617
Quadratic dug22241_ch10a.qxd
11/10/2004
18:29
Page Equations
and Inequalities
Is it possible to measure beauty? For thousands of years artists and philosophers have been challenged to answer this question. The seventeenth-century philosopher John Locke said, Beauty consists of a certain composition of color and gure causing delight in the beholder. Over the centuries many architects, sculptors, and painters have searched for beauty in their work by exploring numerical patterns in various art forms. Today many artists and architects still use the concepts of beauty given to us by the ancient Greeks. One principle, called the Golden Rectangle, concerns the most pleasing proportions of a rectangle. The Golden Rectangle appears in nature as well as in many cultures. Examples of it can be seen in Leonardo da Vincis Proportions of the Human Figure as well as in Indonesian temples and Chinese pagodas. Perhaps one of the best-known examples of the Golden Rectangle is in the faade and
W W L W L W W
10
10.1
Factoring and Completing the Square
10.2 The Quadratic Formula 10.3 Graphing Parabolas 10.4
More on Quadratic Equations Quadratic and Rational Inequalities
10.5
Chapter
oor plan of the Parthenon, built in Athens in the fth century B.C.
In Exercise 91 of Section 10.4 we will see that the principle of the Golden Rectangle is based on a proportion that we can solve using the quadratic formula.
dug22241_ch10a.qxd
11/10/2004
18:29
Page 618
618
Chapter 10 Quadratic Equations and Inequalities
10-2
10.1
In this Section
Review of Factoring Review of the Even-Root Property Completing the Square Miscellaneous Equations Imaginary Solutions
Factoring and Completing the Square
Factoring and the even-root property were used to solve quadratic equations in Chapters 5, 6, and 9. In this section we rst review those methods. Then you will learn the method of completing the square, which can be used to solve any quadratic equation.
Review of Factoring
A quadratic equation has the form ax2 bx c 0, where a, b, and c are real numbers with a 0. In Section 5.6 we solved quadratic equations by factoring and then applying the zero factor property. Zero Factor Property The equation ab 0 is equivalent to the compound equation a 0 or b 0.
Of course we can only use the factoring method when we can factor the quadratic polynomial. To solve a quadratic equation by factoring we use the following strategy.
Strategy for Solving Quadratic Equations by Factoring
1. 2. 3. 4. 5.
Write the equation with 0 on one side. Factor the other side. Use the zero factor property to set each factor equal to zero. Solve the simpler equations. Check the answers in the original equation.
EXAMPLE
1
Solving a quadratic equation by factoring Solve 3x 2 4x 15 by factoring.
Solution Helpful Hint
After you have factored the quadratic polynomial, use FOIL to check that you have factored correctly before proceeding to the next step.
Subtract 15 from each side to get 0 on the right-hand side: 3x 2 4x 15 (3x 5)(x 3) 0 or x3 5 or x 5 3
5 , 3
3x
5 3x x
0 0 0 3
Factor the left-hand side. Zero factor property
The solution set is
3 . Check the solutions in the original equation. Now do Exercises 514
dug22241_ch10a.qxd
11/10/2004
18:29
Page 619
10-3
10.1 Factoring and Completing the Square
619
Review of the Even-Root Property
In Chapter 9 we solved some simple quadratic equations by using the even-root property, which we restate as follows: Even-Root Property Suppose n is a positive even integer. If k If k If k 0, then x n 0, then x n 0, then x n k is equivalent to x k is equivalent to x 0. k has no real solution.
n
k.
By the even-root property x2 4 is equivalent to x 4 has no real solutions. x 0, and x2
2, x2
0 is equivalent to
EXAMPLE
2
Solving a quadratic equation by the even-root property Solve (a 1)2 9.
Solution
By the even-root property x 2 k is equivalent to x (a a 1 a 3 4 or or 1)2 a1 a1 a 9 9 3 2 2, 4 .
Even-root property
k.
Check these solutions in the original equation. The solution set is Now do Exercises 1524
Helpful Hint
The area of an x by x square and two x by 3 rectangles is x2 6x. The area needed to complete the square in this gure is 9:
3 93
Completing the Square
We cannot solve every quadratic by factoring because not all quadratic polynomials can be factored. However, we can write any quadratic equation in the form of Example 2 and then apply the even-root property to solve it. This method is called completing the square. The essential part of completing the square is to recognize a perfect square trinomial when given its rst two terms. For example, if we are given x2 6x, how do we recognize that these are the rst two terms of the perfect square trinomial x2 6x 9? To answer this question, recall that x2 6x 9 is a perfect square trinomial because it is the square of the binomial x 3: (x 3)2 x2 2 3x 32 x2 6x 9
3
3x
x
x2
3x
x
3
Notice that the 6 comes from multiplying 3 by 2 and the 9 comes from squaring the 3. So to nd the missing 9 in x 2 6x, divide 6 by 2 to get 3, then square 3 to get 9. This procedure can be used to nd the last term in any perfect square trinomial in which the coefcient of x 2 is 1.
dug22241_ch10a.qxd
11/10/2004
18:29
Page 620
620
Chapter 10 Quadratic Equations and Inequalities
10-4
Rule for Finding the Last Term The last term of a perfect square trinomial is the square of one-half of the coefcient of the middle term. In symbols, the perfect square trinomial whose rst two terms are x 2 bx is x 2 bx
b 2
2
.
