2 5 9
9
z2 5 }
4
Write original equation.
Divide each side by 4.
Î4
}
9
z56 }
Take square roots of each side.
3
z 5 6}
2
Simplify.
3
3
c The solutions are 2}
and }
.
2
2
APPROXIMATING SQUARE ROOTS In cases where d in the equation x2 5 d
is not a perfect square or a fraction whose numerator and denominator are
not perfect squares, you need to approximate the square root. A calculator
can be used to find an approximation.
EXAMPLE 3
Approximate solutions of a quadratic equation
Solve 3x 2 2 11 5 7. Round the solutions to the nearest hundredth.
Solution
3x2 2 11 5 7
Write original equation.
2
3x 5 18
Add 11 to each side.
2
x 56
Divide each side by 3.
}
x 5 6 Ï6
Take square roots of each side.
x ø 6 2.45
Use a calculator. Round to the nearest hundredth.
c The solutions are about 22.45 and about 2.45.
✓
GUIDED PRACTICE
for Examples 1, 2, and 3
Solve the equation.
1. c 2 2 25 5 0
2. 5w 2 1 12 5 28
3. 2x2 1 11 5 11
4. 25x2 5 16
5. 9m2 5 100
6. 49b2 1 64 5 0
Solve the equation. Round the solutions to the nearest hundredth.
7. x2 1 4 5 14
8. 3k 2 2 1 5 0
9. 2p2 2 7 5 2
10.4 Use Square Roots to Solve Quadratic Equations
653
EXAMPLE 4
Solve a quadratic equation
Solve 6(x 2 4)2 5 42. Round the solutions to the nearest hundredth.
6(x 2 4)2 5 42
Write original equation.
(x 2 4)2 5 7
Divide each side by 6.
}
x 2 4 5 6 Ï7
Take square roots of each side.
}
x 5 4 6 Ï7
Add 4 to each side.
}
}
c The solutions are 4 1 Ï 7 ø 6.65 and 4 2 Ï 7 ø 1.35.
CHECK To check the solutions, first write
the equation so that 0 is on one
side as follows: 6(x 2 4)2 2 42 5 0.
Then graph the related function
y 5 6(x 2 4)2 2 42. The x-intercepts
appear to be about 6.6 and about
1.3. So, each solution checks.
EXAMPLE 5
ANOTHER WAY
For alternative methods
for solving the problem
in Example 5, turn
to page 659 for the
Problem Solving
Workshop.
1.3
6.6
Solve a multi-step problem
SPORTS EVENT During an ice hockey game, a
remote-controlled blimp flies above the crowd
and drops a numbered table-tennis ball. The
number on the ball corresponds to a prize.
Use the information in the diagram to find the
amount of time that the ball is in the air.
Solution
STEP 1 Use the vertical motion model to write
an equation for the height h (in feet)
of the ball as a function of time t (in
seconds).
DETERMINE
VELOCITY
When an object is
dropped, it has an
initial vertical velocity
of 0 feet per second.
h 5 216t 2 1 vt 1 s
Vertical motion model
h 5 216t 2 1 0t 1 45
Substitute for v and s.
STEP 2 Find the amount of time the ball is in the
air by substituting 17 for h and solving for t.
h 5 216t 2 1 45
17 5 216t 2 1 45
228 5 216t
28
16
}5t
INTERPRET
SOLUTION
Because the time
cannot be a negative
number, ignore the
negative square root.
654
2
2
Write model.
Substitute 17 for h.
Subtract 45 from each side.
Divide each side by 216.
Î2816 5 t
Take positive square root.
1.32 ø t
Use a calculator.
}
}
c The ball is in the air for about 1.32 seconds.
Chapter 10 Quadratic Equations and Functions
45 ft
17 ft
Not drawn to scale
✓
GUIDED PRACTICE
for Examples 4 and 5
Solve the equation. Round the solutions to the nearest hundredth, if necessary.
10. 2(x 2 2)2 5 18
11. 4(q 2 3)2 5 28
12. 3(t 1 5)2 5 24
13. WHAT IF? In Example 5, suppose the table-tennis ball is released 58 feet
above the ground and is caught 12 feet above the ground. Find the amount
of time that the ball is in the air. Round your answer to the nearest
hundredth of a second.
10.4
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 25 and 59
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 15, 16, 29, 51, 52, 57, and 60
5 MULTIPLE REPRESENTATIONS
Ex. 62
SKILL PRACTICE
1. VOCABULARY Copy and complete: If b2 5 a, then b is a(n) ? of a.
2.
★ WRITING Describe two methods for solving a quadratic equation of the
form ax2 1 c 5 0.
EXAMPLES
1 and 2
on pp. 652–653
for Exs. 3–16
SOLVING EQUATIONS Solve the equation.
3. 3x2 2 3 5 0
4. 2x2 2 32 5 0
5. 4x2 2 400 5 0
6. 2m2 2 42 5 8
7. 15d2 5 0
8. a2 1 8 5 3
9. 4g 2 1 10 5 11
10. 2w 2 1 13 5 11
11. 9q2 2 35 5 14
12. 25b2 1 11 5 15
13. 3z2 2 18 5 218
14. 5n2 2 17 5 219
15.
★ MULTIPLE CHOICE Which of the following is a solution of the equation
61 2 3n2 5 214?
A 5
16.
B 10
C 25
D 625
★ MULTIPLE CHOICE Which of the following is a solution of the equation
13 2 36x2 5 212?
6
A 2}
5
5
C }
1
B }
6
6
D 5
EXAMPLE 3
APPROXIMATING SQUARE ROOTS Solve the equation. Round the solutions to
on p. 653
for Exs. 17–29
the nearest hundredth.
17. x2 1 6 5 13
18. x2 1 11 5 24
19. 14 2 x2 5 17
20. 2a2 2 9 5 11
21. 4 2 k 2 5 4
22. 5 1 3p2 5 38
23. 53 5 8 1 9m2
24. 221 5 15 2 2z2
25. 7c 2 5 100
26. 5d2 1 2 5 6
27. 4b2 2 5 5 2
28. 9n2 2 14 5 23
29.
★ MULTIPLE CHOICE The equation 17 2 }14 x2 5 12 has a solution between
which two integers?
A 1 and 2
B 2 and 3
C 3 and 4
D 4 and 5
10.4 Use Square Roots to Solve Quadratic Equations
655
ERROR ANALYSIS Describe and correct the error in solving the equation.
30. 2x2 2 54 5 18
31. 7d2 2 6 5 217
2x2 2 54 5 18
7d2 2 6 5 217
2x2 5 72
7d2 5 211
x2 5 36
11
d 2 5 2}
7
}
x 5 Ï 36
d ø 61.25
x56
The solutions are about 21.25
and about 1.25.
The solution is 6.
EXAMPLE 4
SOLVING EQUATIONS Solve the equation. Round the solutions to the nearest
on p. 654
for Exs. 32–40
hundredth.
32. (x 2 7)2 5 6
33. 7(x 2 3)2 5 35
34. 6(x 1 4)2 5 18
35. 20 5 2(m 1 5)2
36. 5(a 2 2)2 5 70
37. 21 5 3(z 1 14)2
1
38. }
(c 2 8)2 5 3
3
39. }
(n 1 1)2 5 33
4
40. }
(k 2 6)2 5 20
2
2
3
SOLVING EQUATIONS Solve the equation. Round the solutions to the nearest
hundredth, if necessary.
41. 3x2 2 35 5 45 2 2x2
42. 42 5 3(x2 1 5)
25 2
44. t}
5 49
45. 11 }
1
3
2
1 w 22 7 2
2
2 20 5 101
43. 11x2 1 3 5 5(4x2 2 3)
46. (4m2 2 6)2 5 81
GEOMETRY Use the given area A of the circle to find the radius r or the
diameter d to the nearest hundredth.
47. A 5 144π in.2
r
48. A 5 21π m 2
49. A 5 34π ft 2
r
d
50. REASONING An equation of the graph shown is
y
1
y5}
(x 2 2)2 1 1. Two points on the parabola have
2
y-coordinates of 9. Find the x-coordinates of these
points.
1
1
51.
x
★ SHORT RESPONSE Solve x2 5 1.44 without using a calculator. Explain
your reasoning.
52.
★ OPEN – ENDED
Give values for a and c so that ax2 1 c 5 0 has
(a) two solutions, (b) one solution, and (c) no solution.
CHALLENGE Solve the equation without graphing.
53. x2 2 12x 1 36 5 64
656
5 WORKED-OUT SOLUTIONS
on p. WS1
54. x2 1 14x 1 49 5 16
★ 5 STANDARDIZED
TEST PRACTICE
55. x2 1 18x 1 81 5 25
PROBLEM SOLVING
EXAMPLE 5
56. FALLING OBJECT Fenway Park is a Major League Baseball park in Boston,
on p. 654
for Exs. 56–57
Massachusetts. The park offers seats on top of the left field wall. A person
sitting in one of these seats accidentally drops his sunglasses on the field.
