104
3.1
Chapter Three QUADRATIC FUNCTIONS
INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS
A baseball is “popped” straight up by a batter. The height of the ball above the ground is given by
the function y = f (t) = −16t2 + 47t + 3, where t is time in seconds after the ball leaves the bat
and y is in feet. See Figure 3.1. Note that the path of the ball is straight up and down, although the
graph of height against time is a curve. The ball goes up fast at first and then more slowly because
of gravity; thus the graph of its height as a function of time is concave down.
y (feet)
y = f (t) = −16t2 + 47t + 3
Initial
height
R
3
t (seconds)
3
Figure 3.1: The height of a ball t seconds after being “popped up”. (Note: This
graph does not represent the ball’s path.)
The baseball height function is an example of a quadratic function, whose general form is
y = ax2 + bx + c. The graph of a quadratic function is a parabola. Notice that the function in
Figure 3.1 is concave down and has a maximum corresponding to the time at which the ball stops
rising and begins to fall back to the earth. The maximum point on the parabola is called the vertex.
The intersection of the parabola with the horizontal axis gives the times when the height of the
ball is zero; these times are the zeros of the height function. The horizontal intercepts of a graph
occur at the zeros of the function. In this section we examine the zeros and concavity of a quadratic
functions, and in the next section we see how to find the vertex.
Finding the Zeros of a Quadratic Function
A natural question to ask is when the ball hits the ground. The graph suggests that y = 0 when t is
approximately 3. In symbols, the question is: For what value(s) of t does f (t) = 0? These are the
zeros of the function.
It is easy to find the zeros of a quadratic function if it can be expressed in factored form,
q(x) = a(x − r)(x − s),
where a, r, and s are constants, a = 0. Then r and s are zeros of the function q.
The baseball function factors as y = −1(16t + 1)(t − 3), so the zeros of f are t = 3 and
1
t = − 16
. In this problem we are interested in positive values of t, so the ball hits the ground 3
seconds after it was hit. (For more on factoring, see the Skills Review on page 120.)
Example 1
Find the zeros of f (x) = x2 − x − 6.
Solution
To find the zeros, set f (x) = 0 and solve for x by factoring:
x2 − x − 6 = 0
(x − 3)(x + 2) = 0.
Thus the zeros are x = 3 and x = −2.
3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS
105
We can also find the zeros of a quadratic function by using the quadratic formula. (See the
Skills Review on page 120 to review the quadratic formula.)
Example 2
Find the zeros of f (x) = x2 − x − 6 by using the quadratic formula.
Solution
We solve the equation x2 − x − 6 = 0. For this equation, a = 1, b = −1, and c = −6. Thus
√
−(−1) ± (−1)2 − 4(1)(−6)
−b ± b2 − 4ac
x=
=
2a
2(1)
√
1 ± 25
=
= 3 or − 2.
2
The zeros are x = 3 and x = −2, the same as we found by factoring.
Since the zeros of a function occur at the horizontal intercepts of its graph, quadratic functions
without horizontal intercepts (such as in the next example) have no zeros.
Example 3
1
Figure 3.2 shows a graph of h(x) = − x2 − 2. What happens if we try to use algebra to find the
2
zeros of h?
y
x
y -intercept
(0, −2)
h(x)
Figure 3.2: Zeros of h(x) = −(x2 /2) − 2?
Solution
To find the zeros, we solve the equation
1
− x2 − 2 = 0
2
1
− x2 = 2
2
x2 = −4
√
x = ± −4.
√
Since −4 is not a real number, there are no real solutions, so h has no real zeros. This corresponds
to the fact that the graph of h in Figure 3.2 does not cross the x-axis.
Concavity and Quadratic Functions
A quadratic function has a graph that is either concave up or concave down.
Example 4
Let f (x) = x2 . Find the average rate of change of f over the consecutive intervals of length 2
starting at x = −4 and ending at x = 4. What do these rates tell you about the concavity of the
graph of f ?
106
Chapter Three QUADRATIC FUNCTIONS
Solution
Between x = −4 and x = −2, we have
(−2)2 − (−4)2
Average rate of change f (−2) − f (−4)
=
= −6.
=
of f
−2 − (−4)
−2 + 4
Between x = −2 and x = 0, we have
02 − (−2)2
Average rate of change f (0) − f (−2)
=
= −2.
=
of f
0 − (−2)
0+2
Between x = 0 and x = 2, we have
22 − 02
Average rate of change f (2) − f (0)
=
=
= 2.
of f
2−0
2−0
Between x = 2 and x = 4, we have
42 − 22
Average rate of change f (4) − f (2)
=
=
= 6.
of f
4−2
4−2
Since the rates of change are increasing, the graph is concave up. See Figure 3.3.
f (x) = x2
Slope = −6
Slope = 6
?
?
Slope = 2
Slope = −2
?
−4
?
−2
2
4
x
Figure 3.3: Rate of change and concavity of f (x) = x2
Example 5
A high-diver jumps off a 10-meter platform. For t in seconds after the diver leaves the platform until
she hits the water, her height h in meters above the water is given by
h = f (t) = −4.9t2 + 8t + 10.
The graph of this function is shown in Figure 3.4.
(a) Estimate and interpret the domain and range of the function, and the intercepts of the graph.
(b) Identify the concavity.
h (meters)
15
f (t)
10
5
1
2
3
t (seconds)
Figure 3.4: Height of diver above water as a function of time
3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS
Solution
107
(a) The diver enters the water when her height above the water is 0. This occurs when
h = f (t) = −4.9t2 + 8t + 10 = 0.
Using the quadratic formula to solve this equation, we find the only positive solution is t =
2.462 seconds. The domain is the interval of time the diver is in the air, which is approximately
0 ≤ t ≤ 2.462. To find the range of f , we look for the largest and smallest outputs for h. From
the graph, the diver’s maximum height appears to occur at about t = 1, so we estimate the
largest output value for f to be about
f (1) = −4.9 · 12 + 8 · 1 + 10 = 13.1 meters.
Thus, the range of f is approximately 0 ≤ f (t) ≤ 13.1. Methods to find the exact maximum
height are found in Section 3.2.
The vertical intercept of the graph is
f (0) = −4.9 · 02 + 8 · 0 + 10 = 10 meters.
The diver’s initial height is 10 meters (the height of the diving platform). The horizontal intercept is the point where f (t) = 0, which we found in part (a). The diver enters the water
approximately 2.462 seconds after leaving the platform.
(b) In Figure 3.4, we see that the graph is bending downward over its entire domain, so it is concave
down. This is reflected in Table 3.1, where the rate of change, Δh/Δt, is decreasing.
