4 Quadratic Equations: Getting Information from a Model
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SECTION 5.4
■
Quadratic Equations: Getting Information from a Model
x 2 + 5x = 24
x 2 + 5x - 24 = 0
1x - 32 1x + 82 = 0
x - 3 = 0 or
x + 8 = 0
x = 3
x = -8
449
Given equation
Subtract 24
Factor
Zero-Product Property
Solve
The solutions are x = 3 and x = - 8.
✓ C H E C K We substitute x = 3 into the original equation:
132 2 + 5132 = 9 + 15 = 24 ✓
We substitute x = - 8 into the original equation:
1- 82 2 + 51- 82 = 64 - 40 = 24 ✓
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NOW TRY EXERCISE 7
■
Do you see why one side of the equation must be 0 in Example 1? Factoring the
equation as x1x + 52 = 24 does not help us find the solutions, since 24 can be
factored in infinitely many ways, such as 6 # 4, 12 # 48, A- 25 B # 1- 602 , and so on.
example 2
Graphing a Quadratic Function
Let f 1x2 = x 2 + 5x - 24.
(a) Find the x-intercepts of the graph of f.
(b) Sketch the graph of f, and label the x- and y-intercepts and the vertex.
Solution
x- and y-intercepts are reviewed
in Algebra Toolkit D.2, page T71.
(a) To find the x-intercepts, we solve the equation
x 2 + 5x - 24 = 0
The equation was solved in Example 1. The x-intercepts are 3 and - 8.
(b) To find the y-intercept, we set x equal to 0:
f 102 = 0 + 5 # 0 - 24 = - 24
So the y-intercept is - 24.
The function f is a quadratic function where a is 1 and b is 5. So the
x-coordinate of the vertex occurs at
x =-
b
2a
5
=-
=-
2#1
5
2
Formula
Replace a by 1 and b by 5
Calculate
450
CHAPTER 5
Quadratic Functions and Models
■
The y-coordinate of the vertex is
y = f a-
b
b
2a
Formula
5
= f a- b
2
a is 1 and b is 5
5 2
5
= a - b + 5 a - b - 24
2
2
Definition of f
=-
121
4
Calculate
The graph is a parabola with vertex at A- 52, - 121
4 B . The parabola opens upward
because a 7 0. The graph is shown in Figure 1.
y
x-intercepts _8, 3
10
0
x
2
y-intercept _24
(
Vertex _ 52 , _ 121
4
)
f i g u r e 1 Graph of f 1x2 = x 2 + 5x - 24
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NOW TRY EXERCISE 53
■
e x a m p l e 3 Finding a Quadratic Function from a Graph
The graph of a quadratic function f is shown in Figure 2. Find an equation that represents the function f.
y
6
_3
0
Solution
1
x
We observe from the graph that the x-intercepts are - 3 and 1. So we can express the
function f in factored form:
f 1x2 = a1x - 12 1x + 32
figure 2
We need to find a. From the graph we see that the y-intercept is 6, so f 102 = 6. We have
f 10 2 = a10 - 12 10 + 32
Replace x by 0
6 = a10 - 12 10 + 32
y-intercept is 6
6 = - 3a
Simplify
a = -2
Divide by - 3 and switch sides
It follows that f 1x2 = - 21x - 12 1x + 32 .
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NOW TRY EXERCISE 17
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SECTION 5.4
2
■
Quadratic Equations: Getting Information from a Model
451
■ Solving Quadratic Equations: The Quadratic Formula
If a quadratic equation does not factor readily, we can solve it by using the quadratic formula. This formula gives the solutions of the general quadratic equation
ax 2 + bx + c = 0 and is derived by using the technique of completing the square
(see Algebra Toolkit C.3 , page T62).
The Quadratic Formula
The solutions of the quadratic equation ax 2 + bx + c = 0, where a
0, are
- b ; 2b - 4ac
2a
2
x=
e x a m p l e 4 Using the Quadratic Formula
Find all solutions of each equation.
(a) 3x 2 - 5x - 1 = 0
(b) 4x 2 + 12x + 9 = 0
(c) x 2 + 2x + 2 = 0
Solution
(a) In this quadratic equation a is 3, b is - 5, and c is - 1.
b = -5
3x 2 - 5x - 1 = 0
a =3
c = -1
By the Quadratic Formula,
x=
- 1- 52 ; 21- 52 2 - 4132 1- 12
2132
=
5 ; 237
6
If approximations are desired, we can use a calculator to obtain
x =
5 + 237
L 1.8471
6
and
x=
5 - 237
- L - 0.1805
6
(b) Using the Quadratic Formula where a is 4, b is 12, and c is 9 gives
x =
- 12 ; 21122 2 - 4 # 4 # 9
2#4
=
- 12 ; 0
3
=8
2
This equation has only one solution, x = - 32.