The height h (in feet) of the sunglasses can be modeled by the function
h 5 216t 2 1 38 where t is the time (in seconds) since the sunglasses
were dropped. Find the time it takes for the sunglasses to reach the field.
Round your answer to the nearest hundredth of a second.
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
57.
★ MULTIPLE CHOICE Which equation can be used to find the time it
takes for an object to hit the ground after it was dropped from a height
of 68 feet?
A 216t 2 5 0
B 216t 2 2 68 5 0 C 216t 2 1 68 5 0 D 216t 2 5 68
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
58. INTERNET USAGE For the period 1995–2001, the number y (in
thousands) of Internet users worldwide can be modeled by the
function y 5 12,697x2 1 55,722 where x is the number of years since
1995. Between which two years did the number of Internet users
worldwide reach 100,000,000?
59. GEMOLOGY To find the weight w (in carats) of round faceted gems,
gemologists use the formula w 5 0.0018D2ds where D is the diameter (in
millimeters) of the gem, d is the depth (in millimeters) of the gem, and
s is the specific gravity of the gem. Find the diameter to the nearest tenth
of a millimeter of each round faceted gem in the table.
Gem
60.
Weight
(carats)
Depth
(mm)
Specific
gravity
Diameter
(mm)
a.
Amethyst
1
4.5
2.65
?
b.
Diamond
1
4.5
3.52
?
c.
Ruby
1
4.5
4.00
?
★ SHORT RESPONSE In deep water, the speed s (in meters per second)
of a series of waves and the wavelength L (in meters) of the waves are
related by the equation 2πs2 5 9.8L.
L
Crest
Crest
The wavelength L is the distance between one crest and the next.
a. Find the speed to the nearest hundredth of a meter per second of a
series of waves with the following wavelengths: 6 meters, 10 meters,
and 25 meters. (Use 3.14 for π.)
b. Does the speed of a series of waves increase or decrease as the
wavelength of the waves increases? Explain.
10.4 Use Square Roots to Solve Quadratic Equations
657
61. MULTI-STEP PROBLEM The Doyle log rule is a formula
used to estimate the amount of lumber that can be
sawn from logs of various sizes. The amount of lumber
L(D 2 4)
16
Diameter
2
V (in board feet) is given by V 5 } where L is
Boards
the length (in feet) of a log and D is the small-end
diameter (in inches) of the log.
a. Solve the formula for D.
b. Use the rewritten formula to find the diameters, to the nearest tenth
of a foot, of logs that will yield 50 board feet and have the following
lengths: 16 feet, 18 feet, 20 feet, and 22 feet.
62.
MULTIPLE REPRESENTATIONS A ride at an amusement
park lifts seated riders 250 feet above the ground. Then
the riders are dropped. They experience free fall until the
brakes are activated at 105 feet above the ground.
a. Writing an Equation Use the vertical motion model to
write an equation for the height h (in feet) of the riders
as a function of the time t (in seconds) into the free fall.
b. Making a Table Make a table that shows the height of the
riders after 0, 1, 2, 3, and 4 seconds. Use the table
to estimate the amount of time the riders experience
free fall.
c. Solving an Equation Use the equation to find the amount
of time, to the nearest tenth of a second, that the riders
experience free fall.
63. CHALLENGE The height h (in feet) of a dropped object on any planet
g
2
can be modeled by h 5 2} t 2 1 s where g is the acceleration (in feet per
second per second) due to the planet’s gravity, t is the time (in seconds)
after the object is dropped, and s is the initial height (in feet) of the
object. Suppose the same object is dropped from the same height on
Earth and Mars. Given that g is 32 feet per second per second on Earth
and 12 feet per second per second on Mars, on which planet will the
object hit the ground first? Explain.
MIXED REVIEW
PREVIEW
Evaluate the power. (p. 2)
Prepare for
Lesson 10.5 in
Exs. 64–67.
5
64. }
122
2
152
9
65. }
142
2
3
66. }
2
122
7
67. }
2
Write an equation of the line with the given slope and y-intercept. (p. 283)
68. slope: 29
y-intercept: 11
69. slope: 7
70. slope: 3
y-intercept: 27
y-intercept: 22
Write an equation of the line that passes through the given point and is
perpendicular to the given line. (p. 319)
71. (1, 21), y 5 2x
658
72. (0, 8), y 5 4x 1 1
EXTRA PRACTICE for Lesson 10.4, p. 947
73. (29, 24), y 5 23x 1 6
ONLINE QUIZ at classzone.com
Using
ALTERNATIVE METHODS
LESSON 10.4
Another Way to Solve Example 5, page 654
MULTIPLE REPRESENTATIONS In Example 5 on page 654, you saw how to
solve a problem about a dropped table-tennis ball by using a square root.
You can also solve the problem by using factoring or by using a table.
PROBLEM
SPORTS EVENT During an ice hockey
game, a remote-controlled blimp flies
above the crowd and drops a numbered
table-tennis ball. The number on the
ball corresponds to a prize. Use the
information in the diagram to find the
amount of time that the ball is in the air.
45 ft
17 ft
Not drawn to scale
METHOD 1
Using Factoring One alternative approach is to use factoring.
STEP 1 Write an equation for the height h (in feet) of the ball as a function of
time t (in seconds) after it is dropped using the vertical motion model.
h 5 216t 2 1 vt 1 s
Vertical motion model
h 5 216t 2 1 0t 1 45
Substitute 0 for v and 45 for s.
STEP 2 Substitute 17 for h to find the time it takes the ball to reach a
height of 17 feet. Then write the equation so that 0 is on one side.
17 5 216t 2 1 45
Substitute 17 for h.
0 5 216t 2 1 28
Subtract 17 from each side.
STEP 3 Solve the equation by factoring. Replace 28 with the closest
USE AN
APPROXIMATION
By replacing 28 with
25, you will obtain
an answer that is an
approximation of the
amount of time that the
ball is in the air.
perfect square, 25, so that the right side of the equation is
factorable as a difference of two squares.
0 5 216t 2 1 25
Use 25 as an approximation for 28.
0 5 2(16t 2 2 25)
Factor out 21.
0 5 2(4t 2 5)(4t 1 5)
Difference of two squares pattern
4t 2 5 5 0 or 4t 1 5 5 0
5
5
t5}
or t 5 2 }
4
4
Zero-product property
Solve for t.
5
c The ball is in the air about }
, or 1.25, seconds.
4
Using Alternative Methods
659
METHOD 2
Using a Table Another approach is to make and use a table.
STEP 1 Make a table that shows the height
h (in feet) of the ball by substituting
values for time t (in seconds) in the function
h 5 216t 2 1 45. Use increments of 1 second.
Time t
(seconds)
Height h
(feet)
0
45
1
29
2
219
Time t
(seconds)
Height h
(feet)
1.0
29.00
1.1
25.64
1.2
21.96
1.3
17.96
1.4
13.64
STEP 2 Identify the time interval in which the
height of the ball is 17 feet. This happens
between 1 and 2 seconds.
STEP 3 Make a second table using increments of
0.1 second to get a closer approximation.
c The ball is in the air about 1.3 seconds.
P R AC T I C E
1. WHAT IF? In the problem on page 659,
suppose the ball is caught at a height of
10 feet. For how many seconds is the ball
in the air? Solve this problem using two
different methods.
2. OPEN-ENDED Describe a problem about a
dropped object. Then solve the problem and
explain what your solution means in this
situation.
3.
GEOMETRY The box below is a
rectangular prism with the dimensions
shown.
x in.
5 in.
5x in.
a. Write an equation that gives the volume
V (in cubic inches) of the box as a
function of x.
b. The volume of the box is 83 cubic inches.
Find the dimensions of the box. Use
factoring to solve the problem.
c. Make a table to check your answer from
part (b).
660
Chapter 10 Quadratic Equations and Functions
4. TRAPEZE You are learning how to perform
on a trapeze. While hanging from a still
trapeze bar, your shoe comes loose and falls
to a safety net that is 6 feet off the ground. If
your shoe falls from a height of 54 feet, how
long does it take your shoe to hit the net?
Choose any method for solving the problem.
Show your steps.
5. ERROR ANALYSIS A student solved the
problem in Exercise 4 as shown below.
Describe and correct the error.
Let t be the time (in seconds) that the
shoe is in the air.
6 5 216t2 1 54
0 5 216t2 1 60
Replace 60 with the closest perfect
square, 64.
0 5 216t2 1 64
0 5 216(t 2 2)(t 1 2)
t 5 2 or t 5 22
It takes about 2 seconds.