Table 3.1
Slope of f (t) = −4.9t2 + 8t + 10
t (sec)
h (meters)
0
10
Rate of change
Δh/Δt
5.55
0.5
12.775
0.65
1.0
13.100
−4.25
1.5
10.975
−9.15
2.0
6.400
Finding a Formula From the Zeros and Vertical Intercept
If we know the zeros and the vertical intercept of a quadratic function, we can use the factored form
to find a formula for the function.
Example 6
Find the equation of the parabola in Figure 3.5 using the factored form.
f (x)
6
1
3
x
Figure 3.5: Finding a formula for a quadratic from the zeros
108
Chapter Three QUADRATIC FUNCTIONS
Since the parabola has x-intercepts at x = 1 and x = 3, its formula can be written as
Solution
f (x) = a(x − 1)(x − 3).
Substituting x = 0, y = 6 gives
6 = a(3)
a = 2.
Thus, the formula is
f (x) = 2(x − 1)(x − 3).
2
Multiplying out gives f (x) = 2x − 8x + 6.
Formulas for Quadratic Functions
The function f in Example 6 can be written in at least two different ways.
The form f (x) = 2x2 − 8x + 6 shows the the parabola opens upward, since the coefficient
2
of x is positive and the constant 6 gives the vertical intercept. The form f (x) = 2(x − 1)(x − 3)
shows that the parabola crosses the x-axis at x = 1 and x = 3. In general, we have the following:
The graph of a quadratic function is a parabola.
The standard form for a quadratic function is
y = ax2 + bx + c,
where a, b, c are constants, a = 0.
The parabola opens upward if a > 0 or downward if a < 0, and it intersects the y-axis
at c.
The factored form, when it exists, is
y = a(x − r)(x − s),
where a, r, s are constants, a = 0.
The parabola intersects the x-axis at x = r and x = s.
In the next section we look at another form for quadratic functions which shows the vertex of
the graph.
Exercises and Problems for Section 3.1
Skill Refresher
Multiply and write the expressions in Problems S1–S2 without parentheses. Gather like terms.
S1.
S2.
2
2
t + 1 50t − 25t + 125 2t
A2 − B 2
S3. u − 2u
S6. (s + 2t)2 − 4p2
S7. 16x2 − 1
S8. y 3 − y 2 − 12y
2
Solve the equations in Exercises S9–S10.
For Exercises S3–S8, factor completely if possible.
2
S5. 3x2 − x − 4
2
S4. x + 3x + 2
S9. x2 + 7x + 6 = 0
S10. 2w2 + w − 10 = 0
3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS
109
Exercises
Are the functions in Exercises 1–7 quadratic? If so, write the
function in the form f (x) = ax2 + bx + c.
10. Solve for x using the quadratic formula and demonstrate
your solution graphically:
(b) 2x2 + 7.2 = 5.1x
(a) 6x − 13 = 3x2
1. f (x) = 2(7 − x)2 + 1
2. L(P ) = (P + 1)(1 − P )
3. g(m) = m(m2 − 2m) + 3 14 −
m3
3
+
√
3m
In Exercises 11–18, find the zeros (if any) of the function algebraically.
4. h(t) = −16(t − 3)(t + 1)
1
5. R(q) = 2 (q 2 + 1)2
q
6. K(x) = 132 + 13x
7. T (n) =
√
5+
√
3n4 −
9. Find the zeros of Q(x) = 5x−x2 +3 using the quadratic
formula.
n4
4
11. y = (2 − x)(3 − 2x)
12. y = 2x2 + 5x + 2
13. y = 4x2 − 4x − 8
14. y = 7x2 + 16x + 4
15. y = 5x2 + 2x − 1
16. y = −17x2 + 23x + 19
√
18. y = x − x − 12
17. y = x4 + 5x2 + 6
8. Find the zeros of Q(r) = 2r 2 − 6r − 36 by factoring.
Problems
In Problems 19–20, find a formula for the quadratic function
whose graph has the given properties.
19. A y-intercept of y = 7 and x-intercepts at x = 1, 4.
30. Using the factored form, find the formula for the parabola
whose zeros are x = −1 and x = 5, and which passes
through the point (−2, 6).
20. A y-intercept of y = 7 and only one zero at x = −2.
31. Write y = 3(0.5x − 4)(4 − 20x) in the form
y = k(x − r)(x − s) and give the values of k, r, s.
21. Use the quadratic formula to find the time at which the
baseball in Figure 3.1 on page 104 hits the ground.
32. A ball is thrown into the air. Its height (in feet) t seconds
later is given by h(t) = 80t − 16t2 .
22. Is there a quadratic function with zeros x = 1, x = 2
and x = 3?
23. Find two quadratic functions with zeros x = 1, x = 2.
24. Determine the concavity of the graph of f (x) = 4 − x2
between x = −1 and x = 5 by calculating average rates
of change over intervals of length 2.
25. Graph a quadratic function which has all the following properties: concave up, y-intercept is −6, zeros at
x = −2 and x = 3.
26. Without a calculator, graph y = 3x2 − 16x − 12 by factoring and plotting zeros.
27. Without a calculator, graph the following function by factoring and plotting zeros:
y = −4cx + x2 + 4c2
for
c > 0.
In Problems 28–29, find a formula for the parabola.
y
28.
y
29.
(0, 5)
(−6, 0)
(−1, 0)
(0, −1)
(3, 0)
x
(2, 0)
x
(a) Evaluate and interpret h(2).
(b) Solve the equation h(t) = 80. Interpret your solutions and illustrate them on a graph of h(t).
33. Let V (t) = t2 −4t+4 represent the velocity of an object
in meters per second.
(a) What is the object’s initial velocity?
(b) When is the object not moving?
(c) Identify the concavity of the velocity graph.
34. A snowboarder slides up from the bottom of a half-pipe
and comes down again, sliding with little resistance on
the snow. Her height above the edge t seconds after starting up the side is −4.9t2 + 14t − 5 meters.
(a) What is her height at t = 0?
(b) After how many seconds does she reach the air? Return to the edge of the pipe?
(c) How long is she in the air?
110
Chapter Three QUADRATIC FUNCTIONS
35. The percentage of schools with interactive videodisc
players1 each year from 1992 to 1996 is shown in Table 3.2. If x is in years since 1992, show that this data set
can be approximated by the quadratic function p(x) =
−0.8x2 + 8.8x + 7.2. What does this model predict for
the year 2004? How good is this model for predicting the
future?
Table 3.2
Year
1992
1993
1994
1995
1996
Percentage
8
14
21
29.1
29.3
37. A relief package is dropped from a moving airplane.
The package has an initial forward horizontal velocity
and follows a quadratic graph path (instead of dropping
straight down). Figure 3.6 shows the height of the package, h, in km, as a function of the horizontal distance, d,
in meters, as it drops.
