Chapter Three
QUADRATIC FUNCTIONS Contents
3.1 Introduction to the Family of Quadratic
Functions . . . . . . . . . . . . . . . . . 116
Finding the Zeros of a Quadratic Function . . 116
Concavity and Rates of Change for Quadratic
Functions . . . . . . . . . . . . . 117
Finding a Formula From the Zeros
and Vertical Intercept . . . 119
Formulas for Quadratic Functions . . . . . . 120
3.2 The Vertex of a Parabola . . . . . . . . . . 122
The Vertex Form of a Quadratic Function . . 123
Finding a Formula Given the Vertex
and Another Point . . . . . 125
Modeling with Quadratic Functions . . . . . 126
REVIEW PROBLEMS . . . . . . . . . . . 129
STRENGTHEN YOUR UNDERSTANDING 130
Skills Refresher for CHAPTER 3:
QUADRATIC EQUATIONS . . . . . . . . 131
Skills for Factoring . . . . . . . . . . . . . 131
Expanding an Expression . . . . . . . . . . 131
Factoring . . . . . . . . . . . . . . . . . 131
Removing a Common Factor . . . . 131
Grouping Terms . . . . . . . . . . 132
Factoring Quadratics . . . . . . . . 132
Perfect Squares and the Difference of
Squares . . . . . . . . . . 132
Solving Quadratic Equations . . . . . . . . 133
Solving by Factoring . . . . . . . . 133
Solving with the Quadratic Formula . 134
Completing the Square . . . . . . . . . . . 136
Visualizing Completing the Square . . 136
Deriving the Quadratic Formula . . . . . . . 137
116 Chapter Three QUADRATIC FUNCTIONS
3.1 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.
3
3
y = f(t) = −16t2 + 47t+ 3
❘
Initialheight
t (seconds)
y (feet)
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 standard form is
y = ax2 + bx+ c.
The graph of a quadratic function is called a parabola. Notice that the function in Figure 3.1 is con-
cave 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 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 is expressed in factored form,
q(x) = a(x− r)(x− s), where a, r, s are constants, a 6= 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 t = −1/16. 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 131.)
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 117
We can also find the zeros of a quadratic function by using the quadratic formula. (See the
Skills Review on page 131 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
x =−b±
√b2 − 4ac
2a=
−(−1)±√
(−1)2 − 4(1)(−6)
2(1)
=1±
√25
2= 3 or − 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 Figure 3.2 shows a graph of h(x) = −1
2x2 − 2. What happens if we try to use algebra to find the
zeros of h?
✛ yintercept(0,−2)
h(x)
x
y
Figure 3.2: Zeros of h(x) = −(x2/2)− 2?
Solution To find the zeros, we solve the equation
−1
2x2 − 2 = 0
−1
2x2 = 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 Rates of Change for Quadratic Functions
A quadratic function has a graph that is either concave up or concave down. Recall that for a graph
that is concave up, rates of change increase as we move right. For a graph that is concave down,
rates of change decrease as we move right.
Example 4 Let f(x) = x2. Find the average rate of change of f over the four consecutive intervals of length 2
between x = −4 and x = 4. What do these rates tell you about the concavity of the graph of f?
118 Chapter Three QUADRATIC FUNCTIONS
Solution Between x = 0 and x = 2, we have
Average rate of change
of f=
f(2)− f(0)
2− 0=
22 − 02
2− 0= 2.
Between x = 2 and x = 4, we have
Average rate of change
of f=
f(4)− f(2)
4− 2=
42 − 22
4− 2= 6.
Similarly, between x = −4 and x = −2, we can show the average rate of change is −6; between
x = −2 and x = 0, the rate of change is −2. See Figure 3.3. Since the rates of change are increasing,
the graph is concave up.
−4 −2 2 4
❄
Slope = −6
❄
Slope = −2
❄
Slope = 2
❄
Slope = 6
f(x) = x2
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.
1 2 3
5
10f(t)
15
t (seconds)
h (meters)
Figure 3.4: Height of diver above water as a function of time
Solution (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 in Section 3.2.
3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS 119
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 inter-
cept is the point estimated earlier where f(t) = 0. This intercept, t = 2.462 seconds, gives the
time after leaving the platform that the diver enters the water.
(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, ∆h/∆t, is decreasing.
Table 3.1 Slope of f(t) = −4.9t2 + 8t+ 10
t (sec) h (meters)Rate of change
∆h/∆t
0 10
5.55
0.5 12.775
0.65
1.0 13.100
−4.25
1.5 10.975
−9.15
2.0 6.400
The previous two examples illustrate that the concavity of the graph of y = ax2 + bx + c is
determined by the sign of the coefficient a:
• If a > 0, the parabola is concave up.