MIXED REVIEW of Problem Solving
STATE TEST PRACTICE
classzone.com
Lessons 10.1–10.4
1. MULTI-STEP PROBLEM A company’s yearly
profits from 1996 to 2006 can be modeled by
the function y 5 x2 2 8x 1 80 where y is the
profit (in thousands of dollars) and x is the
number of years since 1996.
a. In what year did the company experience
its lowest yearly profit?
b. What was the lowest yearly profit?
2. MULTI-STEP PROBLEM Use the rectangle
below.
(14 2 x) ft
2x ft
a. Find the value of x that gives the greatest
possible area of the rectangle.
b. What is the greatest possible area of the
rectangle?
3. EXTENDED RESPONSE You throw a lacrosse
ball twice using a lacrosse stick.
4. OPEN-ENDED Describe a real-world situation
of an object being dropped. Then write an
equation that models the height of the object
as a function of time. Use the equation to
determine the time it takes the object to hit
the ground.
5. SHORT RESPONSE A football player is
attempting a field goal. The path of the
kicked football can be modeled by the
graph of y 5 20.03x2 1 1.8x where x is the
horizontal distance (in yards) traveled by the
football and y is the corresponding height
(in feet) of the football. Will the football pass
over the goal post that is 10 feet above the
ground and 45 yards away? Explain.
6. GRIDDED ANSWER The force F (in newtons)
a rider feels while a train goes around a
2
mv
curve is given by F 5 }
r where m is the
mass (in kilograms) of the rider, v is the
velocity (in meters per second) of the
train, and r is the radius (in meters) of the
curve. A rider with a mass of 75 kilograms
experiences a force of 18,150 newtons, while
going around a curve that has a radius of
8 meters. Find the velocity (in meters per
second) the train travels around the curve.
7. SHORT RESPONSE The opening of the
tunnel shown can be modeled by the graph
of the equation y 5 20.18x2 1 4.4x 2 12
where x and y are measured in feet.
Y
a. For your first throw, the ball is released
8 feet above the ground with an initial
vertical velocity of 35 feet per second.
Use the vertical motion model to write an
equation for the height h (in feet) of the
ball as a function of time t (in seconds).
b. For your second throw, the ball is released
7 feet above the ground with an initial
vertical velocity of 45 feet per second.
Use the vertical motion model to write an
equation for the height h (in feet) of the
ball as a function of time t (in seconds).
c. If no one catches either throw, for which
throw is the ball in the air longer? Explain.
X
a. Find the maximum height of the tunnel.
b. A semi trailer is 7.5 feet wide, and the top
of the trailer is 10.5 feet above the ground.
Given that traffic travels one way on one
lane through the center of the tunnel, will
the semi trailer fit through the opening of
the tunnel? Explain.
Mixed Review of Problem Solving
661
Investigating
g
Algebra
Algebr
ra
ACTIVITY Use before Lesson 10.5
Algebra
classzone.com
10.5 Completing the Square Using Algebra Tiles
M AT E R I A L S • algebra tiles
QUESTION
How can you use algebra tiles to complete the square?
For an expression of the form x2 1 bx, you can add a constant c to the
expression so that the expression x2 1 bx 1 c is a perfect square trinomial.
This process is called completing the square.
EXPLORE
Complete the square
Find the value of c that makes x 2 1 4x 1 c a perfect square trinomial.
STEP 1 Model expression
Use algebra tiles to model the
expression x2 1 4x. You will
need one x2-tile and four
x-tiles for this expression.
DR AW CONCLUSIONS
STEP 2 Rearrange tiles
STEP 3 Complete the square
Arrange the tiles to form a
square. The arrangement will
be incomplete in one of the
corners.
Determine the number of
1-tiles needed to complete the
square. The number of 1-tiles
is the value of c. So, the perfect
square trinomial is x2 1 4x 1 4
or (x 1 2)2.
Use your observations to complete these exercises
1. Copy and complete the table using algebra tiles.
Expression
Number of 1-tiles needed
to complete the square
Expression written
as a square
x2 1 4x
4
x2 1 4x 1 4 5 (x 1 2)2
x2 1 6x
?
?
x2 1 8x
?
?
?
?
2
x 1 10x
2. In the statement x2 1 bx 1 c 5 (x 1 d)2, how are b and d related?
How are c and d related?
3. Use your answer to Exercise 2 to predict the number of 1-tiles
you would need to add to complete the square for the
expression x2 1 18x.
662
Chapter 10 Quadratic Equations and Functions
10.5
Solve Quadratic Equations
by Completing the Square
You solved quadratic equations by finding square roots.
Before
You will solve quadratic equations by completing the square.
Now
So you can solve a problem about snowboarding, as in Ex. 50.
Why?
For an expression of the form x2 1 bx, you can add a constant c to the
expression so that the expression x2 1 bx 1 c is a perfect square trinomial.
This process is called completing the square.
Key Vocabulary
• completing the
square
• perfect square
trinomial, p. 601
For Your Notebook
KEY CONCEPT
Completing the Square
Words To complete the square for the expression x2 1 bx, add the
square of half the coefficient of the term bx.
1 b2 2
Algebra x2 1 bx 1 }
EXAMPLE 1
2
1
b
5 x1}
2
2
2
Complete the square
Find the value of c that makes the expression x 2 1 5x 1 c a perfect square
trinomial. Then write the expression as the square of a binomial.
STEP 1 Find the value of c. For the expression to be a perfect square
trinomial, c needs to be the square of half the coefficient of bx.
2
5
25
c 5 1}
2 5}
2
4
Find the square of half the coefficient of bx.
STEP 2 Write the expression as a perfect square trinomial. Then write
the expression as the square of a binomial.
25
x2 1 5x 1 c 5 x2 1 5x 1 }
4
5 2
5 1x 1 }
22
✓
GUIDED PRACTICE
25
4
Substitute } for c.
Square of a binomial
for Example 1
Find the value of c that makes the expression a perfect square trinomial.
Then write the expression as the square of a binomial.
1. x2 1 8x 1 c
2. x2 2 12x 1 c
3. x2 1 3x 1 c
10.5 Solve Quadratic Equations by Completing the Square
663
SOLVING EQUATIONS The method of completing the square can be used
to solve any quadratic equation. To use completing the square to solve a
quadratic equation, you must write the equation in the form x2 1 bx 5 d.
EXAMPLE 2
Solve a quadratic equation
Solve x 2 2 16x 5 215 by completing the square.
Solution
x2 2 16x 5 215
Write original equation.
x2 2 16x 1 (28)2 5 215 1 (28)2
AVOID ERRORS
When completing the
square to solve an
equation, be sure
you add the term
b2
} to both sides of the
2
1 2
2
1 216
2 2
Add } , or (28) 2 , to each side.
(x 2 8)2 5 215 1 (28)2
Write left side as the square of
a binomial.
(x 2 8)2 5 49
Simplify the right side.
x 2 8 5 67
equation.
Take square roots of each side.
x5867
Add 8 to each side.
c The solutions of the equation are 8 1 7 5 15 and 8 2 7 5 1.
CHECK
You can check the solutions in the original equation.
If x 5 15:
If x 5 1:
(15)2 2 16(15) 0 215
(1)2 2 16(1) 0 215
215 5 215 ✓
EXAMPLE 3
215 5 215 ✓
Solve a quadratic equation in standard form
Solve 2x 2 1 20x 2 8 5 0 by completing the square.
Solution
2x2 1 20x 2 8 5 0
Write original equation.
2x2 1 20x 5 8
Add 8 to each side.
2
x 1 10x 5 4
Divide each side by 2.
2
1 10
2 2
x2 1 10x 1 52 5 4 1 52
AVOID ERRORS
Be sure that the
coefficient of x2 is 1
before you complete
the square.
Add } , or 52 , to each side.
(x 1 5)2 5 29
Write left side as the square of a binomial.
}
x 1 5 5 6Î29
Take square roots of each side.
}
x 5 25 6 Ï 29
Subtract 5 from each side.
}
}
c The solutions are 25 1 Ï29 ø 0.39 and 25 2 Ï 29 ø 210.39.
✓
GUIDED PRACTICE
for Examples 2 and 3
Solve the equation by completing the square. Round your solutions to the
nearest hundredth, if necessary.
4. x2 2 2x 5 3
664
Chapter 10 Quadratic Equations and Functions
5. m2 1 10m 5 28
6. 3g 2 2 24g 1 27 5 0
EXAMPLE 4
Solve a multi-step problem
CRAFTS You decide to use chalkboard paint to
create a chalkboard on a door. You want the
chalkboard to have a uniform border as shown. You
have enough chalkboard paint to cover 6 square feet.
Find the width of the border to the nearest inch.
7IDTHOF
BORDER
XFT
FT
Solution
#HALKBOARD
STEP 1 Write a verbal model. Then write an
equation. Let x be the width (in feet) of
the border.