(a) From what height was the package released?
(b) How far away from the spot above which it was released does the package hit the ground?
(c) Write a formula for h(d). [Hint: The package starts
falling at the highest point on the graph].
h, height (km)
36. Let f (x) = x2 and g(x) = x2 + 2x − 8.
5
(a) Graph f and g in the window −10 ≤ x ≤ 10,
−10 ≤ y ≤ 10. How are the two graphs similar?
How are they different?
(b) Graph f and g in the window −10 ≤ x ≤ 10,
−10 ≤ y ≤ 100. Why do the two graphs appear
more similar on this window than on the window
from part (a)?
(c) Graph f and g in the window −20 ≤ x ≤ 20,
−10 ≤ y ≤ 400, the window −50 ≤ x ≤ 50,
−10 ≤ y ≤ 2500, and the window −500 ≤ x ≤
500, −2500 ≤ y ≤ 250,000. Describe the change
in appearance of f and g on these three successive
windows.
3.2
4
3
I
Airplane →
Package
dropped
here
2
1
d, horizontal
4430 distance (meters)
Figure 3.6
THE VERTEX OF A PARABOLA
In Section 3.1, we looked at the example of a baseball popped upward by a batter. The height of the
ball above the ground is given by the quadratic function y = f (t) = −16t2 + 47t + 3, where t is
time in seconds after the ball leaves the bat, and y is in feet. See Figure 3.7.
The point on the graph with the largest y value appears to be approximately (1.5, 37.5). (We
find the exact value in Example 4 on page 114.) This means that the baseball reaches its maximum
y (feet)
40
vertex ≈ (1.5, 37.5)
y = f (t) = −16t2 + 47t + 3
Initial
height
R
3
3
t (seconds)
Figure 3.7: Height of baseball at time t
1 Data
from R. Famighetti, ed., The World Almanac and Book of Facts: 1999 (New Jersey: Funk and Wagnalls, 1998).
3.2 THE VERTEX OF A PARABOLA
111
height of about 37.5 feet about 1.5 seconds after being hit. The maximum point on the parabola
is called the vertex. For quadratic functions the vertex shows where the function reaches either its
maximum value or, in the case of a concave-up parabola, its minimum value.
The vertex of a parabola can be easily determined exactly if the quadratic function is written in
the form y = a(x − h)2 + k.
Example 1
(a) Sketch f (x) = (x + 3)2 − 4, and indicate the vertex.
(b) Estimate the coordinates of the vertex from the graph.
(c) Explain how the formula for f can be used to obtain the minimum of f .
Solution
(a) Figure 3.8 shows a sketch of f (x); the vertex gives the minimum value of the function.
(b) The vertex of f appears to be about at the point (−3, −4).
(c) Note that (x + 3)2 is always positive or zero, so (x + 3)2 takes on its smallest value when
x + 3 = 0, that is, at x = −3. At this point (x + 3)2 − 4 takes on its smallest value,
f (−3) = (−3 + 3)2 − 4 = 0 − 4 = −4.
Thus, we see that the vertex is exactly at the point (−3, −4).
f (x)
12
−7
−4
1
x
Figure 3.8: A graph of f (x) = (x + 3)2 − 4
Notice that if we select x-values that are equally spaced to the left and the right of the vertex, the
y-values of the function are equal. For example, f (−2) = f (−4) = −3. The graph is symmetric
about a vertical line that passes through the vertex. This line is called the axis of symmetry. The
function in Example 1 has axis of symmetry x = −3.
The Vertex Form of a Quadratic Function
Writing the function f in Example 1 in the form
f (x) = (x + 3)2 − 4
enables us to find the vertex of the graph and the location and value for the minimum of the function.
In general, we have the following:
112
Chapter Three QUADRATIC FUNCTIONS
The vertex form of a quadratic function is
y = a(x − h)2 + k,
where a, h, k are constants, a = 0.
The graph of this quadratic function has vertex (h, k) and axis of symmetry x = h.
A quadratic function can always be expressed in both standard form and vertex form. For example, for the function f (x) = (x + 3)2 − 4 in Example 1, we convert to standard form by expanding
(x + 3)2 and gather like terms, so that f can be written as
f (x) = x2 + 6x + 5.
This form shows that the parabola opens upward and that the vertical intercept is 5.
In this case we can also factor f and write it as
f (x) = (x + 1)(x + 5).
This form shows that the parabola crosses the x-axis at x = −1 and x = −5. In fact, all three forms
share the same constant a, which tells us whether the parabola opens upward or downward. In the
function f , we see that a = 1.
To convert from vertex form to standard form, we multiply out the squared term. To convert
from standard form to vertex form, we complete the square.2
Example 2
Put each quadratic function into vertex form by completing the square and then graph it.
(a) s(x) = x2 − 6x + 8
(b) t(x) = −4x2 − 12x − 8
Solution
(a) To complete the square, find the square of half of the coefficient of the x-term, (−6/2)2 = 9.
Add and subtract this number after the x-term:
s(x) = x2 − 6x + 9 −9 + 8,
Perfect square
so
s(x) = (x − 3)2 − 1.
The vertex of s is (3, −1) and the axis of symmetry is the vertical line x = 3. The parabola
opens upward. See Figure 3.9.
x = −3/2
3
3
−3
6
x
−3
−3
6
x
−3
x=3
Figure 3.9: s(x) = x2 − 6x + 8
2A
Figure 3.10: t(x) = −4x2 − 12x − 8
more detailed explanation of this method is in the Skills Review on page 125.
3.2 THE VERTEX OF A PARABOLA
113
(b) To complete the square, first factor out −4, the coefficient of x2 , giving
t(x) = −4(x2 + 3x + 2).
Now add and subtract the square of half the coefficient of the x-term, (3/2)2 = 9/4, inside the
parentheses. This gives
9 9
t(x) = −4 x2 + 3x + − + 2
4 4
Perfect square
2
3
1
t(x) = −4
x+
−
2
4
2
3
t(x) = −4 x +
+ 1.
2
The vertex of t is (−3/2, 1), the axis of symmetry is x = −3/2, and the parabola opens downward. See Figure 3.10.
Finding a Formula Given the Vertex and Another Point
If we know the vertex of a quadratic function and one other point, we can use the vertex form to
find its formula.
Example 3
Find the formula for the quadratic function graphed in Figure 3.11.
y
m(x)
(0, 5)
(−3, 2)
x
x = −3
Figure 3.11: Finding a formula for a quadratic from the vertex
Solution
Since the vertex is given, we use the form m(x) = a(x − h)2 + k to find a, h, and k. The vertex is
(−3, 2), so h = −3 and k = 2. Thus,
m(x) = a(x − (−3))2 + 2,
so
m(x) = a(x + 3)2 + 2.