EXAMPLE
3
Finding the last term Find the perfect square trinomial whose rst two terms are given. 4 b) x 2 5x c) x 2 x a) x 2 8x 7
d) x 2
3 x 2
Solution
a) One-half of 8 is 4, and 4 squared is 16. So the perfect square trinomial is x2 b) One-half of 5 is
5 , and 2 5 2
8x
16.
4
squared is 25. So the perfect square trinomial is 5x
4 , 49
x2 c) Since 1
2 4 7 2 7
25 . 4
and 2 squared is
7
the perfect square trinomial is 4 . 49
x2 d) Since 1
2
3 2
4 x 7
9 , 16
3 4
and
3 4
2
the perfect square trinomial is 9 . 16 Now do Exercises 2532
x2
3 x 2
Another essential step in completing the square is to write the perfect square trinomial as the square of a binomial. Recall that a2 and a2 2ab b2 (a b)2. 2ab b2 (a b)2
EXAMPLE
4
Factoring perfect square trinomials Factor each trinomial.
a) x 2 c) z 2 12x 36 4 4 z 3 9 b) y 2 7y 49 4
dug22241_ch10a.qxd
11/10/2004
18:29
Page 621
10-5
10.1 Factoring and Completing the Square
621
Helpful Hint
To square a binomial use the following rule (not FOIL): Square the rst term. Add twice the product of the terms. Add the square of the last term.
Solution
a) The trinomial x 2 b 6. So 12x x2 36 is of the form a 2 12x 36 (x 2ab 6)2. 2ab 72 . 2 2ab 22 . 3 b2 with a z and b 2 with a y and b2 with a x and
Check by squaring x 6. b) The trinomial y 2 7y 49 is of the form a 2 4 b 7. So
2
y2 Check by squaring y c) The trinomial z 2 b
2 . 3 4 3z 7 2. 4 9
7y
49 4
y
is of the form a 2 4 z 3 4 9 z
So z2
Now do Exercises 3340 In Example 5 we use the skills that we practiced in Examples 2, 3, and 4 to solve the quadratic equation ax2 bx c 0 with a 1 by the method of completing the square.
EXAMPLE
5
Completing the square with a 1 Solve x 2 6x 5 0 by completing the square.
Calculator Close-Up
The solutions to x
2
Solution
The perfect square trinomial whose rst two terms are x 2 6x is x
2
6x
5
0
6x
9.
correspond to the x-intercepts for the graph of y x
2
6x
5.
So we move 5 to the right-hand side of the equation, then add 9 to each side to create a perfect square on the left side: x2 x2 6x 6x (x x 3 x 2 1 or or 9 4 4 2 5 5 5
Subtract 5 from each side.
So we can check our solutions by graphing and using the TRACE feature as shown here.
6
9
Add 9 to each side to get a perfect square trinomial. Factor the left-hand side. Even-root property
8
2
3)2 x3 x3 x
Check in the original equation:
6
( 1)2 and ( 5)2 The solution set is 1, 5.
6( 1) 6( 5)
5 5
0 0
Now do Exercises 4148
dug22241_ch10a.qxd
11/10/2004
18:30
Page 622
622
Chapter 10 Quadratic Equations and Inequalities
10-6
CAUTION All of the perfect square trinomials that we have used so far had a leading
coefcient of 1. If a 1, then we must divide each side of the equation by a to get an equation with a leading coefcient of 1. The strategy for solving a quadratic equation by completing the square is stated in the following box.
Study Tip
Most instructors believe that what they do in class is important. If you miss class, then you miss what is important to your instructor and what is most likely to appear on the test.
Strategy for Solving Quadratic Equations by Completing the Square
1. If a 1, then divide each side of the equation by a. 2. Get only the x2 and the x terms on the left-hand side. 3. Add to each side the square of 1 the coefcient of x.
2
4. 5. 6. 7.
Factor the left-hand side as the square of a binomial. Apply the even-root property. Solve for x. Simplify.
EXAMPLE
6
Completing the square with a 1 Solve 2x 2 3x 2 0 by completing the square.
Solution
For completing the square, the coefcient of x 2 must be 1. So we rst divide each side of the equation by 2: 2x2 x2 x2 x2 3x 2 2 3 x1 2 3 x 2 9 3 x 16 2 32 x 4 3 x 4 3 x 4 x 0 2 0 1 1 25 16 25 16 5 4 8 4 9 16
Divide each side by 2. Simplify. Get only x2 and x terms on the left-hand side. One-half of 3 is 3, and
2 4 32 4 9 . 16
Calculator Close-Up
Note that the x-intercepts for the graph of the function y 2x2
1 ,0 2
Factor the left-hand side. Even-root property
3x :
6
2
are ( 2, 0) and
x
3 4 x
4
2
5 4 2 4
or 1 2 or
2 2, 1 .