Area of
chalkboard
(square feet)
WRITE EQUATION
The width of the border
is subtracted twice
because it is at the top
and the bottom of the
door, as well as at the
left and the right.
5
Length of
chalkboard
5
(7 2 2x)
6
(feet)
FT
p
Width of
chalkboard
p
(3 2 2x)
(feet)
STEP 2 Solve the equation.
6 5 (7 2 2x)(3 2 2x)
6 5 21 2 20x 1 4x
Write equation.
2
Multiply binomials.
215 5 4x2 2 20x
Subtract 21 from each side.
15
2}
5 x2 2 5x
4
Divide each side by 4.
2
1 52 2
15
25
25
2}
1}
5 x2 2 5x 1 }
4
4
15
25
5
2}
1}
5 1x 2 }
2
4
4
2
Write right side as the square of a binomial.
2
5 2
2
1
5
2
}5 x2}
Î2
5
5
} 6 Î} 5 x
2
2
25
4
Add 2} , or }, to each side.
4
2
Simplify left side.
}
5
5
6 }
5x2}
Take square roots of each side.
2
}
5
2
Add } to each side.
Î2
}
Î2
}
5
5
5
5
The solutions of the equation are }
1 }
ø 4.08 and }
2 }
ø 0.92.
2
2
It is not possible for the width of the border to be 4.08 feet because the
width of the door is 3 feet. So, the width of the border is 0.92 foot. Convert
0.92 foot to inches.
12 in.
0.92 ft p }
5 11.04 in.
1 ft
Multiply by conversion factor.
c The width of the border should be about 11 inches.
✓
GUIDED PRACTICE
for Example 4
7. WHAT IF? In Example 4, suppose you have enough chalkboard paint to
cover 4 square feet. Find the width of the border to the nearest inch.
10.5 Solve Quadratic Equations by Completing the Square
665
10.5
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 19 and 47
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 24, 25, 42, and 49
5 MULTIPLE REPRESENTATIONS
Ex. 47
SKILL PRACTICE
1. VOCABULARY Copy and complete: The process of writing an expression
of the form x2 1 bx as a perfect square trinomial is called ? .
2.
EXAMPLE 1
on p. 663
for Exs. 3–11
★ WRITING Give an example of an expression that is a perfect square
trinomial. Explain why the expression is a perfect square trinomial.
COMPLETING THE SQUARE Find the value of c that makes the expression
a perfect square trinomial. Then write the expression as the square of a
binomial.
3. x2 1 6x 1 c
4. x2 1 12x 1 c
5. x2 2 4x 1 c
6. x2 2 8x 1 c
7. x2 2 3x 1 c
8. x2 1 5x 1 c
1
10. x2 2 }
x1c
4
11. x2 2 }
x1c
9. x2 1 2.4x 1 c
2
3
EXAMPLES
2 and 3
SOLVING EQUATIONS Solve the equation by completing the square. Round
on p. 664
for Exs. 12–27
12. x2 1 2x 5 3
13. x2 1 10x 5 24
14. c 2 2 14c 5 15
15. n2 2 6n 5 72
16. a2 2 8a 1 15 5 0
17. y 2 1 4y 2 21 5 0
11
18. w 2 2 5w 5 }
19. z2 1 11z 5 2}
4
your solutions to the nearest hundredth, if necessary.
21
4
2
★
1
★
5
4
2
22. v 2 7v 1 1 5 0
23. m 1 3m 1 } 5 0
MULTIPLE CHOICE What are the solutions of 4x2 1 16x 5 9?
9
1 9
B 2}
,}
A 2}
, 2}
2
2
25.
3
2
21. k 2 8k 2 7 5 0
24.
2
20. g 2 2 }
g57
9
1
C }
, 2}
2 2
2
1 9
D }
,}
2
2 2
MULTIPLE CHOICE What are the solutions of x2 1 12x 1 10 5 0?
}
A 26 6 Ï46
}
B 26 6 Ï 26
}
}
C 6 6 Ï 26
D 6 6 Ï 46
ERROR ANALYSIS Describe and correct the error in solving the given
equation.
26. x2 2 14x 5 11
27. x2 2 2x 2 4 5 0
x2 2 14x 5 11
x2 2 2x 2 4 5 0
x2 2 14x 1 49 5 11
x2 2 2x 5 4
(x 2 7) 2 5 11
x2 2 2x 1 1 5 4 1 1
}
x 2 7 5 6Ï11
(x 1 1) 2 5 5
}
x 5 7 6 Ï11
}
x 1 1 5 6Ï 5
}
x 5 1 6 Ï5
666
Chapter 10 Quadratic Equations and Functions
SOLVING EQUATIONS Solve the equation by completing the square. Round
your solutions to the nearest hundredth, if necessary.
28. 2x2 2 8x 2 14 5 0
29. 2x2 1 24x 1 10 5 0
30. 3x2 2 48x 1 39 5 0
31. 4y 2 1 4y 2 7 5 0
32. 9n2 1 36n 1 11 5 0
33. 3w 2 2 18w 2 20 5 0
34. 3p2 2 30p 2 11 5 6p
35. 3a2 2 12a 1 3 5 2a2 2 4
36. 15c 2 2 51c 2 30 5 9c 1 15
37. 7m2 1 24m 2 2 5 m2 2 9 38. g 2 1 2g 1 0.4 5 0.9g 2 1 g
39. 11z2 2 10z 2 3 5 29z2 1 }
3
4
GEOMETRY Find the value of x. Round your answer to the nearest
hundredth, if necessary.
40. Area of triangle 5 108 m 2
41. Area of rectangle 5 288 in.2
3x in.
xm
(2x 1 10) in.
(x 1 6) m
42.
★
1 2
b 2
WRITING How many solutions does x2 1 bx 5 c have if c < 2 } ? Explain.
2
43. CHALLENGE The product of two consecutive negative integers is 210. Find
the integers.
44. CHALLENGE The product of two consecutive positive even integers is 288.
Find the integers.
PROBLEM SOLVING
EXAMPLE 4
45. LANDSCAPING You are building a rectangular brick patio
on p. 665
for Exs. 45–46
surrounded by crushed stone in a rectangular courtyard as
shown. The crushed stone border has a uniform width x (in
feet). You have enough money in your budget to purchase
patio bricks to cover 140 square feet. Solve the equation
140 5 (20 2 2x)(16 2 2x) to find the width of the border.
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
46. TRAFFIC ENGINEERING The distance d (in feet) that it takes a car to come
to a complete stop on dry asphalt can be modeled by d 5 0.05s2 1 1.1s
where s is the speed of the car (in miles per hour). A car has 78 feet to come
to a complete stop. Find the maximum speed at which the car can travel.
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
47.
MULTIPLE REPRESENTATIONS For the period 198522001, the average
salary y (in thousands of dollars) per season of a Major League Baseball
player can be modeled by y 5 7x2 2 4x 1 392 where x is the number of
years since 1985.
a. Solving an Equation Write and solve an equation to find the year
when the average salary was $1,904,000.
b. Drawing a Graph Use a graph to check your solution to part (a).
10.5 Solve Quadratic Equations by Completing the Square
667
48. MULTI-STEP PROBLEM You have 80 feet of fencing to make
a rectangular horse pasture that covers 750 square feet. A
barn will be used as one side of the pasture as shown.
a. Write equations for the perimeter and area of
the pasture.
b. Use substitution to solve the system of equations
W
W
from part (a). What are the possible dimensions of
the pasture?
49.
★
SHORT RESPONSE You purchase stock for $16 per share, and you sell
the stock 30 days later for $23.50 per share. The price y (in dollars) of a
share during the 30 day period can be modeled by y 5 20.025x2 1 x 1 16
where x is the number of days after the stock is purchased. Could you have
sold the stock earlier for $23.50 per share? Explain.
50. SNOWBOARDING During a “big air” competition,
snowboarders launch themselves from a half pipe,
perform tricks in the air, and land back in the
half pipe.
Initial vertical
velocity = 24 ft/sec
a. Model Use the vertical motion model to write
an equation that models the height h (in feet) of
a snowboarder as a function of the time t
(in seconds) she is in the air.
16.4 ft
b. Apply How long is the snowboarder in the air if
she lands 13.2 feet above the base of the half pipe?
Round your answer to the nearest tenth of a
second.
"MHFCSB
Cross section of a half pipe
at classzone.com
51. CHALLENGE You are knitting a rectangular scarf. The pattern you have
created will result in a scarf that has a length of 60 inches and a width of
4 inches. However, you happen to have enough yarn to cover an area of
480 square inches. You decide to increase the dimensions of the scarf so
that all of your yarn will be used. If the increase in the length is 10 times
the increase in the width, what will the dimensions of the scarf be?