(c) Using the Quadratic Formula where a is 1, b is 2, and c is 2 gives
x =
- 2 ; 22 2 - 4 # 2 - 2 ; 2- 4 - 2 ; 22- 1
=
=
= - 1 ; 2- 1
2
2
2
Since the square of any real number is nonnegative, 1- 1 is undefined in the
real number system. The equation has no real solution.
■
NOW TRY EXERCISES 27, 31, AND 33
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452
2
CHAPTER 5
■
Quadratic Functions and Models
■ The Discriminant
The quantity b 2 - 4ac that appears under the square root sign in the Quadratic
Formula is called the discriminant of the equation ax 2 + bx + c = 0 and is given
the symbol D. If D is negative, then 2b 2 - 4ac is undefined, so the quadratic equation has no real solution. If D is zero, then the equation has exactly one real solution.
If D is positive, then the equation has two distinct real solutions. The following box
summarizes these observations.
The Discriminant
The discriminant of the quadratic equation ax 2 + bx + c = 0, where a
is D = b 2 - 4ac.
0,
1. If D 6 0, then the equation has no real solutions.
2. If D = 0, then the equation has exactly one real solution.
3. If D 7 0, then the equation has two distinct real solutions.
We will see in the next example that we can use the discriminant to determine
whether the graph of a quadratic function has two x-intercepts, one x-intercept, or no
x-intercepts.
e x a m p l e 5 Using the Discriminant to Find the Number of x-Intercepts
A quadratic function f 1x2 = ax 2 + bx + c is given. Find the discriminant of the
equation ax 2 + bx + c = 0, and use it to determine the number of x-intercepts of the
graph of f. Sketch a graph of f to confirm your answer.
(a) f 1x2 = x 2 - 6x + 8
(b) f 1x2 = x 2 - 6x + 9
(c) f 1x2 = x 2 - 6x + 10
Solution
(a) The equation is x 2 - 6x + 8 = 0, so a is 1, b is - 6, and c is 8. The discriminant is
D = b 2 - 4ac = 1- 62 2 - 4 # 1 # 8 = 32 7 0
Since the discriminant is positive, there are two distinct solutions, and hence
there are two x-intercepts for the graph of f. The graph of f in Figure 3(a)
confirms this.
(b) The equation is x 2 - 6x + 9 = 0, so a is 1, b is - 6, and c is 9. The discriminant is
D = b 2 - 4ac = 1- 62 2 - 4 # 1 # 9 = 0
Since the discriminant is zero, there is exactly one solution and hence one
x-intercept for the graph of f. The graph in Figure 3(b) confirms this.
SECTION 5.4
■
Quadratic Equations: Getting Information from a Model
453
(c) The equation is x 2 - 6x + 10 = 0, so a is 1, b is - 6, and c is 10. The
discriminant is
D = b 2 - 4ac = 1- 62 2 - 4 # 1 # 10 = 36 - 40 = - 4 6 0
Since the discriminant is negative, there are no solutions, and hence there are
no x-intercepts for the graph of f. The graph in Figure 3(c) confirms this.
3
3
6
0
_1.5
3
6
0
_1.5
(a) f(x)=x™-6x+8
6
0
_1.5
(b) f(x)=x™-6x+9
(c) f(x)=x™-6x+10
figure 3
■
2
NOW TRY EXERCISES 43, 45, AND 47
■
■ Modeling with Quadratic Functions
We now study real-world phenomena that are modeled by quadratic functions, and
we solve quadratic equations to get information from the model.
e x a m p l e 6 Dimensions of a Lot
A rectangular building lot is 8 feet longer than it is wide.
(a) Find a function that models the area of the lot for any width.
(b) If the lot has area of 2900 square feet, find the dimensions of the lot.