To find a, use the y-intercept (0, 5). Substitute x = 0 and y = m(0) = 5 into the formula for m(x)
and solve for a:
5 = a(0 + 3)2 + 2
3 = 9a
1
a= .
3
114
Chapter Three QUADRATIC FUNCTIONS
Thus, the formula is
1
(x + 3)2 + 2.
3
If we want the formula in standard form, we multiply out:
1
m(x) = x2 + 2x + 5.
3
m(x) =
Modeling with Quadratic Functions
In applications, it is often useful to find the maximum or minimum value of a quadratic function. In
the next example, we return to the baseball example that started this section.
Example 4
For t in seconds, the height of a baseball in feet is given by the formula
y = f (t) = −16t2 + 47t + 3.
Using algebra, find the maximum height reached by the baseball and the time that height is reached.
Solution
The maximum height is at the vertex, so we complete the square to write the function in vertex
form:
3
47
y = f (t) = −16 t2 − t −
16
16
2 2
47
47
47
3
2
= −16 t − t + −
− −
−
16
32
32
16
2
47
192
2209
= −16
t−
−
−
32
1024 1024
2
47
2401
= −16
t−
−
32
1024
2
47
2401
= −16 t −
.
+
32
64
2401
Thus, the vertex is at the point ( 47
32 , 64 ). This means that the ball reaches it maximum height of
37.516 feet at t = 1.469 seconds. Note that these values are in line with the the graphical estimate
(1.5, 37.5) from the beginning of the section.
Example 5
A city decides to make a park by fencing off a section of riverfront property. Funds are allotted for
80 meters of fence. The area enclosed will be a rectangle, but only three sides will be enclosed by
fence—the other side will be bounded by the river. What is the maximum area that can be enclosed?
Solution
Two sides are perpendicular to the bank of the river and have equal length, which we call h. The
other side is parallel to the bank of the river; its length is b. See Figure 3.12. The area, A, of the park
is the product of the lengths of adjacent sides, so A = bh.
Since the fence is 80 meters long, we have
2h + b = 80
b = 80 − 2h.
3.2 THE VERTEX OF A PARABOLA
115
Thus,
A = bh = (80 − 2h)h
A = −2h2 + 80h.
The function A = −2h2 + 80h is quadratic. Since the coefficient of h2 is negative, the parabola
opens downward and we have a maximum at the vertex. The factored form of the quadratic is
A = −2h(h − 40), so the zeros are h = 0 and h = 40. The vertex of the parabola occurs on its axis
of symmetry, midway between the zeros at h = 20. Substituting h = 20 gives the maximum area:
A = −2 · 20(20 − 40) = −40(−20) = 800 meters2 .
River
h
h
b
Figure 3.12: A park next to a river
Exercises and Problems for Section 3.2
Skill Refresher
For Exercises S1–S4, complete the square for each expression.
S1. y 2 − 12y
S2. s2 + 6s − 8
S3. c2 + 3c − 7
S4. 4s2 + s + 2
In Exercises S8–S10, solve using factoring, completing the
square, or the quadratic formula.
S8. −3t2 + 4t + 9 = 0
S9. n2 + 4n − 3 = 2
S10. 2q 2 + 4q − 5 = 8
In Exercises S5–S7, solve by completing the square.
S5. r 2 − 6r + 8 = 0
S6. n2 = 3n + 18
S7. 5q 2 − 8 = 2q
Exercises
For the quadratic functions in Exercises 1–2, state the coordinates of the vertex, the axis of symmetry, and whether the
parabola opens upward or downward.
1. f (x) = 3(x − 1)2 + 2
5. Sketch the quadratic functions given in standard form.
Identify the values of the parameters a, b, and c. Label
the zeros, axis of symmetry, vertex, and y-intercept.
(a)
g(x) = x2 + 3
(b) f (x) = −2x2 +4x+16
2. g(x) = −(x + 3)2 − 4
6. Show that the function y = −x2 + 7x − 13 has no real
zeros.
3. Find the vertex and axis of symmetry of the graph of
v(t) = t2 + 11t − 4.
7. Find the value of k so that the graph of y = (x − 3)2 + k
passes through the point (6, 13).
4. Find the vertex and axis of symmetry of the graph of
w(x) = −3x2 − 30x + 31.
8. The parabola y = ax2 + k has vertex (0, −2) and passes
through the point (3, 4). Find its equation.
116
Chapter Three QUADRATIC FUNCTIONS
In Exercises 9–12, find a formula for the parabola.
y
9.
y
10.
(4, 7)
For Exercises 13–16, convert the quadratic functions to vertex form by completing the square. Identify the vertex and the
axis of symmetry.
13. f (x) = x2 + 8x + 3
(0, 4)
14. g(x) = −2x2 + 12x + 4
x
15. p(t) = 2t2 − 0.12t + 0.1
(5, 5)
(3, 3)
16. w(z) = −3z 2 + 9z − 2
x
In Exercises 17–19, write the quadratic function in its standard, vertex, and factored forms.
y
11.
y
12.
2
x
−5
(−4, 0)
x2 − x
2
5t2 − 20
(5, 0) 18. f (t) =
2
x
19. g(s) = (s − 5)(2s + 3)
(2, 36)
17. y = −6 +
(3, −5)
Problems
20. Using the vertex form, find a formula for the parabola
with vertex (2, 5) that passes through the point (1, 2).
In Problems 21–24, find a formula for the quadratic function
whose graph has the given properties.
21. A vertex at (4, 2) and a y-intercept of y = 6.
22. A vertex at (4, 2) and a y-intercept of y = −4.
23. A vertex at (4, 2) and zeros at x = −3, 11.
24. A vertex at (−7, −3) and contains the point (−3, −7).
25. Let f be a quadratic function whose graph is a concave
up parabola with a vertex at (1, −1), and a zero at the
origin.
(a)
(b)
(c)
(d)
Graph y = f (x).
Determine a formula for f (x).
Determine the range of f .
Find any other zeros.
26. Find the vertex of y = 0.03x2 + 1.8x + 2 exactly. Graph
the function, labeling all intercepts.
27. Find the vertex of y = 26+0.4x−0.01x2 exactly. Graph
the function, labeling all intercepts.
3 Adapted
28. If we know a quadratic function f has a zero at x = −1
and vertex at (1, 4), do we have enough information to
find a formula for this function? If your answer is yes,
find it; if not, give your reasons.