• If a < 0, the parabola is concave down.
Example 6 Determine the concavity of the graphs of the following quadratic functions:
(a) y = −3x2 − 3x+ 2(b) y = −3(−2x+ 1)(x− 4)
Solution (a) Since the coefficient of x2 is negative, this function has a graph which is concave down.
(b) Expanding we have
y = −3(−2x+ 1)(x− 4) = −3(−2x2 + 8x+ x− 4) = 6x2 − 27x+ 12.
Since the coefficient of x2 is positive, the graph of this function is concave up.
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 7 Find the equation of the parabola in Figure 3.5 using the factored form.
1 3
6
x
f(x)
Figure 3.5: Finding a formula for a quadratic from the zeros
120 Chapter Three QUADRATIC FUNCTIONS
Solution Since the parabola has x-intercepts at x = 1 and x = 3, its formula can be written as
f(x) = a(x− 1)(x− 3).
To find a, we substitute x = 0, y = 6, giving
6 = a · 3a = 2.
Thus, the formula is
f(x) = 2(x− 1)(x− 3).
Multiplying out gives f(x) = 2x2 − 8x+ 6.
Formulas for Quadratic Functions
The function f in Example 7 can be written in at least two different ways. The standard form
f(x) = 2x2 − 8x+ 6 shows that the parabola opens upward, since the coefficient of x2 is positive;
the constant 6 gives the vertical intercept. The factored 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 6= 0.
The parabola is concave up (opens upward) if a > 0 or concave down (opens downward)
if a < 0, and 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 6= 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 how to find the
vertex of the graph.
Exercises and Problems for Section 3.1
Skill Refresher
Multiply and write the expressions in Exercises S1–S2 with-
out parentheses. Gather like terms.
S1.(t2 + 1
)50t−
(25t2 + 125
)2t
S2.(A2 −B2
)2
For Exercises S3–S8, factor completely if possible.
S3. u2 − 2u S4. x2 + 3x+ 2
S5. 3x2 − x− 4 S6. (s+ 2t)2 − 4p2
S7. 16x2 − 1 S8. y3 − y2 − 12y
Solve the equations in Exercises S9–S10.
S9. x2 + 7x+ 6 = 0 S10. 2w2 + w − 10 = 0
3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS 121
Exercises
Are the functions in Exercises 1–8 quadratic? If so, write the
function in the form f(x) = ax2 + bx+ c.
1. g(t) = 3(t− 2)2 + 7
2. f(x) = (2x− 3)(5− x)
3. w(n) = n(n− 3)(n− 2)− n2(n− 8)
4. h(t) = −16(t− 3)(t+ 1)
5. R(q) =1
q2(q2 + 1)2
6. K(x) = 132 + 13x
7. T (n) =√5 +
√3n4 −
√
n4
4
8. r(v) =v2 +
√2
3+
v − 3
5+ πv2
9. Find the zeros of Q(r) = 2r2 − 6r − 36 by factoring.
10. Find the zeros of Q(x) = 5x−x2+3 using the quadratic
formula.
11. Solve for x using the quadratic formula and demonstrate
your solution graphically:
(a) 6x− 1
3= 3x2 (b) 2x2 + 7.2 = 5.1x
In Exercises 12–19, find the zeros (if any) of the function al-
gebraically.
12. y = (2− x)(3− 2x) 13. y = 9x2 + 6x+ 1
14. y = 6x2 − 17x+ 12 15. N(t) = t2 − 7t+ 10
16. y = 89x2 + 55x+ 34 17. Q(r) = 2r2 − 6r − 36
18. y = x4 + 5x2 + 6 19. y = x−√x− 12
In Exercises 20–23, is the graph of the quadratic function con-
cave up or concave down?
20. y = 2x− 3x2 + 1 21. y = 5(x− 3)2
22. y = 3(1− 2x)(x− 4) 23. y = −2(5x−1)(4−x)
Problems
24. Use the quadratic formula to find the time at which the
baseball in Figure 3.1 on page 116 hits the ground.
In Problems 25–26, find a formula for the quadratic function
whose graph has the given properties.
25. A y-intercept of y = 7 and the only zero at x = −2.
26. A y-intercept of y = 7 and x-intercepts at x = 1, 4.
27. Is there a quadratic function with zeros x = 1, x = 2and x = 3?
28. Graph a quadratic function which has all the follow-
ing properties: concave up, y-intercept is −6, zeros at
x = −2 and x = 3.