Solution
(a) We want a function A that models the area of the lot in terms of the width. Let
w = width of lot
We translate the information given in the problem into the language of algebra
(see Figure 4).
w
w+8
figure 4
In Words
In Algebra
Width of lot
Length of lot
w
w +8
Now we set up the model:
area of lot = width of lot * length of lot
A1w2 = w1w + 82
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CHAPTER 5
■
Quadratic Functions and Models
(b) Suppose the lot has an area of 2900 square feet. Then
A1w 2 = w1w + 82
Model
2900 = w1w + 82
Area of lot is 2900
2
2900 = w + 8w
0 = w 2 + 8w - 2900
Subtract 2900
0 = 1w - 502 1w + 582
Factor
w = - 58
Zero-Product Property
w = 50
or
Expand
Since the width of a lot must be a positive number, we conclude that the width
is 50 feet. The length of the lot is w + 8 = 50 + 8 = 58 feet.
■
■
NOW TRY EXERCISE 65
e x a m p l e 7 Ticket Sales
A hockey team plays in an arena that has a seating capacity of 15,000 spectators.
With the ticket price set at $14, average attendance at recent games has been 9500.
A market survey indicates that for each dollar the ticket price is lowered, the average attendance increases by 1000.
(a) Find a function that models the revenue in terms of ticket price.
(b) At what ticket price is the revenue $100,000?
(c) What ticket price is so high that no one attends and so no revenue is generated?
Solution
(a) In Example 5 of Section 5.3 we found the following function that models the
revenue R in terms of the ticket price x:
R1x2 = - 1000x 2 + 23,500x
(b) We want to find the ticket price for which R1x2 = 100,000.
R1x2 = - 1000x 2 + 23,500x
100,000 = - 1000x 2 + 23,500x
2
100 = - x + 23.5x
150,000
y=100,000
Model
Revenue is 100,000
x 2 - 23.5x + 100 = 0
Divide by 1000
Subtract 100
The expression on the left-hand side of this equation does not factor easily, so
we use the Quadratic Formula where a is - 1, b is 23.5, and c is - 100.
x =
_2
_30,000 x=5.58
26
=
x=17.92
A graph of the revenue function R
shows that revenue is $100,000
when the ticket price is about $5.58
and $17.92.
- 23.5 ; 21- 23.52 2 - 41- 12 1- 1002
21- 12
- 23.5 ; 2152.25
-2
Using a calculator, we find
x L 17.92
or
x L 5.58
So the revenue is $100,000 when the ticket price is $17.92 or $5.58.
SECTION 5.4
■
Quadratic Equations: Getting Information from a Model
455
(c) We want to find the ticket price for which R1x2 = 0.
150,000
R1x2 = - 1000x 2 + 23,500x
Model
2
26
_2
0 = - 1000x + 23,500x
Revenue is 0
0 = - x 2 + 23.5x
Divide by 1000
_30,000
A graph of the revenue function R
shows that R1x2 = 0 when the ticket
price x is 0 or $23.50 (the
x -intercepts of the graph).
0 = x1- x + 23.52
Factor
x =0
Zero-Product Property
x = 23.5
or
According to this model, a ticket price of $23.50 is just too high; at that price,
no one attends the hockey game. (Of course the revenue is also zero if the
ticket price is zero.)
■
NOW TRY EXERCISE 67
■
e x a m p l e 8 Model Rocket
A model rocket is shot straight upward with an initial speed of 800 ft/s, and the
height of the rocket is given by
h = 800t - 16t 2
where h is measured in feet and t in seconds.
(a) How many seconds does it take for the rocket to reach a height of 8000 feet?
(b) How long will it take the rocket to hit the ground?
Solution
y=8000
11,000
(a) We want to find the time t such that h = 8000.
h = 800t - 16t 2
8000 = 800t - 16t 2
10 = t - 0.02t
_2
_1000
52
t=13.8
t=36.2
A graph of the height function h
shows that the rocket reaches a
height of 8000 ft at times of about
13.8 s and 36.2 s.
0.02t 2 - t + 10 = 0
Height is 8000
Divide by 800
Subtract 10 and switch sides
The expression on the left-hand side does not factor easily, so we use the
Quadratic Formula where a is 0.02, b is - 1, and c is 10.
t =
=
11,000
2
Model
1 ; 21- 12 2 - 410.022 1102
210.022
1 ; 20.20
0.04
Using a calculator, we find
t L 13.8
or
t L 36.2
So the rocket is at a height of 8000 feet after about 13.8 seconds and again
after 36.2 seconds.
(b) At ground level the height is 0, so we must solve the equation
_2
_1000
A graph of the height function h
shows that h1t2 = 0 when t is
about 0 or 50 seconds (the
t -intercepts of the graph).