29. If you have a string of length 50 cm, what are the dimensions of the rectangle of maximum area that you can
enclose with your string? Explain your reasoning. What
about a string of length k cm?
30. A football player kicks a ball at an angle of 37◦ above the
ground with an initial speed of 20 meters/second. The
height, h, as a function of the horizontal distance traveled, d, is given by:
h = 0.75d − 0.0192d2 .
(a) Graph the path the ball follows.
(b) When the ball hits the ground, how far is it from the
spot where the football player kicked it?
(c) What is the maximum height the ball reaches during
its flight?
(d) What is the horizontal distance the ball has traveled
when it reaches its maximum height?3
from R. Halliday, D. Resnick, and K. Krane, Physics (New York: Wiley, 1992), p. 58.
REVIEW EXERCISES AND PROBLEMS FOR CHAPTER THREE
31. A ballet dancer jumps in the air. The height, h(t), in feet,
of the dancer at time t, in seconds since the start of the
jump, is given by4
h(t) = −16t2 + 16T t,
where T is the total time in seconds that the ballet dancer
is in the air.
117
(a) Why does this model apply only for 0 ≤ t ≤ T ?
(b) When, in terms of T , does the maximum height of
the jump occur?
(c) Show that the time, T , that the dancer is in the air is
related to H, the maximum height of the jump, by
the equation
H = 4T 2 .
CHAPTER SUMMARY
• General Formulas for Quadratic Functions
Standard form:
y = ax2 + bx + c, a = 0
Factored form:
y = a(x − r)(x − s)
Vertex form:
y = a(x − h)2 + k
• Graphs of Quadratic Functions
Graphs are parabolas
Vertex (h, k)
Axis of symmetry, x = h
Effect of parameter a
Opens upward (concave up) if a > 0, minimum at
(h, k)
Opens downward (concave down) if a < 0, maximum at (h, k)
Factored form displays zeros at x = r and x = s
• Solving Quadratic Equations
Factoring
Quadratic formula
x=
−b ±
√
b2 − 4ac
2a
Completing the square
REVIEW EXERCISES AND PROBLEMS FOR CHAPTER THREE
Exercises
Are the functions in Exercises 1–4 quadratic? If so, write the
function in the form f (x) = ax2 + bx + c.
1. f (x) = (2x − 3)(5 − x)
In Exercises 13–18, find a possible formula for the parabola
with the given conditions.
2. g(t) = 3(t − 2)2 + 7
3. w(n) = n(n − 3)(n − 2) − n2 (n − 8)
√
v2 + 2
v−3
4. r(v) =
+
+ πv 2
3
5
In Exercises 5–10, find the zeros (if any) of the function algebraically.
5. y = 9x2 + 6x + 1
6. y = 6x2 − 17x + 12
7. y = 89x2 + 55x + 34
8. y = 3x2 − 2x + 6
9. N (t) = t2 − 7t + 10
10. Q(r) = 2r 2 − 6r − 36
11. Show that y = x2 − x + 41 has no real zeros.
4 K.
12. Find the vertex and axis of symmetry for the parabola
whose equation is y = 3x2 − 6x + 5.
Laws, The Physics of Dance (Schirmer, 1984).
13. The vertex is (1, −2) and the y-intercept is y = −5.
14. The vertex is (4, −2) and the y-intercept is y = −3.
15. The vertex is (7, 3) and the parabola contains the point
(3, 7).
16. The parabola goes through the origin and its vertex is
(1, −1).
17. The x-intercepts are at x = −1 and x = 2 and (−2, 16)
is on the function’s graph.
18. The parabola has only one x-intercept at x = 1/2 and
has a y-intercept at 3.
118
Chapter Three QUADRATIC FUNCTIONS
In Exercises 19–22, find a formula for the parabola.
y
21.
22.
(2, 0)
y
x
y
19.
(−1, 0)
20.
1
(−6, 9)
9
y
x
(3, 0)
−15
(0, −3)
(0, −4)
(10, 8)
(6, 5)
x
x
Problems
For each parabola in Problems 23–26, state the coordinates of
the vertex, the axis of symmetry, the y-intercept, and whether
the curve is concave up or concave down.
23. y = 2(x − 3/4)2 − 2/3
24. y = −1/2(x + 6)2 − 4
25. y = (x − 0.6)2
26. y = −0.3x2 − 7
In Problems 27–30, write each function in factored form or
vertex form and then state its vertex and zeros.
y = 0.3x2 − 0.6x − 7.2
y = 2x2 − 4x − 2
y = −3x2 + 24x − 36
y = 2x2 + (7/3)x + 1
For x between x = −2 and x = 4, determine the concavity of the graph of f (x) = (x−1)2 +2 by calculating
average rates of change over intervals of length 2.
32. A tomato is thrown vertically into the air at time t = 0.
Its height, d(t) (in feet), above the ground at time t (in
seconds) is given by
27.
28.
29.
30.
31.
d(t) = −16t2 + 48t.
(a) Graph d(t).
(b) Find t when d(t) = 0. What is happening to the
tomato the first time d(t) = 0? The second time?
(c) When does the tomato reach its maximum height?
(d) What is the maximum height that the tomato
reaches?
33. When slam-dunking, a basketball player seems to hang
in the air at the height of his jump. The height h(t), in
feet above the ground, of a basketball player at time t, in
seconds since the start of a jump, is given by
h(t) = −16t2 + 16T t,
where T is the total time in seconds that it takes to complete the jump. For a jump that takes 1 second to complete, how much of this time does the basketball player
spend at the top 25% of the trajectory? [Hint: Find the
maximum height reached. Then find the times at which
the height is 75% of this maximum.]
CHECK YOUR UNDERSTANDING
Are the statements in Problems 1–15 true or false? Give an
explanation for your answer.
1. The quadratic function f (x) = x(x + 2) is in factored
form.
2. If f (x) = (x + 1)(x + 2), then the zeros of f are 1 and
2.
3. A quadratic function whose graph is concave up has a
maximum.
4. All quadratic equations have the form f (x) = ax2 .
5. If the height above the ground of an object at time t is
given by s(t) = at2 + bt + c, then s(0) tells us when the
object hits the ground.
6. To find the zeros of f (x) = ax2 + bx + c, solve the
equation ax2 + bx + c = 0 for x.
7. Every quadratic equation has two real solutions.
8. There is only one quadratic function with zeros at x =
−2 and x = 2.
CHECK YOUR UNDERSTANDING
9. A quadratic function has exactly two zeros.
10. The graph of every quadratic equation is a parabola.
11. The maximum or minimum point of a parabola is called
its vertex.
12. If a parabola is concave up its vertex is a maximum point.
119
13. If the equation of a parabola is written as y = a(x −
h)2 + k, then the vertex is located at the point (−h, k).