29. Can you graph a quadratic function which has all the fol-
lowing properties: concave down, y-intercept is −6, ze-
ros at x = −2 and x = 3?
30. 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.
31. The graph of a quadratic function, f(x), passes through
the points (1, 2), (3, 4) and (5, 2). Is the graph of f(x)concave up or concave down? Give a reason for your an-
swer.
32. The quadratic function, f(x), has no zeros and its graph
passes through the point (1, 1). Is the graph of f(x) con-
cave up or concave down? Give a reason for your answer.
33. Without a calculator, graph the following function by fac-
toring and plotting zeros:
y = −4cx+ x2 + 4c2 for c > 0.
34. Without a calculator, graph y = 3x2 − 16x− 12 by fac-
toring and plotting zeros.
In Problems 35–36, find a formula for the parabola.
35.
(−1, 0)
(0,−1)
(3, 0)x
y 36.
(−6, 0)
(0, 5)
(2, 0)x
y
37. 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).
38. Find two different quadratic functions with zeros x = 1,
x = 2.
122 Chapter Three QUADRATIC FUNCTIONS
39. 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.
40. A ball is thrown into the air. Its height (in feet) t seconds
later is given by h(t) = 80t− 16t2.
(a) Evaluate and interpret h(2).(b) Solve the equation h(t) = 80. Interpret your solu-
tions and illustrate them on a graph of h(t).
41. 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 top edge of the pipe t sec-
onds 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 top
edge? Return to the edge of the pipe?
(c) How long is she in the air?
42. Let V (t) = t2 − 4t + 4 represent the velocity after tseconds 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.
43. Table 3.2 gives the approximate number of cell phone
subscribers, S, in millions, in the US.1 If x is in years
since 2005, show that this data can be approximated by
the quadratic function p(x) = −1.9x2 +24.4x+208.8.
What does this model predict for the year 2015? How
good is this model for predicting the future?
Table 3.2
Year 2005 2006 2007 2008 2009
S (millions) 207.9 233.0 250.0 262.7 276.6
44. Let f(x) = x2 and g(x) = x2 + 2x− 8.
(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.
45. 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 pack-
age, 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 re-
leased 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.]
4430
5
4
3
2
1
Airplane →
d, horizontaldistance (meters)
h, height (km)
■Packagedroppedhere
Figure 3.6
3.2 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). (See
Example 5 on page 126 for the exact values.) This means that the baseball reaches its maximum
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 determined exactly if the quadratic function is written in the
1http://www.infoplease.com/ipa/A0933563.html. Last accessed January 13, 2014.
3.2 THE VERTEX OF A PARABOLA 123
3
3
40 vertex ≈ (1.5, 37.5)y = f(t) = −16t2 + 47t+ 3
❘
Initialheight
t (seconds)
y (feet)
Figure 3.7: Height of baseball at time t
form
y = a(x− h)2 + k.
This form arises when we shift the parabola y = x2 horizontally by h and vertically by k, moving
the vertex from (0, 0) to (h, 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 .
(d) Explain how vertical and horizontal shifts can be used to obtain the coordinates of the vertex.
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) and the minimum is −4.
(d) The function f(x) = (x+3)2− 4 is a transformation of the function f(x) = x2, with an inside
change of −3 and an outside change of −4. This indicates that f(x) = x2 has been shifted 3units to the left and 4 units down. Thus, the vertex is (−3,−4).
−7
−3
1−4
12
x
f(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
enabled us to find the vertex of the graph and the location and value for the minimum of the function.
124 Chapter Three QUADRATIC FUNCTIONS
In general, we have the following:
The vertex form of a quadratic function is
y = a(x− h)2 + k, where a, h, k are constants, a 6= 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.
Example 2 (a) Convert the quadratic f(x) = (x+ 3)2 − 4 in Example 1 to standard form.
(b) Explain how the vertical intercept can be found from the standard form.
(c) Find the zeros of f .
(d) Explain how the axis of symmetry of the parabola is related to the zeros.
Solution (a) Expanding (x+ 3)2 and gathering like terms converts f to standard form:
f(x) = x2 + 6x+ 9− 4 = x2 + 6x+ 5.
Since the coefficient of x2 is a = 1, the parabola opens upward.
(b) The vertical intercept is f(0) = 5. In the standard form y = ax2 + bx+ c, the y-intercept is the
constant c.(c) To find the zeros, we factor f and write it as
f(x) = (x+ 1)(x+ 5).
This form shows that the zeros are x = −1 and x = −5.