MIXED REVIEW
PREVIEW
Evaluate the expression for the given value of x. (p. 74)
Prepare for
Lesson 10.6 in
Exs. 52–57.
52. 3 1 x 2 6; x 5 8
53. 11 2 (2x) 1 15; x 5 21
54. 2x 1 18 2 20; x 5 210
55. 32 2 x 2 5; x 5 5
56. x 1 14.7 2 16.2; x 5 2.3
57. 29.2 2 (211.4) 2 x; x 5 24.5
Solve the proportion. (p. 168)
4
8
58. }
5}
m23
3
5
3
59. }
a5}
a15
c12
6
2c 2 3
5
60. } 5 }
Solve the equation.
668
61. (x 2 4)(x 1 9) 5 0 (p. 575)
62. x2 2 15x 1 26 5 0 (p. 583)
63. 3x2 1 10x 1 7 5 0 (p. 593)
64. 4x2 2 20x 1 25 5 0 (p. 600)
EXTRA PRACTICE for Lesson 10.5, p. 947
ONLINE QUIZ at classzone.com
Extension
Use after Lesson 10.5
Graph Quadratic Functions in
Vertex Form
GOAL Graph quadratic functions in vertex form.
Key Vocabulary
• vertex form
In Lesson 10.2, you graphed quadratic functions in standard form.
Quadratic functions can also be written in vertex form, y 5 a(x 2 h)2 1 k
where a Þ 0. In this form, the vertex of the graph can be easily determined.
For Your Notebook
KEY CONCEPT
Graph of Vertex Form y 5 a(x 2 h)2 1 k
The graph of y 5 a(x 2 h)2 1 k is the graph of y 5 ax2 translated h units
horizontally and k units vertically.
Characteristics of the graph of
y 5 a(x 2 h)2 1 k:
y
y 5 a(x 2 h) 2 1 k
• The vertex is (h, k).
(h, k)
y 5 ax 2
• The axis of symmetry is x 5 h.
• The graph opens up if a > 0, and
k
(0, 0)
x
h
the graph opens down if a < 0.
EXAMPLE 1
Graph a quadratic function in vertex form
Graph y 5 2(x 1 2)2 1 3.
Solution
STEP 1 Identify the values of a, h, and k: a 5 21, h 5 22, and k 5 3.
Because a < 0, the parabola opens down.
STEP 2 Draw the axis of symmetry, x 5 22.
STEP 3 Plot the vertex (h, k) 5 (22, 3).
y
(22, 3)
3
STEP 4 Plot four points. Evaluate the function
for two x-values less than the
x-coordinate of the vertex.
1 x
2
x 5 23: y 5 2(23 1 2) 1 3 5 2
x 5 25: y 5 2(25 1 2)2 1 3 5 26
x 5 22
Plot the points (23, 2) and (25, 26) and
their reflections, (21, 2) and (1, 26), in
the axis of symmetry.
STEP 5 Draw a parabola through the plotted points.
Extension: Graph Quadratic Functions in Vertex Form
669
EXAMPLE 2
Graph a quadratic function
Graph y = x 2 2 8x 1 11.
Solution
STEP 1 Write the function in vertex form by completing the square.
y 5 x2 2 8x 1 11
y 1 ■ 5 (x2 2 8x 1 ■ ) 1 11
Write original function.
y 1 16 5 (x2 2 8x 1 16) 1 11
Add }
y 1 16 5 (x 2 4)2 1 11
Write x2 2 8x 1 16 as a square of
a binomial.
y 5 (x 2 4)2 2 5
Prepare to complete the square.
2
1 28
2 2
5 (24) 2 5 16 to each side.
Subtract 16 from each side.
STEP 2 Identify the values of a, h, and k: a 5 1, h 5 4, and k 5 25.
Because a > 0, the parabola opens up.
STEP 3 Draw the axis of symmetry, x 5 4.
y
STEP 4 Plot the vertex (h, k) 5 (4, 25).
x54
STEP 5 Plot four more points. Evaluate the
1
function for two x-values less than
the x-coordinate of the vertex.
1
x
2
x 5 3: y 5 (3 2 4) 2 5 5 24
x 5 1: y 5 (1 2 4)2 2 5 5 4
Plot the points (3, 24) and (1, 4)
and their reflections, (5, 24) and
(7, 4), in the axis of symmetry.
(4, 25)
STEP 6 Draw a parabola through the plotted points.
PRACTICE
EXAMPLE 1
on p. 669
for Exs. 1–6
EXAMPLE 2
on p. 670
for Exs. 7–12
Graph the quadratic function. Label the vertex and axis of symmetry.
1. y 5 (x 1 2)2 2 5
2. y 5 2(x 2 4)2 1 1
3. y 5 x2 1 3
4. y 5 3(x 2 1)2 2 2
5. y 5 22(x 1 5)2 2 2
1
6. y 5 2 }
(x 1 4)2 1 4
2
Write the function in vertex form, then graph the function. Label the vertex
and axis of symmetry.
7. y 5 x2 2 12x 1 36
8. y 5 x2 1 8x 1 15
9. y 5 2x2 1 10x 2 21
10. y 5 2x2 2 12x 1 19
11. y 5 23x2 2 6x 2 1
12. y 5 2}x2 2 6x 2 21
1
2
13. Write an equation in vertex form of the
parabola shown. Use the coordinates of the
vertex and the coordinates of a point on the
graph to write the equation.
y
(210, 5)
(22, 5)
1
(26, 1)
21
670
Chapter 10 Quadratic Equations and Functions
x
10.6
Before
Now
Why?
Key Vocabulary
• quadratic formula
Solve Quadratic Equations
by the Quadratic Formula
You solved quadratic equations by completing the square.
You will solve quadratic equations using the quadratic formula.
So you can solve a problem about film production, as in Example 3.
By completing the square for the quadratic equation ax2 1 bx 1 c 5 0, you
can develop a formula that gives the solutions of any quadratic equation in
standard form. This formula is called the quadratic formula. (The quadratic
formula is developed on page 727.)
For Your Notebook
KEY CONCEPT
The Quadratic Formula
The solutions of the quadratic equation ax2 1 bx 1 c 5 0 are
2b 6 Îb2 2 4ac
2a
}
x 5}} where a Þ 0 and b2 2 4ac ≥ 0.
★
EXAMPLE 1
Standardized Test Practice
What are the solutions of 3x 2 1 5x 5 8?
8
ANOTHER WAY
Instead of solving the
equation, you can check
the answer choices in
the equation.
8
8
B 21 and }
A 21 and 2}
3
8
D 1 and }
C 1 and 2}
3
3
3
Solution
3x2 1 5x 5 8
Write original equation.
2
3x 1 5x 2 8 5 0
Write in standard form.
}
6 Ïb 2 4ac
x 5 2b
}}
2
Quadratic formula
2a
}}
Ï
x 5 }}
25 6 52 2 4(3)(28)
2(3)
Substitute values in the quadratic formula:
a 5 3, b 5 5, and c 5 28.
}
25 6 Ï121
5}
Simplify.
6 11
5 25
}
Simplify the square root.
6
6
8
1 11
25 2 11
The solutions of the equation are 25
} 5 1 and } 5 2}.
3
6
c The correct answer is C.
6
A B C D
10.6 Solve Quadratic Equations by the Quadratic Formula
671
EXAMPLE 2
Solve a quadratic equation
Solve 2x 2 2 7 5 x.
2x2 2 7 5 x
Write original equation.
2x2 2 x 2 7 5 0
Write in standard form.
}
6 Ïb 2 4ac
x 5 2b
}}
2
Quadratic formula
2a
}}
Ï
5 }}}
2(21) 6 (21)2 2 4(2)(27)
2(2)
Substitute values in the quadratic
formula: a 5 2, b 5 21, and c 5 27.
}
6 Ï57
5 1}
Simplify.
4
}
}
1 Ï57
2 Ï57
c The solutions are 1}
ø 2.14 and 1}
ø 21.64.
4
"MHFCSB
CHECK
✓
4
at classzone.com
Write the equation in standard
form, 2x2 2 x 2 7 5 0. Then graph
the related function y 5 2x2 2 x 2 7.
The x-intercepts are about 21.6 and 2.1.
So, each solution checks.
GUIDED PRACTICE
21.6
2.1
for Examples 1 and 2
Use the quadratic formula to solve the equation. Round your solutions to
the nearest hundredth, if necessary.
1. x2 2 8x 1 16 5 0
EXAMPLE 3
2. 3n2 2 5n 5 21
3. 4z2 5 7z 1 2
Use the quadratic formula
FILM PRODUCTION For the period 197122001, the number y of films produced
in the world can be modeled by the function y 5 10x2 2 94x 1 3900 where x is
the number of years since 1971. In what year were 4200 films produced?