14. If the equation of a parabola is written as y = a(x −
h)2 + k, then the axis of symmetry is found at x = h.
15. If the equation of a parabola is y = ax2 + bx + c and
a < 0, then the parabola opens downward.
120
SKILLS REFRESHER FOR CHAPTER THREE
SKILLS REFRESHER FOR CHAPTER 3: QUADRATIC
EQUATIONS
SKILLS FOR FACTORING
Expanding an Expression
The distributive property for real numbers a, b, and c tells us that
a(b + c) = ab + ac,
and
(b + c)a = ba + ca.
We use the distributive property and the rules of exponents to multiply algebraic expressions involving parentheses. This process is sometimes referred to as expanding the expression.
Example 1
Solution
Multiply the following expressions and simplify.
√
1
(a) 3x2 x + x−3
(b) (2t)2 − 5 t
6
2 1 −3
2
1
1 −3
2
x
(a) 3x x + x
= 3x (x) + 3x
= 3x3 + x−1
6
6
2
√
√
√
(b) (2t)2 − 5 t = (2t)2 ( t) − 5 t = 4t2 t1/2 − 5t1/2 = 4t5/2 − 5t1/2
If there are two terms in each factor, then there are four terms in the product:
(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd.
The following special cases of the above product occur frequently. Learning to recognize their forms
aids in factoring.
(a + b)(a − b) = a2 − b2
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
Example 2
Solution
Expand the following and simplify by gathering like terms.
√
√
(a) (5x2 + 2)(x − 4)
(b) (2 r + 2)(4 r − 3)
(c)
2
1
3− x
2
(a) 5x2 + 2 (x − 4) = 5x2 (x) + 5x2 (−4) + (2)(x) + (2)(−4) = 5x3 − 20x2 + 2x − 8
√
√
√
√
√
√
(b) (2 r + 2)(4 r − 3) = (2)(4)( r)2 + (2)(−3)( r) + (2)(4)( r) + (2)(−3) = 8r + 2 r − 6
2
2
1
1
1
1
x + − x = 9 − 3x + x2
(c) 3 − x = 32 − 2 (3)
2
2
2
4
Factoring
To write an expanded expression in factored form, we “un-multiply” the expression. Some techniques for factoring are given in this section. We can check factoring by multiplying the factors.
SKILLS FOR FACTORING
121
Removing a Common Factor
It is sometimes useful to factor out the same factor from each of the terms in an expression. This is
basically the distributive law in reverse:
ab + ac = a(b + c).
One special case is removing a factor of −1, which gives
−a − b = −(a + b)
Another special case is
(a − b) = −(b − a)
Example 3
Solution
Factor the following:
2 2
4
(a)
x y + xy
3
3
(b) (2p + 1)p3 − 3p(2p + 1)
(c) −
s2 t
st2
−
8w
16w
2
2 2
4
x y + xy = xy(x + 2)
3
3
3
(b) (2p + 1)p3 − 3p(2p + 1) = (p3 − 3p)(2p + 1) = p(p2 − 3)(2p + 1)
(Note that the expression
(2p +1) was one of the factors common to both terms.)
st2
st
s2 t
t
−
=−
(c) −
s+
8w
16w
8w
2
(a)
Grouping Terms
Even though all the terms may not have a common factor, we can sometimes factor by first grouping
the terms and then removing a common factor.
Example 4
Solution
Factor x2 − hx − x + h.
x2 − hx − x + h = x2 − hx − (x − h) = x(x − h) − (x − h) = (x − h)(x − 1)
Factoring Quadratics
One way to factor quadratics is to mentally multiply out the possibilities.
Example 5
Factor t2 − 4t − 12.
Solution
If the quadratic factors, it will be of the form
t2 − 4t − 12 = (t + ?)(t + ?).
We are looking for two numbers whose product is −12 and whose sum is −4. By trying combinations, we find
t2 − 4t − 12 = (t − 6)(t + 2).
122
SKILLS REFRESHER FOR CHAPTER THREE
Example 6
Factor 4 − 2M − 6M 2 .
Solution
By a similar method to the previous example, we find 4 − 2M − 6M 2 = (2 − 3M )(2 + 2M ).
Perfect Squares and the Difference of Squares
Recognition of the special products (x + y)2 , (x − y)2 and (x + y)(x − y) in expanded form is
useful in factoring. Reversing the results given on page 120, we have
a2 + 2ab + b2 = (a + b)2 ,
a2 − 2ab + b2 = (a − b)2 ,
a2 − b2 = (a − b)(a + b).
When we see squared terms in an expression to be factored, it is often useful to look for one of these
forms. The difference of squares identity (the third one listed above) is especially useful.
Example 7
Factor: (a) 16y 2 − 24y + 9
(b) 25S 2 R4 − T 6
Solution
(a) 16y 2 − 24y + 9 = (4y − 3)2
2 2 (b) 25S 2 R4 − T 6 = 5SR2 − T 3 = 5SR2 − T 3 5SR 2 + T 3 (c) x2 (x − 2) + 16(2 − x) = x2 (x − 2) − 16(x − 2) = (x − 2) x2 − 16 = (x − 2)(x − 4)(x + 4)
(c)
x2 (x − 2) + 16(2 − x)
Solving Quadratic Equations
Example 8
Solution
Give exact and approximate solutions to x2 = 3.
√
The exact solutions
√ are x = ± 3; approximate ones are x ≈ ±1.73,2 or x ≈ ±1.732, or x ≈
±1.73205. (since 3 = 1.732050808 . . .). Notice that the equation x = 3 has only two exact
solutions, but many possible approximate solutions, depending on how much accuracy is required.
Solving by Factoring
Some equations can be put into factored form such that the product of the factors is zero. Then we
solve by using the fact that if a · b = 0, then either a or b (or both) is zero.
Example 9
Solve (x + 1)(x + 3) = 15.
Solution
Although it is true that if a · b = 0, then a = 0 or b = 0, it is not true that a · b = 15 means that
a = 15 or b = 15, or that a and b are 3 and 5. To solve this equation, we expand the left-hand side
and rearrange so that the right-hand side is equal to zero:
x2 + 4x + 3 = 15,
x2 + 4x − 12 = 0.
Then, factoring gives
(x − 2)(x + 6) = 0.
Thus x = 2 and x = −6 are solutions.
Example 10
Solve 2(x + 3)2 = 5(x + 3).
SKILLS FOR FACTORING
Solution
123
You might be tempted to divide both sides by (x + 3). However, if you do this you will overlook
one of the solutions. Instead, write
2(x + 3)2 − 5(x + 3) = 0
(x + 3) (2(x + 3) − 5) = 0
(x + 3)(2x + 6 − 5) = 0
(x + 3)(2x + 1) = 0.