(d) The axis of symmetry is x = −3, which is the vertical line through the midpoint of the zeros:
(−1− 5)/2 = −3.
As we see in Example 2, if a quadratic can be written in factored form, its axis of symmetry is
at the midpoint of the zeros. To convert from vertex to standard form, we expand the squared term
and gather like terms. In the next example, we convert from from standard form to vertex form by
completing the square.2
Example 3 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︸ ︷︷ ︸
Perfect square
−9 + 8,
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.
2A more detailed explanation of this method is in the Skills Review on page 136.
3.2 THE VERTEX OF A PARABOLA 125
−3 6
−3
3
x = 3
x
Figure 3.9: s(x) = x2 − 6x+ 8
−3 6
−3
3
x = −3/2
x
Figure 3.10: t(x) = −4x2−12x−8
(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
t(x) = −4
(
x2 + 3x+9
4︸ ︷︷ ︸
Perfect square
−9
4+ 2
)
t(x) = −4
((
x+3
2
)2
− 1
4
)
t(x) = −4
(
x+3
2
)2
+ 1.
The vertex of t is (−3/2, 1), the axis of symmetry is x = −3/2, and the parabola opens down-
ward. 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 4 Find the formula for the quadratic function graphed in Figure 3.11.
(−3, 2)
(0, 5)
x = −3
m(x)
x
y
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.
126 Chapter Three QUADRATIC FUNCTIONS
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
a =1
3.
Thus, the formula is
m(x) =1
3(x+ 3)2 + 2.
If we want the formula in standard form, we multiply out:
m(x) =1
3x2 + 2x+ 5.
Modeling with Quadratic Functions
In applications, it is often useful to find the maximum or minimum value of a quadratic function.
The next example returns to the baseball that started this chapter. Problem 25 on page 128 asks you
to derive the vertex form of the baseball height function; here we see what this form can tell us.
Example 5 For t in seconds, the height of a baseball in feet is given both in standard and in vertex form by
y = f(t) = −16t2 + 47t+ 3 = −16
(
t− 47
32
)2
+2401
64.
Find the maximum height reached by the baseball and the time at which that height is reached.
Solution From the vertex form, we see the vertex is at the point(47
32, 2401
64
)= (1.469, 37.516). This means
that the ball reaches it maximum height of 37.516 feet at t = 1.469 seconds. Note that these values
are close to the graphical estimates (1.5, 37.5) made at the beginning of this section.
Example 6 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 a
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.
h h
b
River
Figure 3.12: A park next to a river
3.2 THE VERTEX OF A PARABOLA 127
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.
Exercises and Problems for Section 3.2
Skill Refresher
For Exercises S1–S4, complete the square for each expression.
S1. y2 − 12y S2. s2 + 6s− 8
S3. c2 + 3c− 7 S4. 4s2 + s+ 2
In Exercises S5–S7, solve by completing the square.
S5. r2 − 6r + 8 = 0 S6. n2 = 3n+ 18
S7. 5q2 − 8 = 2q
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. 2q2 + 4q − 5 = 8
Exercises
For the quadratic functions in Exercises 1–2, state the coor-
dinates of the vertex, the axis of symmetry, and whether the
parabola opens upward or downward.
1. f(x) = 3(x− 1)2 + 2
2. g(x) = −(x+ 3)2 − 4
3. Find the vertex and axis of symmetry of the graph of
v(t) = t2 + 11t− 4.
4. Find the vertex and axis of symmetry for the parabola
whose equation is y = 3x2 − 6x+ 5.
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
In Exercises 6–9, find a formula for the parabola.
6.
1(−1, 0) (3, 0)
(0,−3)
x
y 7.
−15
9(−6, 9)
x
y
8.
(0,−4)
(2, 0)x
y 9.
(10, 8)
(6, 5)x
y
For Exercises 10–13, convert the quadratic functions to ver-
tex form by completing the square. Identify the vertex and the
axis of symmetry.
10. f(x) = x2 + 8x+ 3
11. g(x) = −2x2 + 12x+ 4
12. p(t) = 2t2 − 0.12t+ 0.1
13. w(z) = −3z2 + 9z − 2
In Exercises 14–16, write the quadratic function in its stan-
dard, vertex, and factored forms.
14. y = −6 +x2 − x
2
15. f(t) =5t2 − 20
2
16. g(s) = (s− 5)(2s+ 3)
128 Chapter Three QUADRATIC FUNCTIONS
In Exercises 17–24, write a formula and graph the transforma-
tion of m(n) = 1
2n2.