Solution
y 5 10x2 2 94x 1 3900
Write function.
2
4200 5 10x 2 94x 1 3900
Substitute 4200 for y.
0 5 10x2 2 94x 2 300
Write in standard form.
}}
INTERPRET
SOLUTIONS
The solution 23 can
be ignored because
23 represents the year
1968, which is not in
the given time period.
672
Substitute values in the quadratic
formula: a 5 10, b 5 294, and c 5 2300.
Ï
x 5 }}}
2(294) 6 (294)2 2 4(10)(2300)
2(10)
}
94 6 Ï20,836
5 }}
Simplify.
20
}
}
94 1 Ï20,836
94 2 Ï 20,836
The solutions of the equation are }}
ø 12 and }}
ø 23.
20
20
c There were 4200 films produced about 12 years after 1971, or in 1983.
Chapter 10 Quadratic Equations and Functions
✓
GUIDED PRACTICE
for Example 3
4. WHAT IF? In Example 3, find the year when 4750 films were produced.
For Your Notebook
CONCEPT SUMMARY
Methods for Solving Quadratic Equations
Method
Lesson(s)
Factoring
9.4–9.8
Graphing
10.3
Use when approximate solutions are
adequate.
Finding square roots
10.4
Use when solving an equation that can
be written in the form x2 5 d.
Completing the square
10.5
Can be used for any quadratic equation
ax2 1 bx 1 c 5 0 but is simplest to apply
when a 5 1 and b is an even number.
Quadratic formula
10.6
Can be used for any quadratic equation.
EXAMPLE 4
When to Use
Use when a quadratic equation can be
factored easily.
Choose a solution method
Tell what method you would use to solve the quadratic equation. Explain
your choice(s).
a. 10x2 2 7 5 0
b. x2 1 4x 5 0
c. 5x2 1 9x 2 4 5 0
Solution
a. The quadratic equation can be solved using square roots because
the equation can be written in the form x2 5 d.
b. The equation can be solved by factoring because the expression
x2 1 4x can be factored easily. Also, the equation can be solved by
completing the square because the equation is of the form
ax2 1 bx 1 c 5 0 where a 5 1 and b is an even number.
c. The quadratic equation cannot be factored easily, and completing
the square will result in many fractions. So, the equation can be
solved using the quadratic formula.
✓
GUIDED PRACTICE
for Example 4
Tell what method you would use to solve the quadratic equation. Explain
your choice(s).
5. x2 1 x 2 6 5 0
6. x2 2 9 5 0
7. x2 1 6x 5 5
10.6 Solve Quadratic Equations by the Quadratic Formula
673
10.6
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 19 and 47
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 12, 25, and 50
5 MULTIPLE REPRESENTATIONS
Ex. 49
SKILL PRACTICE
1. VOCABULARY What formula can be used to solve any quadratic
equation?
2.
★
WRITING What method(s) would you use to solve 2x2 1 8x 5 1?
Explain your choice(s).
EXAMPLES
1 and 2
on pp. 671–672
for Exs. 3–27
SOLVING QUADRATIC EQUATIONS Use the quadratic formula to solve the
equation. Round your solutions to the nearest hundredth, if necessary.
3. x2 1 5x 2 104 5 0
4. 4x2 2 x 2 18 5 0
5. 6x2 2 2x 2 28 5 0
6. m2 1 3m 1 1 5 0
7. 2z2 1 z 1 14 5 0
8. 22n2 2 5n 1 16 5 0
9. 4w 2 1 20w 1 25 5 0
12.
10. 2t 2 1 3t 2 11 5 0
11. 26g 2 1 9g 1 8 5 0
★
MULTIPLE CHOICE What are the solutions of 10x2 2 3x 2 1 5 0?
1
1
1
1
1
1
1
1
A 2}
and 2}
B 2}
and }
C }
and 2}
D }
and }
5
5
5
5
2
2
2
2
SOLVING QUADRATIC EQUATIONS Use the quadratic formula to solve the
equation. Round your solutions to the nearest hundredth, if necessary.
13. x2 2 5x 5 14
14. 3x2 2 4 5 11x
15. 9 5 7x2 2 2x
16. 2m2 1 9m 1 7 5 3
17. 210 5 r 2 2 10r 1 12
18. 3g 2 2 6g 2 14 5 3g
19. 6z2 5 2z2 1 7z 1 5
20. 8h2 1 8 5 6 2 9h
21. 4t 2 2 3t 5 5 2 3t 2
22. 24y 2 2 3y 1 3 5 2y 1 4
23. 7n 1 5 5 23n2 1 2
24. 5w 2 1 4 5 w 1 6
25.
★
MULTIPLE CHOICE What are the solutions of x2 1 14x 5 2x 2 11?
A 22 and 222
B 21 and 211
C 1 and 11
D 2 and 22
ERROR ANALYSIS Describe and correct the error in solving the equation.
26. 7x2 2 5x 2 1 5 0
27. 22x2 1 3x 5 1
}}
}}
25 6 Ï (25)2 2 4(7)(21)
x 5 }}
2(7)
}
25 6 Ï 53
5}
14
23 6 Ï32 2 4(22)(1)
x 5 }}
2(22)
}
23 6 Ï17
5}
24
x ø 20.88 and x ø 0.16
x ø 20.28 and x ø 1.78
EXAMPLE 4
CHOOSING A METHOD Tell what method(s) you would use to solve the
on p. 673
for Exs. 28–33
quadratic equation. Explain your choice(s).
674
28. 3x2 2 27 5 0
29. 5x2 5 25
30. 2x2 2 12x 5 0
31. m2 1 5m 1 6 5 0
32. z2 2 4z 1 1 5 0
33. 210g 2 1 13g 5 4
Chapter 10 Quadratic Equations and Functions
SOLVING QUADRATIC EQUATIONS Solve the quadratic equation using any
method. Round your solutions to the nearest hundredth, if necessary.
34. 22x2 5 232
35. x2 2 8x 5 216
36. x2 1 2x 2 6 5 0
37. x2 5 12x 2 36
38. x2 1 4x 5 9
39. 24x2 1 x 5 217
40. 11x2 2 1 5 6x2 1 2
41. 22x2 1 5 5 3x2 2 10x
42. (x 1 13)2 5 25
GEOMETRY Use the given area A of the rectangle to find the value of x.
Then give the dimensions of the rectangle.
43. A 5 91 m 2
44. A 5 209 ft 2
(x 1 2) m
(4x 2 5) ft
(2x 1 3) m
(4x 1 3) ft
45. CHALLENGE The solutions of the quadratic equation ax2 1 bx 1 c 5 0 are
}
}
2a
2a
2b 2Ïb2 2 4ac
1Ï b2 2 4ac
x 5 2b
}} and x 5 }}. Find the mean of the solutions.
How is the mean of the solutions related to the graph of y 5 ax2 1 bx 1 c?
Explain.
PROBLEM SOLVING
EXAMPLE 3
on p. 672
for Exs. 46–47
46. ADVERTISING For the period 199022000, the amount of money y (in
billions of dollars) spent on advertising in the U.S. can be modeled by
the function y 5 0.93x2 1 2.2x 1 130 where x is the number of years since
1990. In what year was 164 billion dollars spent on advertising?
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
47. CELL PHONES For the period 198522001, the number y (in millions) of
cell phone service subscribers in the U.S. can be modeled by the function
y 5 0.7x2 2 4.3x 1 5.5 where x is the number of years since 1985. In what
year were there 16,000,000 cell phone service subscribers?
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
48. MULTI-STEP PROBLEM A football is punted from a height of 2.5 feet above
the ground and with an initial vertical velocity of 45 feet per second.
Not drawn to scale
5.5 ft
2.5 ft
a. Use the vertical motion model to write an equation that gives
the height h (in feet) of the football as a function of the time t (in
seconds) after it has been punted.
b. The football is caught 5.5 feet above the ground as shown in the
diagram. Find the amount of time that the football is in the air.
10.6 Solve Quadratic Equations by the Quadratic Formula
675
49.
MULTIPLE REPRESENTATIONS For the period 1997–2002, the number
y (in thousands) of 16- and 17-year-olds employed in the United States
can be modeled by the function y 5 246.7x2 1 169x 1 2650 where x is the
number of years since 1997.
a. Solving an Equation Write and solve an equation to find the year
during which 2,500,000 16- and 17-year-olds were employed.
b. Drawing a Graph Graph the function on a graphing calculator. Use
the trace feature to find the year when 2,500,000 16- and 17-year-olds
were employed. Use the graph to check your answer from part (a).
50.
★ SHORT RESPONSE NASA creates a weightless
WEIGHTLESS
ENVIRONMENT
H
environment by flying a plane in a series of
FT
parabolic paths. The height h (in feet) of a plane
after t seconds in a parabolic flight path can be
modeled by the graph of h 5 211t 2 1 700t 1 21,000.