Thus, x = −1/2 and x = −3 are solutions.
Note that if we had divided by (x + 3) at the start, we would have lost the solution x = −3,
which was obtained by setting x + 3 = 0.
Solving with the Quadratic Formula
Instead of factoring, we can solve the equation ax2 + bx + c = 0 by using the quadratic formula:
√
−b ± b2 − 4ac
.
x=
2a
The quadratic formula is derived by completing the square for y = ax2 + bx + c. See page 125.
Example 11
Solve 11 + 2x = x2 .
Solution
The equation is
−x2 + 2x + 11 = 0.
The expression on the left does not factor using integers, so we use
√
√
√
√
−2 + 4 − 4(−1)(11)
−2 + 48
−2 + 16 · 3
−2 + 4 3
=
=
=
= 1 − 2 3,
x=
2(−1)
−2
−2
−2
√
√
√
√
−2 − 4 − 4(−1)(11)
−2 − 48
−2 − 16 · 3
−2 − 4 3
=
=
=
= 1 + 2 3.
x=
2(−1)
−2
−2
−2
√
√
The exact solutions are x = 1 − 2 3 and x = 1 + 2 3.
√
√
The decimal approximations to these numbers x = 1 − 2 3 = −2.464 and x = 1 + 2 3 =
4.464 are approximate solutions to this equation. The approximate solutions could also be found
directly from a graph or calculator.
Exercises to Skills for Factoring
For Exercises 1–15, expand and simplify.
1. 2(3x − 7)
2. −4(y + 6)
3. 12(x + y)
4. −7(5x − 8y)
5. x(2x + 5)
6. 3z(2x − 9z)
7. −10r(5r + 6rs)
8. x(3x − 8) + 2(3x − 8)
9. 5z(x − 2) − 3(x − 2)
124
SKILLS REFRESHER FOR CHAPTER THREE
10. (x + 1)(x + 3)
51. x2 + y 2
11. (x − 2)(x + 6)
52. a4 − a2 − 12
12. (5x − 1)(2x − 3)
53. (t + 3)2 − 16
13. (y + 1)(z + 3)
54. x2 + 4x + 4 − y 2
14. (12y − 5)(8w + 7)
55. a3 − 2a2 + 3a − 6
15. (5z − 3)(x − 2)
56. b3 − 3b2 − 9b + 27
Multiply and write the expressions in Problems 16–22 without
parentheses. Gather like terms.
16. −(x − 3) − 2(5 − x)
60. y 2 − 3xy + 2x2
3x − 2x2 4 + (5 + 4x)(3x − 4)
61. x2 e−3x + 2xe−3x
19. P (p − 3q)2
62. t2 e5t + 3te5t + 2e5t
20. 4(x − 3)2 + 7
√
2
21. −
2x + 1
58. hx2 + 12 − 4hx − 3x
59. r(r − s) − 2(s − r)
17. (x − 5)6 − 5(1 − (2 − x))
18.
57. c2 d2 − 25c2 − 9d2 + 225
63. P (1 + r)2 + P (1 + r)2 r
64. x2 − 6x + 9 − 4z 2
22. u u−1 + 2u 2u
For Exercises 23–67, factor completely if possible.
65. dk + 2dm − 3ek − 6em
66. πr 2 − 2πr + 3r − 6
23. 2x + 6
24. 3y + 15
67. 8gs − 12hs + 10gm − 15hm
25. 5z − 30
26. 4t − 6
Solve the equations in Exercises 68–93.
27. 10w − 25
28. 3u4 − 4u3
68. y 2 − 5y − 6 = 0
7
2
3 2
69. 4s2 + 3s − 15 = 0
3
2
+
=8
70.
x
2x
3
71.
+1=5
x−1
√
72. y − 1 = 13
29. 3u + 12u
30. 12x y − 18x
31. 14r 4 s2 − 21rst
32. x2 + 3x − 2
33. x2 − 3x + 2
34. x2 − 3x − 2
35. x2 + 2x + 3
36. x2 − 2x − 3
37. x2 − 2x + 3
38. x2 + 2x − 3
73. −16t2 + 96t + 12 = 60
39. 2x2 + 5x + 2
40. 2x2 − 10x + 12
74. g 3 − 4g = 3g 2 − 12
41. x2 + 3x − 28
42. x3 − 2x2 − 3x
75. 8 + 2x − 3x2 = 0
43. x3 + 2x2 − 3x
44. ac + ad + bc + bd
2
45. x + 2xy + 3xz + 6yz
46. x2 − 1.4x − 3.92
47. a2 x2 − b2
76. 2p3 + p2 − 18p − 9 = 0
77. N 2 − 2N − 3 = 2N (N − 3)
78.
1 3
t =t
64
79. x2 − 1 = 2x
80. 4x2 − 13x − 12 = 0
48. πr 2 + 2πrh
81. 60 = −16t2 + 96t + 12
49. B 2 − 10B + 24
82. n5 + 80 = 5n4 + 16n
50. c2 + x2 − 2cx
83. 5a3 + 50a2 = 4a + 40
COMPLETING THE SQUARE
84. y 2 + 4y − 2 = 0
2
7
85.
+ 2
=0
z−3
z − 3z
x2 + 1 − 2x2
=0
86.
(x2 + 1)2
1
87. 4 − 2 = 0
L
1
1
88. 2 +
−
=0
q+1
q−1
√
89. r 2 + 24 = 7
1
90. √
= −2
3
x
√
1
91. 3 x = x
2
v
92. 10 =
7π
(3x + 4)(x − 2)
=0
93.
(x − 5)(x − 1)
100.
101.
102.
125
x2 + y 2 = 36
y =x−3
y = 4 − x2
y − 2x = 1
y = x3 − 1
y = ex
103. Let be the line of slope 3 passing through the origin. Find the points of intersection of the line and the
parabola whose equation is y = x2 . Sketch the line and
the parabola, and label the points of intersection.
Determine the points of intersection for Problems 104–105.
104.
y
y = x−1
x2 + y 2 = 25
x
In Exercises 94–97, solve for the indicated variable.
94. T = 2π
l
, for l.
g
95. Ab5 = C, for b.
96. |2x + 1| = 7, for x.
2
97.
105.
y
y = x2
2
x − 5mx + 4m
= 0, for x
x−m
Solve the systems of equations in Exercises 98–102.
98.
99.
y = 2x − x2
y = −3
y = 15 − 2x
x
y = 1/x
y = 4x
COMPLETING THE SQUARE
An example of changing the form of an expression is the conversion of ax2 + bx + c into the form
a(x − h)2 + k. We make this conversion by completing the square, a method for producing a perfect
square within a quadratic expression. A perfect square is an expression of the form:
(x + n)2 = x2 + 2nx + n2 .