17. y = m(n) + 1 18. y = m(n+ 1)
19. y = m(n)− 3.7 20. y = m(n− 3.7)
21. y = m(n) +√13 22. y = m(n+ 2
√2)
23. y = m(n+ 3) + 7 24. y = m(n− 17)− 159
Problems
25. Find the vertex form of f(t) = −16t2 + 47t+ 3.
26. Show that y = x2 − x+ 41 has no real zeros.
27. Find the value of k so that the graph of y = (x−3)2+kpasses through the point (6, 13).
28. The parabola y = ax2+k has vertex (0,−2) and passes
through the point (3, 4). Find its equation.
29. Using the vertex form, find a formula for the parabola
with vertex (2, 5) that passes through the point (1, 2).
In Problems 30–34, find a formula for the quadratic function
whose graph has the given properties.
30. Vertex at (4, 2) and y-intercept of y = 6.
31. Vertex at (4, 2) and y-intercept of y = −4.
32. Vertex at (4, 2) and zeros at x = −3, 11.
33. Vertex at (7, 3) and passing through the point (3, 7).
34. A single x-intercept at x = 1/2 and a y-intercept at 3.
35. Let f be a quadratic function whose graph is concave up
with a vertex at (1,−1), and a zero at the origin.
(a) Graph y = f(x).(b) Determine a formula for f(x).(c) Determine the range of f .
(d) Find any other zeros.
36. Find the vertex of y = 26+0.4x−0.01x2 exactly. Graph
the function, labeling all intercepts.
37. Find the vertex of y = 0.03x2+1.8x+2 exactly. Graph
the function, labeling all intercepts.
38. If we know a quadratic function f has a zero at x = −1and 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.
39. Figure 3.13 shows f(x) = x2. Define g by shifting the
graph of f to the right 2 units and down 1 unit; see Fig-
ure 3.14. Find a formula for g in terms of f . Find a for-
mula for g in terms of x.
−2 4−1
4 f(x) = x2
x
y
Figure 3.13
−2 4−1
4
g
x
y
Figure 3.14
40. The function g(x) is obtained by shifting the graph of
y = x2. If g(3) = 16, give a possible formula for gwhen
(a) g is the result of applying only a horizontal shift to
y = x2.
(b) g is the result of applying only a vertical shift to
y = x2.
(c) g is the result of applying a horizontal shift right 2units and an appropriate vertical shift of y = x2.
41. Graph y = x2 − 10x + 25 and y = x2. Use a shift
transformation to explain the relationship between the
two graphs.
42. Let f(x) = x2 and let g(x) = (x− 3)2 + 2.
(a) Give the formula for g in terms of f , and describe
the relationship between f and g in words.
(b) Is g a quadratic function? If so, find its standard form
and the parameters a, b, and c.
(c) Graph g, labeling all important features.
43. If you have a string of length 50 cm, what are the di-
mensions of the rectangle of maximum area that you can
enclose with your string? Explain your reasoning. What
about a string of length k cm?
44. 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 by3
h(t) = −16t2 + 16Tt,
where T is the total time in seconds that the ballet dancer
is in the air.
(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?
3K. Laws, The Physics of Dance (Schirmer, 1984).
REVIEW EXERCISES AND PROBLEMS FOR CHAPTER THREE 129
(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 2.45. A football player kicks a ball at an angle of 37◦ above the
ground with an initial speed of 20 meters/second. The
height in meters, h, as a function of the horizontal dis-
tance 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?4
CHAPTER SUMMARY
• General Formulas for Quadratic Functions
Standard form:
y = ax2 + bx+ c, a 6= 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 aOpens upward (concave up) if a > 0, minimum at
(h, k)Opens downward (concave down) if a < 0, maxi-
mum 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 1–3 quadratic? If so, write the
function in standard form.
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 4–9, find the zeros (if any) of the function alge-
braically.
4. y = 2x2 + 5x+ 2 5. y = 4x2 − 4x− 8
6. y = 7x2 + 16x+ 4 7. y = 5x2 + 2x− 1
8. y = −17x2 +23x+19 9. y = 3x2 − 2x+ 6
10. Show that the function y = −x2 + 7x − 13 has no real
zeros.
11. Find the vertex and axis of symmetry of the graph of
w(x) = −3x2 − 30x+ 31.
In Exercises 12–16, find a possible formula for the parabola
with the given conditions.
12. The vertex is (1,−2) and the y-intercept is y = −5.
13. The vertex is (4,−2) and the y-intercept is y = −3.
14. The vertex is (−7,−3) and it passes through the point
(−3,−7).