The passengers experience a weightless environment
when the height of the plane is greater than or equal
to 30,800 feet. Find the period of weightlessness on such
a flight. Explain.
T
51. CHALLENGE Mineral deposits have formed a uniform coating that is
4 millimeters thick on the inside of a water pipe. The cross-sectional
area of the pipe has decreased by 10%. What was the original diameter
of the pipe (to the nearest tenth of a millimeter)?
MIXED REVIEW
PREVIEW
Evaluate the expression.
Prepare for
Lesson 10.7 in
Exs. 52–55.
52. 9x2 when x 5 2 (p. 8)
2 5w
53. 6}
when w 5 10 (p. 103)
2w
}
}
54. 2 1 Ïx when x 5 121 (p. 110)
55. 8 2 Ïx when x 5 49 (p. 110)
Graph the equation.
56. x 5 8 (p. 215)
2
59. y 5 27x (p. 628)
57. 3x 2 y 5 2 (p. 225)
2
60. y 5 8x 2 2 (p. 628)
2
58. y 5 2}
x 2 6 (p. 244)
5
61. y 5 2x2 2 6x 1 5 (p. 635)
QUIZ for Lessons 10.4–10.6
Solve the equation using square roots. (p. 652)
1. 3x2 2 48 5 0
2. 26x2 5 224
3. x2 1 5 5 16
Solve the equation by completing the square. (p. 663)
4. x2 1 2x 1 6 5 0
5. x2 1 10x 2 12 5 0
7. x2 2 12x 5 30
8. x2 2 5x 5 2}
9
4
6. x2 2 8x 5 26
9. x2 1 x 5 27.75
Solve the equation using the quadratic formula. (p. 671)
10. x2 1 4x 1 1 5 0
676
11. 23x2 1 3x 5 21
EXTRA PRACTICE for Lesson 10.6, p. 947
12. 4x2 2 11x 5 3
ONLINE QUIZ at classzone.com
Investigating
g
Algebra
Algebr
ra
ACTIVITY Use before Lesson 10.7
10.7 The Discriminant
QUESTION
How can you determine the number of solutions of a
quadratic equation?
}
6 Ïb 2 4ac
2
In the quadratic formula, x 5 2b
}}, the expression b 2 4ac is called
2a
the discriminant.
2
EXPLORE
Determine how the discriminant is related to the number of
solutions of a quadratic equation
STEP 1 Find the number of solutions
y
y 5 x 2 2 6x 1 12
Find the number of solutions of the equations below
by finding the number of x-intercepts of the graphs
of the related functions.
0 5 x2 2 6x 2 7
0 5 x2 2 6x 1 9
0 5 x2 2 6x 1 12
2
y 5 x 2 2 6x 1 9
8 x
STEP 2 Find the value of b2 2 4ac
For each equation in Step 1, determine whether the value of
b2 2 4ac is positive, negative, or zero.
STEP 3 Make a table
Organize your results from Steps 1 and 2 in a table as shown.
Equation
Number of solutions
Value of b2 2 4ac
0 5 x2 2 6x 2 7
?
?
0 5 x2 2 6x 1 9
?
?
?
?
2
0 5 x 2 6x 1 12
y 5 x 2 2 6x 2 7
STEP 4 Make a conjecture
Make a generalization about the value of the discriminant and
the number of solutions of a quadratic equation.
DR AW CONCLUSIONS
Use your observations to complete these exercises
1. Repeat Steps 123 using the following equations: x2 1 4x 2 5 5 0,
x2 1 4x 1 4 5 0, and x2 1 4x 1 6 5 0. Is your conjecture still true?
2. Notice that the expression b2 2 4ac is under the radical sign in the
quadratic formula. Use this observation to explain why the value of
b2 2 4ac determines the number of solutions of a quadratic equation.
10.7 Interpret the Discriminant
677
10.7
Before
Now
Why?
Key Vocabulary
• discriminant
Interpret the
Discriminant
You used the quadratic formula.
You will use the value of the discriminant.
So you can solve a problem about gymnastics, as in Ex. 49.
In the quadratic formula, the expression b2 2 4ac is called the discriminant
of the associated equation ax2 1 bx 1 c 5 0.
}
6 Ï b2 2 4ac
x 5 2b
}}
2a
discriminant
Because the discriminant is under the radical symbol, the value of the
discriminant can be used to determine the number of solutions of a
quadratic equation and the number of x-intercepts of the graph of the
related function.
For Your Notebook
KEY CONCEPT
Using the Discriminant of ax 2 1 bx 1 c 5 0
READING
Recall that in this
course, solutions
refers to real-number
solutions.
Value of the
discriminant
b2 2 4ac > 0
b2 2 4ac 5 0
b2 2 4ac < 0
Number of
solutions
Two solutions
One solution
No solution
Graph of
y 5 ax2 1 bx 1 c
y
y
x
x
Two x-intercepts
EXAMPLE 1
One x-intercept
x
No x-intercept
Use the discriminant
Equation
ax2 + bx + c = 0
Discriminant
b2 – 4ac
Number of
solutions
a. 2x2 1 6x 1 5 5 0
62 2 4(2)(5) 5 24
No solution
b. x2 2 7 5 0
02 2 4(1)(27) 5 28
Two solutions
2
c. 4x 2 12x 1 9 5 0
678
y
Chapter 10 Quadratic Equations and Functions
2
(212) 2 4(4)(9) 5 0
One solution
EXAMPLE 2
Find the number of solutions
Tell whether the equation 3x 2 2 7 5 2x has two solutions, one solution, or
no solution.
Solution
STEP 1 Write the equation in standard form.
3x2 2 7 5 2x
2
3x 2 2x 2 7 5 0
Write equation.
Subtract 2x from each side.
STEP 2 Find the value of the discriminant.
b2 2 4ac 5 (22)2 2 4(3)(27)
5 88
Substitute 3 for a, 22 for b, and 27 for c.
Simplify.
c The discriminant is positive, so the equation has two solutions.
✓
GUIDED PRACTICE
for Examples 1 and 2
Tell whether the equation has two solutions, one solution, or no solution.
1. x2 1 4x 1 3 5 0
EXAMPLE 3
2. 2x2 2 5x 1 6 5 0
3. 2x2 1 2x 5 1
Find the number of x-intercepts
Find the number of x-intercepts of the graph of y 5 x 2 1 5x 1 8.
Solution
Find the number of solutions of the equation 0 5 x2 1 5x 1 8.
b2 2 4ac 5 (5)2 2 4(1)(8)
5 27
Substitute 1 for a, 5 for b, and 8 for c.
Simplify.
c The discriminant is negative, so the equation has no solution. This means
that the graph of y 5 x2 1 5x 1 8 has no x-intercepts.
CHECK
✓
You can use a graphing calculator to
check the answer. Notice that the graph
of y 5 x2 1 5x 1 8 has no x-intercepts.
GUIDED PRACTICE
for Example 3
Find the number of x-intercepts of the graph of the function.
4. y 5 x2 1 10x 1 25
5. y 5 x2 2 9x
6. y 5 2x2 1 2x 2 4
10.7 Interpret the Discriminant
679
EXAMPLE 4
Solve a multi-step problem
FOUNTAINS The Centennial Fountain in Chicago shoots a water arc that can
be modeled by the graph of the equation y 5 20.006x2 1 1.2x 1 10 where x is
the horizontal distance (in feet) from the river’s north shore and y is the height
(in feet) above the river. Does the water arc reach a height of 50 feet? If so,
about how far from the north shore is the water arc 50 feet above the water?
y
50
North shore
x
50
Solution
STEP 1 Write a quadratic equation. You want to know whether the water arc
reaches a height of 50 feet, so let y 5 50. Then write the quadratic
equation in standard form.
y 5 20.006x2 1 1.2x 1 10
2
50 5 20.006x 1 1.2x 1 10
0 5 20.006x2 1 1.2x 2 40
Write given equation.
Substitute 50 for y.
Subtract 50 from each side.
STEP 2 Find the value of the discriminant of 0 5 20.006x2 1 1.2x 2 40.
b2 2 4ac 5 (1.2)2 2 4(20.006)(240)
5 0.48
a 5 20.006, b 5 1.2, c 5 240
Simplify.
STEP 3 Interpret the discriminant. Because the discriminant is positive,
the equation has two solutions. So, the water arc reaches a height of
50 feet at two points on the water arc.
STEP 4 Solve the equation 0 5 20.006x2 1 1.2x 2 40 to find the distance
from the north shore where the water arc is 50 feet above the water.
}
6 Ïb 2 4ac
x 5 2b
}}
2
2a
Quadratic formula
}
USE A SHORTCUT
Because the value of
b2 – 4ac was calculated
in Step 2, you can
substitute 0.48 for
b2 – 4ac.