In order to complete the square in an expression, we must find that number n, which is half the
coefficient of x. Before giving a general procedure, let’s work through an example.
Example 1
Complete the square to rewrite x2 − 10x + 4 in the form a(x − h)2 + k.
126
SKILLS REFRESHER FOR CHAPTER THREE
Step 1: We divide the coefficient of x by 2, giving 12 (−10) = −5.
Step 2: We square the result of step 1, giving (−5)2 = 25.
Step 3: Now add and subtract the 25 after the x-term:
Solution
x2 − 10x + 4 = x2 − 10x + 25 − 25 + 4
= (x2 − 10x + 25) −25 + 4.
Perfect square
Step 4: Notice that we have created a perfect square, x2 − 10x + 25. The next step is to factor the
perfect square and combine the constant terms, −25 + 4, giving the final result:
x2 − 10x + 4 = (x − 5)2 − 21.
Thus, a = +1, h = +5, and k = −21.
Visualizing The Process of Completing The Square
We can visualize how to find the constant that needs to be added to x2 + bx in order to obtain
a perfect square by thinking of x2 + bx as the area of a rectangle. For example, the rectangle in
Figure 3.13 has area x(x + b) = x2 + bx. Now imagine cutting the rectangle into pieces as in
Figure 3.14 and trying to rearrange them to make a square, as in Figure 3.15. The corner piece,
whose area is (b/2)2 , is missing. By adding this piece to our expression, we “complete” the square:
x2 + bx + (b/2)2 = (x + b/2)2 .
x + b/2 b/2
x
6
x
x
x
x+b
Figure 3.13: Rectangle with
sides x and x + b
x
x + b/2
x
b/2
b/2
?
b/2
Figure 3.14: Cut off 2 strips
of width b/2
b/2
Figure 3.15: Rearrange to see a square
with missing corner of area (b/2)2
The procedure we followed can be summarized as follows:
To complete the square in the expression x2 + bx + c, divide the coefficient of x by 2, giving
b/2. Then add and subtract (b/2)2 = b2 /4 and factor the perfect square:
2
b
b2
+ c.
x + bx + c = x +
−
2
4
2
To complete the square in the expression ax2 + bx + c, factor out a first.
The next example has a coefficient a with a = 1. After factoring out the coefficient, we follow
the same steps as in Example 1.
127
COMPLETING THE SQUARE
Example 2
Complete the square in the formula h(x) = 5x2 + 30x − 10.
Solution
We first factor out 5:
h(x) = 5(x2 + 6x − 2).
Now we complete the square in the expression x2 + 6x − 2.
Step 1: Divide the coefficient of x by 2, giving 3.
Step 2: Square the result: 32 = 9.
Step 3: Add the result after the x term, then subtract it:
h(x) = 5(x2 + 6x + 9 −9 − 2).
Perfect square
Step 4: Factor the perfect square and simplify the rest:
h(x) = 5 (x + 3)2 − 11 .
Now that we have completed the square, we can multiply by the 5:
h(x) = 5(x + 3)2 − 55.
Deriving the Quadratic Formula
We derive a general formula to find the zeros of q(x) = ax2 + bx + c, with a = 0, by completing
the square. To find the zeros, set q(x) = 0:
ax2 + bx + c = 0.
Before we complete the square, we factor out the coefficient of x2 :
c
b
a x2 + x +
= 0.
a
a
Since a = 0, we can divide both sides by a:
c
b
x2 + x + = 0.
a
a
To complete the square, we add and then subtract ((b/a)/2)2 = b2 /(4a2 ):
b2
b
b2
c
x2 + x + 2 − 2 + = 0.
a
4a
4a
a
Perfect square
We factor the perfect square and simplify the constant term, giving:
2 2
b − 4ac
b
−
=0
x+
2a
4a2
since
−b2
4a2
+
c
a
=
−b2
4a2
+
4ac
4a2
=−
b2 −4ac
4a2
128
SKILLS REFRESHER FOR CHAPTER THREE
2
b2 − 4ac
b
2 −4ac
=
to both sides
adding b 4a
x+
2
2a
4a2
√
b2 − 4ac
b
± b2 − 4ac
=±
x+
=
taking the square root
2
2a
4a
2a
√
b2 − 4ac
−b
±
x=
subtracting b/2a
2a √ 2a
−b ± b2 − 4ac
.
x=
2a
Exercises to Skills for Completing the Square
For Exercises 1–8, complete the square for each expression.
1. x2 + 8x
22. y = −3x2 − x − 2
In Exercises 23–29, solve by completing the square.
2
2. w + 7w
3. 2r 2 + 20r
4. 3t2 + 24t − 13
5. a2 − 2a − 4
23. g 2 = 2g + 24
24. p2 − 2p = 6
25. d2 − d = 2
26. 2r 2 + 4r − 5 = 0
27. 2s2 = 1 − 10s
28. 7r 2 − 3r − 6 = 0
29. 5p2 + 9p = 1
6. n2 + 4n − 5
7. 3r 2 + 9r − 4
In Exercises 30–35, solve by using the quadratic formula.
8. 12g 2 + 8g + 5
2
In Exercises 9–12, rewrite in the form a(x − h) + k.
9. x2 − 2x − 3
10. 10 − 6x + x2
11. −x2 + 6x − 2
12. 3x2 − 12x + 13
In Exercises 13–22, complete the square to find the vertex of
the parabola.
30. n2 − 4n − 12 = 0
31. 2y 2 + 5y = −2
32. 6k2 + 11k = −3
33. w2 + w = 4
34. z 2 + 4z = 6
35. 2q 2 + 6q − 3 = 0
In Exercises 36–46, solve using factoring, completing the
square, or the quadratic formula.
36. r 2 − 2r = 8
2
37. s2 + 3s = 1
2
14. y = x − x + 4
38. z 3 + 2z 2 = 3z + 6
15. y = −x2 − 8x + 2
39. 25u2 + 4 = 30u
13. y = x + 6x + 3
16. y = x2 − 3x − 3
17. y = −x2 + x − 6
18. y = 3x2 + 12x
19. y = −4x2 + 8x − 6
2
40. v 2 − 4v − 9 = 0
41. 3y 2 = 6y + 18
42. 2p2 + 23 = 14p
43. 2w3 + 24 = 6w2 + 8w
44. 4x2 + 16x − 5 = 0
20. y = 5x − 5x + 7
45. 49m2 + 70m + 22 = 0
21. y = 2x2 − 7x + 3
46. 8x2 − 1 = 2x