15. The parabola goes through the origin and its vertex is
(1,−1).
16. The x-intercepts are at x = −1 and x = 2 and (−2, 16)is on the function’s graph.
4Adapted from R. Halliday, D. Resnick, and K. Krane, Physics (New York: Wiley, 1992), p. 58.
130 Chapter Three QUADRATIC FUNCTIONS
In Exercises 17–20, find a formula for the parabola.
17.
(0, 4)
(4, 7)
x
y 18.
(5, 5)
(3, 3)x
y
19.
2
−5(3,−5)
y
x
20.
(−4, 0)
(2, 36)
(5, 0)x
y
Problems
For each parabola in Problems 21–24, state the coordinates of
the vertex, the axis of symmetry, the y-intercept, and whether
the curve is concave up or concave down.
21. y = 2(x−3/4)2−2/3 22. y = −1/2(x+6)2 − 4
23. y = (x− 0.6)2 24. y = −0.3x2 − 7
In Problems 25–28, write each function in factored form or
vertex form and then state its vertex and zeros.
25. y = 0.3x2 − 0.6x− 7.2
26. y = 2x2 − 4x− 2
27. y = −3x2 + 24x− 36
28. y = 2x2 + (7/3)x+ 1
29. For x between x = −2 and x = 4, determine the con-
cavity of the graph of f(x) = (x−1)2+2 by calculating
average rates of change over intervals of length 2.
30. 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
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?
31. 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 + 16Tt,
where T is the total time in seconds that it takes to com-
plete the jump. For a jump that takes 1 second to com-
plete, 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.]
STRENGTHEN 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.
9. A quadratic function has exactly two zeros.
10. The graph of every quadratic function 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.
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.
131
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 involv-
ing parentheses. This process is sometimes referred to as expanding the expression.
Example 1 Multiply the following expressions and simplify.
(a) 3x2
(
x+1
6x−3
)
(b)((2t)2 − 5
)√t
Solution (a) 3x2
(
x+1
6x−3
)
=(3x2)(x) +
(3x2)(1
6x−3
)
= 3x3 +1
2x−1.
(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 Expand the following and simplify by gathering like terms.
(a) (5x2 + 2)(x− 4) (b) (2√r + 2)(4
√r − 3) (c)
(
3− 1
2x
)2
Solution (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.
(c)
(
3− 1
2x
)2
= 32 − 2 (3)
(1
2x
)
+
(
−1
2x
)2
= 9− 3x+1
4x2.
Factoring
To write an expanded expression in factored form, we “un-multiply” the expression. Some tech-
niques for factoring are given in this section. We can check factoring by multiplying the factors.
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).
132 SKILLS REFRESHER FOR CHAPTER THREE
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 Factor the following:
(a)2
3x2y +
4
3xy (b) (2p+ 1)p3 − 3p(2p+ 1) (c) −s2t
8w− st2
16w
Solution (a)2
3x2y +
4
3xy =
2
3xy(x+ 2).
(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.)
(c) −s2t
8w− st2
16w= − st
8w
(
s+t
2
)
.
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 Factor x2 − hx− x+ h.
Solution 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 is 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 combina-
tions, we find
t2 − 4t− 12 = (t− 6)(t+ 2).
Example 6 Factor 4− 2M − 6M2.
Solution By a similar method to the previous example, we find 4− 2M − 6M2 = (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 131, we have
a2 + 2ab+ b2 = (a+ b)2,
a2 − 2ab+ b2 = (a− b)2,
a2 − b2 = (a− b)(a+ b).
SKILLS FOR FACTORING 133
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 previously) is especially useful.
Example 7 Factor: (a) 16y2 − 24y + 9 (b) 25S2R4 − T 6 (c) x2(x− 2) + 16(2− x)
Solution (a) 16y2 − 24y + 9 = (4y − 3)2.
(b) 25S2R4 − T 6 =(5SR2
)2 −(T 3)2
=(5SR2 − T 3
) (5SR2 + 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).
Solving Quadratic Equations
Example 8 Give exact and approximate solutions to x2 = 3.
Solution The exact solutions are x = ±√3; approximate ones are x ≈ ±1.73, or x ≈ ±1.732, or x ≈
±1.73205 (since√3 = 1.732050808 . . .). Notice that the equation x2 = 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 the solutions.
Example 10 Solve 2(x+ 3)2 = 5(x+ 3).
Solution 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.
134 SKILLS REFRESHER FOR CHAPTER THREE
Solving with the Quadratic Formula
Instead of factoring, we can solve the equation ax2 + bx+ c = 0 by using the quadratic formula:
x =−b±
√b2 − 4ac
2a.