✓
21.2 6 Ï0.48
5}
2(20.006)
x ø 42 or x ø 158
Substitute values in the quadratic formula.
Use a calculator.
c The water arc is 50 feet above the water about 42 feet from the north shore
and about 158 feet from the north shore.
GUIDED PRACTICE
for Example 4
7. WHAT IF? In Example 4, does the water arc reach a height of 70 feet?
If so, about how far from the north shore is the water arc 70 feet above
the water?
680
Chapter 10 Quadratic Equations and Functions
10.7
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 9 and 47
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 18, 19, 40, 41, and 47
SKILL PRACTICE
1. VOCABULARY Write the quadratic formula and circle the expression that
represents the discriminant.
2.
★
WRITING Explain how the discriminant of ax2 1 bx 1 c 5 0 is related
to the graph of y 5 ax2 1 bx 1 c.
EXAMPLES
1 and 2
on pp. 678–679
for Exs. 3–21
USING THE DISCRIMINANT Tell whether the equation has two solutions, one
solution, or no solution.
3. x2 1 x 1 1 5 0
4. 2x2 2 5x 2 6 5 0
5. 22x2 1 8x 2 4 5 0
6. 3m2 2 6m 1 7 5 0
7. 9v 2 2 6v 1 1 5 0
8. 23q2 1 8 5 0
9. 25p2 2 16p 5 0
10. 2h2 1 3 5 4h
11. 10 5 x2 2 5x
1 2
12. }
z 125z
4
13. 23g 2 2 4g 5 }
14. 8r 2 1 10r 2 1 5 4r
15. 3n2 1 3 5 10n 2 3n2
16. 8x2 1 9 5 4x2 2 4x 1 8
17. w 2 2 7w 1 29 5 4 2 7w
3
4
18.
★
MULTIPLE CHOICE What is the value of the discriminant of the
equation 5x2 2 7x 2 2 5 0?
A 29
19.
★
B 9
C 59
D 89
MULTIPLE CHOICE How many solutions does 2x2 1 4x 5 8 have?
A None
B One
C Two
D Three
ERROR ANALYSIS Describe and correct the error in finding the number of
solutions of the equation.
20. 4x2 1 12x 1 9 5 0
21. 3x2 2 7x 2 4 5 29
b2 2 4ac 5 122 2 4(4)(9)
b2 2 4ac 5 (27) 2 2 4(3)(24)
5 144 2 144
5 49 2 (248)
50
5 97
The equation has two solutions.
The equation has two solutions.
EXAMPLE 3
FINDING THE NUMBER OF x-INTERCEPTS Find the number of x-intercepts of
on p. 679
for Exs. 22–30
the graph of the function.
22. y 5 x2 2 2x 2 4
23. y 5 2x2 2 x 2 1
24. y 5 4x2 1 4x 1 1
25. y 5 2x2 2 5x 1 5
26. y 5 x2 2 6x 1 9
27. y 5 6x2 1 x 1 2
28. y 5 213x2 1 2x 1 6
1 2
29. y 5 }
x 2 3x 1 9
2 2
30. y 5 }
x 2 5x 1 12
4
3
REASONING Give a value of c for which the equation has (a) two solutions,
(b) one solution, and (c) no solution.
31. x2 2 2x 1 c 5 0
32. x2 2 8x 1 c 5 0
33. 4x2 1 12x 1 c 5 0
10.7 Interpret the Discriminant
681
USING THE DISCRIMINANT Tell whether the vertex of the graph of the
function lies above, below, or on the x-axis. Explain your reasoning.
34. y 5 x2 2 3x 1 2
35. y 5 3x2 2 6x 1 3
36. y 5 6x2 2 2x 1 4
37. y 5 215x2 1 10x 2 25
38. y 5 23x2 2 4x 1 8
39. y 5 9x2 2 24x 1 16
40.
★
OPEN – ENDED
41.
★
EXTENDED RESPONSE Use the rectangular prism shown.
Write a function of the form y 5 ax2 1 bx 1 c whose
graph has one x-intercept.
a. The surface area of the prism is 314 square meters. Write
an equation that you can solve to find the value of w.
8m
b. Use the discriminant to determine the number of values of
w in the equation from part (a).
c. Solve the equation. Do the value(s) of w make sense in the
context of the problem? Explain.
(w 1 4) m
CHALLENGE Find all values of k for which the equation has (a) two solutions,
(b) one solution, and (c) no solution.
42. 2x2 1 x 1 3k 5 0
43. x2 2 4kx 1 36 5 0
44. kx2 1 5x 2 16 5 0
PROBLEM SOLVING
EXAMPLE 4
45. BIOLOGY The amount y (in milliliters per gram of body mass per hour)
of oxygen consumed by a parakeet during flight can be modeled by the
function y 5 0.06x2 2 4x 1 87 where x is the speed (in kilometers per
hour) of the parakeet.
on p. 680
for Exs. 45–46
a. Use the discriminant to show that it is possible for a parakeet to
consume 25 milliliters of oxygen per gram of body mass per hour.
b. Find the speed(s) at which the parakeet consumes 25 milliliters of
oxygen per gram of body mass per hour. Round your solution(s) to
the nearest tenth.
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
46. FOOD For the period 195021999, the average amount y (in pounds per
person per year) of butter consumed in the United States can be modeled
by y 5 0.0051x2 2 0.37x 1 11 where x is the number of years since 1950.
According to the model, did the butter consumption in the United States
ever reach 5 pounds per person per year? If so, in what year(s)?
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
47.
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★
SHORT RESPONSE The frame of the tent shown is
defined by a rectangular base and two parabolic arches
that connect the opposite corners of the base. The graph
of y 5 20.18x2 1 1.6x models the height y (in feet) of one
of the arches x feet along the diagonal of the base. Can
a child that is 4 feet tall walk under one of the arches
without having to bend over? Explain.
5 WORKED-OUT SOLUTIONS
on p. WS1
★ 5 STANDARDIZED
TEST PRACTICE
wm
48. SCIENCE Between the months of April and September, the number y of
hours of daylight per day in Seattle, Washington, can be modeled by
y 5 20.00046x2 1 0.076x 1 13 where x is the number of days since April 1.
a. Do any of the days between April and September in Seattle have
17 hours of daylight? If so, how many?
b. Do any of the days between April and September in Seattle have
14 hours of daylight? If so, how many?
49. MULTI-STEP PROBLEM During a trampoline competition, a trampolinist
leaves the mat when her center of gravity is 6 feet above the ground. She
has an initial vertical velocity of 32 feet per second.
a. Use the vertical motion model to write an equation
that models the height h (in feet) of the center of
gravity of the trampolinist as a function of the
time t (in seconds) into her jump.
b. Does her center of gravity reach a height of 24 feet
during the jump? If so, at what time(s)?
c. On another jump, the trampolinist leaves the mat
h ft
when her center of gravity is 6 feet above the ground
and with an initial vertical velocity of 35 feet per
second. Does her center of gravity reach a height of
24 feet on this jump? If so, at what time(s)?
50. CHALLENGE Last year, a manufacturer sold backpacks for $24 each.
At this price, the manufacturer sold about 1000 backpacks per week.
A marketing analyst predicts that for every $1 reduction in the price of
the backpack, the manufacturer will sell 100 more backpacks per week.
a. Write a function that models the weekly revenue R (in dollars) that
the manufacturer will receive for x reductions of $1 in the price of
the backpack.
b. Is it possible for the manufacturer to receive a weekly revenue
of $28,000? $30,000? What is the maximum weekly revenue that
the manufacturer can receive? Explain your answers using the
discriminants of quadratic equations.
MIXED REVIEW
PREVIEW
Graph the function.
Prepare for
Lesson 10.8 in
Exs. 51–56.
51. y 5 5x 2 10 (p. 225)
1
52. y 5 }
x (p. 244)
53. y 5 }x 2 5 (p. 244)
54. y 5 5x (p. 520)
55. y 5 (0.2) x (p. 531)
56. y 5 6x2 2 3 (p. 628)
57. a 1 5 5 2 (p. 134)
58. f 2 6 5 13 (p. 134)
59. 4z 2 3 5 27 (p. 141)
60. 9w 1 4 5 241 (p. 141)
61. 2b 2 b 2 6 5 8 (p. 148)
62. 5 1 2(x 2 4) 5 9 (p. 148)
3
4
4
Solve the equation.
Solve the equation by factoring. (p. 593)
63. 2x2 2 3x 2 5 5 0
64. 4n2 1 2n 2 6 5 0
EXTRA PRACTICE for Lesson 10.7, p. 947
65. 5a2 1 21a 1 4 5 0
ONLINE QUIZ at classzone.com
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