52
h = 800t - 16t 2
Model
0 = - 16t 2 + 800t
Height is 0
0 = 16t1- t + 50 2
Factor
t =0
Zero-Product Property
or
t = 50
456
CHAPTER 5
■
Quadratic Functions and Models
According to this model, the rocket hits the ground after 50 seconds. (Of
course the rocket is also at ground level at time 0.)
■
NOW TRY EXERCISE 63
■
Notice that the model in Example 8 is valid only for 0 … t … 50. For values of
t outside this interval the height h would be negative (that is, the rocket would be
traveling below ground level).
Check your knowledge of factoring quadratic expressions by doing the following
problems. You can review this topic in Algebra Toolkit B.2 on page T33.
1. Factor each expression by factoring out common factors.
(a) x 2 + x
(c) 21x + 12 2 + 1x + 12
(b) 4t 2 - 6t
(d) 12r + 32 13r + 42 - 513r + 4 2
2. Factor the given expression using the Difference of Squares Formula.
(a) x 2 - 36
(c) 1u + 12 2 - 36
(b) 4t 2 - 16
(d) 41w - 22 2 - 49
3. Factor the perfect square.
(a) x 2 + 8x + 16
(c) 9r 2 - 24r + 16
(b) t 2 - 12t + 36
(d) 1u + 22 2 + 101u + 22 + 25
4. Factor each expression by trial and error.
(a) x 2 + 5x + 6
(c) 2s 2 - s - 3
(b) t 2 - t - 12
(d) 6u 2 - 7u - 3
5.4 Exercises
CONCEPTS
Fundamentals
1. The Quadratic Formula gives us the solutions of the equation ax 2 + bx + c = 0.
______.
(a) State the Quadratic Formula: x = ________
(b) In the equation 12 x 2 - x - 4 = 0, a = _______, b = _______, and
c = _______. So the solution of the equation is x = ______.
2. To solve the quadratic equation x 2 - 4x - 5 = 0, we can do the following.
(a) Factor the equation as 1x - ____ 2 # 1x + ____2 = 0, so the solutions are _______
and _______.
______, so the solutions are
(b) Use the Quadratic Formula to get x = ________
_______ and _______.
3. Let f 1x2 = ax 2 + bx + c, and let D = b 2 - 4ac.
(a) If D 7 0, then the number of x-intercepts for the graph of f is _______.
(b) If D = 0, then the number of x-intercepts for the graph of f is _______.
(c) If D 6 0, then the number of x-intercepts for the graph of f is _______.
4. The graph of a quadratic function f 1x 2 = ax 2 + bx + c is shown at the top of the next page.
(a) The x-intercepts are _______.
SECTION 5.4
Quadratic Equations: Getting Information from a Model
457
(b) The discriminant of the equation ax 2 + bx + c = 0 is _______ (positive, 0, or
negative).
y
(c) The solution(s) to the equation ax 2 + bx + c = 0 is (are) _______.
1
0
■
1
3
5
x
Think About It
5. Consider the quadratic function y = 1x - 22 1x - 42 .
(a) The graph is a parabola that opens _______.
(b) The x-intercepts are _______ and _______.
(c) From the location of the vertex between the x-intercepts, we see that the
_4
x-coordinate of the vertex is _____________.
Graph for Exercise 4
6. Consider the quadratic function y = 1x - m2 1x - n2 .
(a) The graph is a parabola that opens _____________.
(b) The x-intercepts are _______ and _______.
(c) From the location of the vertex between the x-intercepts, we see that the
x-coordinate of the vertex is _______.
(d) Using the formula we found in part (c), we find that the x-coordinate of the vertex
of y = 1x - 32 1x - 52 is _______.
SKILLS
■
7–16
Solve the equation by factoring.
2
8. x 2 + 3x = 4
7. x + x = 12
9. t 2 - 7t + 12 = 0
10. t 2 + 8t + 12 = 0
11. 3s 2 - 5s - 2 = 0
12. 4x 2 - 4x - 15 = 0
13. 2y 2 + 7y + 3 = 0
14. 4w 2 = 4w + 3
15. 6x 2 + 5x = 4
16. 3x 2 + 1 = 4x
17–20 ■ A graph of a quadratic function f is given.
(a) Find the x-intercepts.
(b) Find an equation that represents the function f (as in Example 3).
17.
y
15
18.
y
6
2
_1 0
5
0
3
x
5
y
19.
x
3
y
20.