The quadratic formula is derived by completing the square for y = ax2 + bx+ c. See page 136.
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
x =−2 +
√
4− 4(−1)(11)
2(−1)=
−2 +√48
−2=
−2 +√16 · 3
−2=
−2 + 4√3
−2= 1− 2
√3,
x =−2−
√
4− 4(−1)(11)
2(−1)=
−2−√48
−2=
−2−√16 · 3
−2=
−2− 4√3
−2= 1 + 2
√3.
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 for 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)
10. (x+ 1)(x+ 3)
11. (x− 2)(x+ 6)
12. (5x− 1)(2x− 3)
13. (y + 1)(z + 3)
14. (12y − 5)(8w + 7)
15. (5z − 3)(x− 2)
Multiply and write the expressions in Problems 16–22 without
parentheses. Gather like terms.
16. −(x− 3)− 2(5− x)
17. (x− 5)6− 5(1− (2− x))
18.(3x− 2x2
)4 + (5 + 4x)(3x− 4)
19. P (p− 3q)2 20. 4(x− 3)2 + 7
21. −(√
2x+ 1)2
22. u(u−1 + 2u
)2u
For Exercises 23–67, factor completely if possible.
23. 2x+ 6 24. 3y + 15
25. 5z − 30 26. 4t− 6
27. 10w − 25 28. 3u4 − 4u3
29. 3u7 + 12u2 30. 12x3y2 − 18x
31. 14r4s2 − 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
39. 2x2 + 5x+ 2 40. 2x2 − 10x+ 12
41. x2 + 3x− 28 42. x3 − 2x2 − 3x
43. x3 + 2x2 − 3x
SKILLS FOR FACTORING 135
44. ac+ ad+ bc+ bd
45. x2 + 2xy + 3xz + 6yz
46. x2 − 1.4x− 3.92
47. a2x2 − b2
48. πr2 + 2πrh
49. B2 − 10B + 24
50. c2 + x2 − 2cx
51. x2 + y2
52. a4 − a2 − 12
53. (t+ 3)2 − 16
54. x2 + 4x+ 4− y2
55. a3 − 2a2 + 3a− 6
56. b3 − 3b2 − 9b+ 27
57. c2d2 − 25c2 − 9d2 + 225
58. hx2 + 12− 4hx− 3x
59. r(r − s)− 2(s− r)
60. y2 − 3xy + 2x2
61. x2e−3x + 2xe−3x
62. t2e5t + 3te5t + 2e5t
63. P (1 + r)2 + P (1 + r)2r
64. x2 − 6x+ 9− 4z2
65. dk + 2dm− 3ek − 6em
66. πr2 − 2πr + 3r − 6
67. 8gs− 12hs+ 10gm− 15hm
Solve the equations in Exercises 68–93.
68. y2 − 5y − 6 = 0
69. 4s2 + 3s− 15 = 0
70.2
x+
3
2x= 8
71.3
x− 1+ 1 = 5
72.√y − 1 = 13
73. −16t2 + 96t+ 12 = 60
74. g3 − 4g = 3g2 − 12
75. 8 + 2x− 3x2 = 0
76. 2p3 + p2 − 18p− 9 = 0
77. N2 − 2N − 3 = 2N(N − 3)
78.1
64t3 = t
79. x2 − 1 = 2x
80. 4x2 − 13x− 12 = 0
81. 60 = −16t2 + 96t+ 12
82. n5 + 80 = 5n4 + 16n
83. 5a3 + 50a2 = 4a+ 40
84. y2 + 4y − 2 = 0
85.2
z − 3+
7
z2 − 3z= 0
86.x2 + 1− 2x2
(x2 + 1)2= 0
87. 4− 1
L2= 0
88. 2 +1
q + 1− 1
q − 1= 0
89.√r2 + 24 = 7
90.13√x
= −2
91. 3√x =
1
2x
92. 10 =
√v
7π
93.(3x+ 4)(x− 2)
(x− 5)(x− 1)= 0
In Exercises 94–97, solve for the indicated variable.
94. T = 2π
√
l
g, for l.
95. Ab5 = C, for b.
96. |2x+ 1| = 7, for x.
97.x2 − 5mx+ 4m2
x−m= 0, for x
Solve the systems of equations in Exercises 98–102.
98.
{
y = 2x− x2
y = −399.
{y = 1/xy = 4x
100.
{
x2 + y2 = 36y = x− 3
101.
{
y = 4− x2
y − 2x = 1
102.
{
y = x3 − 1y = ex
103. Let ℓ be the line of slope 3 passing through the ori-
gin. Find the points of intersection of the line ℓ 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.
x2 + y2 = 25 y = x− 1
x
y 105.
y = x2
y = 15− 2x
x
y
136 SKILLS REFRESHER FOR CHAPTER THREE
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.
Solution Step 1: We divide the coefficient of x by 2, giving 1
2(−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:
x2 − 10x+ 4 = x2 − 10x+ 25− 25 + 4
= (x2 − 10x+ 25)︸ ︷︷ ︸
Perfect square
−25 + 4.
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 3.15 has area x(x + b) = x2 + bx. Now imagine cutting the rectangle into pieces as in
Figure 3.16 and trying to rearrange them to make a square, as in Figure 3.17. 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
x+ b
Figure 3.15: Rectangle with
sides x and x+ b
x
x b/2 b/2
Figure 3.16: Cut off 2 strips
of width b/2
✻
❄
x+ b/2
✲✛ x+ b/2
x x
x b/2
b/2 b/2
Figure 3.17: Rearrange to see a square
with missing corner of area (b/2)2
The procedure we followed can be summarized as follows:
COMPLETING THE SQUARE 137
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:
x2 + bx+ c =
(
x+b
2
)2
− b2
4+ c.
To complete the square in the expression ax2 + bx+ c, factor out a first.
The next example has a coefficient a with a 6= 1. After factoring out the coefficient, we follow
the same steps as in Example 1.
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︸ ︷︷ ︸
Perfect square
−9− 2).
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 6= 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:
a
(
x2 +b
ax+
c
a
)
= 0.
Since a 6= 0, we can divide both sides by a:
x2 +b
ax+
c
a= 0.
138 SKILLS REFRESHER FOR CHAPTER THREE
To complete the square, we add and then subtract ((b/a)/2)2= b2/(4a2):
x2 +b
ax+
b2
4a2︸ ︷︷ ︸
Perfect square
− b2
4a2+
c
a= 0.
We factor the perfect square and simplify the constant term, giving:
(
x+b
2a
)2
−(b2 − 4ac
4a2
)
= 0 since−b
2
4a2+ c
a=
−b2
4a2+ 4ac
4a2= −
(b2−4ac
4a2
)
(
x+b
2a
)2
=b2 − 4ac
4a2adding
b2−4ac
4a2to both sides
x+b
2a= ±
√
b2 − 4ac
4a2=
±√b2 − 4ac
2ataking the square root
x =−b
2a±
√b2 − 4ac
2asubtracting b/2a
x =−b±
√b2 − 4ac
2a.
Exercises for Skills for Completing the Square
For Exercises 1–8, complete the square for each expression.
1. x2 + 8x 2. w2 + 7w
3. 2r2 + 20r 4. 3t2 + 24t− 13
5. a2 − 2a− 4 6. n2 + 4n− 5
7. 3r2 + 9r − 4 8. 12g2 + 8g + 5
In Exercises 9–12, rewrite in the form a(x− h)2 + 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.
13. y = x2 + 6x+ 3 14. y = x2 − x+ 4
15. y = −x2 − 8x+ 2 16. y = x2 − 3x− 3
17. y = −x2 + x− 6 18. y = 3x2 + 12x
19. y = −4x2 + 8x− 6 20. y = 5x2 − 5x+ 7
21. y = 2x2 − 7x+ 3 22. y = −3x2 − x− 2
In Exercises 23–29, solve by completing the square.
23. g2 = 2g + 24 24. p2 − 2p = 6
25. d2 − d = 2 26. 2r2 + 4r − 5 = 0
27. 2s2 = 1− 10s 28. 7r2 − 3r − 6 = 0
29. 5p2 + 9p = 1
In Exercises 30–35, solve by using the quadratic formula.
30. n2 − 4n− 12 = 0 31. 2y2 + 5y = −2
32. 6k2 + 11k = −3 33. w2 + w = 4
34. z2 + 4z = 6 35. 2q2 + 6q − 3 = 0
In Exercises 36–46, solve using factoring, completing the
square, or the quadratic formula.
36. r2 − 2r = 8
37. s2 + 3s = 1
38. z3 + 2z2 = 3z + 6
39. 25u2 + 4 = 30u
40. v2 − 4v − 9 = 0
41. 3y2 = 6y + 18
42. 2p2 + 23 = 14p
43. 2w3 + 24 = 6w2 + 8w
44. 4x2 + 16x− 5 = 0
45. 49m2 + 70m+ 22 = 0
46. 8x2 − 1 = 2x