8
12
4
_3
0
1
x
_2
0
1 x
458
CHAPTER 5
■
Quadratic Functions and Models
21–26
■
Solve the equation by both factoring and using the Quadratic Formula.
2
21. x - 2x - 15 = 0
22. x 2 + 5x - 6 = 0
23. x 2 - 7x + 10 = 0
24. x 2 + 30x + 200 = 0
25. 2x 2 + x - 3 = 0
26. 3x 2 + 7x + 4 = 0
27–38
■
Find all real solutions of the equation.
2
27. t + 3t + 1 = 0
28. 2t 2 - 8t + 4 = 0
29. y 2 + 12y - 27 = 0
30. 8y 2 - 6y - 9 = 0
31. s 2 - 32s +
32. s 2 - 6s + 1 = 0
9
16
=0
33. w 2 = 31w - 12
34. 2 + 2z + 3z 2 = 0
35. x 2 - 15x + 1 = 0
36. 5x 2 - 7x + 5 = 0
37. 10y 2 - 16y + 5 = 0
38. 25y 2 + 70y + 49 = 0
39–42
■
Solve the quadratic equation algebraically and graphically, correct to three
decimal places.
39. x 2 - 0.011x - 0.064 = 0
40. x 2 - 2.450x + 1.500 = 0
41. x 2 - 2.450x + 1.501 = 0
42. x 2 - 1.800x + 0.810 = 0
43–48 ■ A quadratic function f 1x2 = ax 2 + bx + c is given.
(a) Find the discriminant of the equation ax 2 + bx + c = 0. How many real solutions
does this equation have?
(b) Use the answer to part (a) to determine the number of x-intercepts for the graph of
the function f 1x2 = ax 2 + bx + c, and then graph the function to confirm your
answer.
43. f 1x2 = x 2 - 6x + 1
44. f 1x2 = x 2 - 6x + 9
47. f 1x2 = 32x 2 + 40x + 13
48. f 1x2 = 9x 2 - 4x +
46. f 1x2 = 0.1x 2 - 0.38x + 0.361
45. f 1x2 = x 2 + 2.20x + 1.21
4
9
49–52 ■ A graph of a quadratic function f 1x 2 = ax 2 + bx + c is shown.
(a) Find the x-intercept(s), if there are any.
(b) Is the discriminant D = b 2 - 4ac positive, negative, or 0?
(c) Find the solution(s) to the equation ax 2 + bx + c = 0.
(d) Find an equation that represents the function f.
49.
y
2
y
50.
3
_1 0
1
x
(_1, 1)
_1
1
0
1 x
SECTION 5.4
51.
■
Quadratic Equations: Getting Information from a Model
y
52.
459
y
1
0
2
6
x
3
_3
1
0
1
3
x
53–60 ■ A quadratic function f 1x2 = ax 2 + bx + c is given.
(a) Find the x-intercepts of the graph of f.
(b) Sketch the graph of f and label the x- and y-intercepts and the vertex.
53. f 1x2 = x 2 + 2x - 1
54. f 1x2 = x 2 - 8x + 8
57. f 1x2 = - x 2 - 3x + 3
58. f 1x2 = 1 - x - x 2
56. f 1x2 = 5x 2 + 30x + 4
55. f 1x2 = 3x 2 - 6x + 1
59. f 1x2 = 3x 2 - 12x + 13
60. f 1x2 = 2x 2 + 8x + 11
61. Height of a Ball If a ball is thrown directly upward with a velocity of 12 m/s, its
height (in meters) after t seconds is modeled by f 1t 2 = 12t - 4.9t 2.
(a) When does the ball reach a height of 5 meters?
(b) Does the ball reach a height of 8 meters?
(c) When does the ball hit the ground?
(d) Identify the points on the graph that correspond to your solutions to parts (a)–(c).
CONTEXTS
y
2
0
x
62. Path of a Ball A ball is thrown across a playing field from a height of 5 ft above the
ground at an angle of 45° to the horizontal at a speed of 20 ft/s. It can be deduced from
physical principles that the path of the ball is modeled by the function
y
10
f 1x 2 = -
5
0
1
5
10
15
20 x
32 2
x +x+5
1202 2
where x is the distance in feet that the ball has traveled horizontally.
(a) At what horizontal distance x is the ball 7 ft high?
(b) Does the ball reach a height of 10 ft?
(c) At what horizontal distance x does the ball hit the ground?
(d) Identify the points on the graph in the margin that correspond to your solutions to
parts (a)–(c).