2
Check these values in the original equation. The solution set is
6
Now do Exercises 4950
dug22241_ch10a.qxd
11/10/2004
18:30
Page 623
Math at Work
Financial Matters
In the United States over 1 million new homes are sold annually, with a median price of about $200,000. Over 17 million new cars are sold each year with a median price over $20,000. Americans are constantly saving and borrowing. Nearly everyone will need to know a monthly payment or what their savings will total over time. The answers to these questions are in the following table.
What $P Left at Compound Interest Will Grow to P(1 i)nt
What $R Deposited Periodically Will Grow to R (1 i)nt i 1
Periodic Payment That Will Pay off a Loan of $P P i 1 (1 i)
nt
In each case n is the number of periods per year, r is the annual percentage rate (APR), t is the number of years, and i is the interest rate per period i r . For periodic payments or n deposits these expressions apply only if the compounding period equals the payment period. So lets see what these expressions do. A person inherits $10,000 and lets it grow at 4% APR compounded daily for 20 years. Use the rst expression with n 365, i
0.04 , and t 365
20 to get 10,000 1
0.04 365
365 20
or
$22,254.43, which is the amount after 20 years. More often, people save money with periodic deposits. Suppose you deposit $100 per month at 4% compounded monthly for 20 years. Use the second expression with R 100, i
0.04 ,n 12
12, and t
20 to get 100 (1
0.04 12)12 20 0.04 12
1
or $36,677.46, which is the amount
after 20 years. Suppose that you get a 20-year $200,000 mortgage at 7% APR compounded monthly to buy an average house. Try using the third expression to calculate the monthly payment of $1550.60. See the accompanying gure.
20-year $200,000 mortgage Monthly payment ($) 2000 1500 1000 500 0 2 4 6 8 APR (percent) 10
In Examples 5 and 6 the solutions were rational numbers, and the equations could have been solved by factoring. In Example 7 the solutions are irrational numbers, and factoring will not work.
dug22241_ch10a.qxd
11/10/2004
18:30
Page 624
624
Chapter 10 Quadratic Equations and Inequalities
10-8
EXAMPLE
7
A quadratic equation with irrational solutions Solve x 2 3x 6 0 by completing the square.
Solution
Because a 1, we rst get the x 2 and x terms on the left-hand side: x2 x2 x2 3x 3x 3x x x 3 2 6 9 4
2
0 6 6 33 4 33 4 3 2 3 2
33 3 , 33 2
Add 6 to each side.
9 4
One-half of
3 is
3 , 2
and
3 2
2
9 . 4
6
9 4
24 4
9 4
33 4
3 2 x x
Even-root property
33 2 33
Add 3 to each side.
2
The solution set is
3 2
. Now do Exercises 5158
Miscellaneous Equations
Examples 8 and 9 show equations that are not originally in the form of quadratic equations. However, after simplifying these equations, we get quadratic equations. Even though completing the square can be used on any quadratic equation, factoring and the square root property are usually easier and we can use them when applicable. In Examples 8 and 9 we will use the most appropriate method.
EXAMPLE
8
An equation containing a radical Solve x 3 153 x.
Solution
Square both sides of the equation to eliminate the radical: 3 (x 3)2 x 2 6x 9 x 2 7x 144 (x 9)(x 16) or x 16 or x x 153 x ( 153 x)2 153 x 0 0 0 16
The original equation Square each side. Simplify.
Factor. Zero factor property
x
9 x
0 9
dug22241_ch10a.qxd
11/10/2004
18:30
Page 625
10-9
10.1 Factoring and Completing the Square
625
Calculator Close-Up
You can provide graphical support for the solution to Example 8 by graphing y1 and y2 153 x. x 3
Because we squared each side of the original equation, we must check for extraneous roots. Let x 9 in the original equation: 9 3 12 Let x 16 in the original equation: 16 3 13 153 169 ( 16)
Incorrect because 169 13
153 144
9
Correct
It appears that the only point of intersection occurs when x 9.
50
150
200
Because
16 is an extraneous root, the solution set is 9 . Now do Exercises 5962
50
EXAMPLE
9
An equation containing rational expressions
Solve 1
x 3 x 2 5 . 8
Solution
The least common denominator (LCD) for x, x 1 x 8x (x 2) 1 x 8x (x 8x 2) 16 32x 5x 2 5x 2 (5x 5x 2 x 0 2 5 or or 42x 42x 2)(x x 3 x 3 x 2 24x 16 16 16 8) 8 x 2 5 8 8x(x 5x 2 5x 0 0 0 0 8
5 Factor. Multiply each side by for easier factoring. 1
2
2, and 8 is 8x(x
2).
2)
5 8
Multiply each side by the LCD.
10x 10x
Check these values in the original equation. The solution set is 2, 8 . Now do Exercises 6366
Imaginary Solutions
In Chapter 9 we found imaginary solutions to quadratic equations using the even-root property. We can get imaginary solutions also by completing the square.
dug22241_ch10a.qxd
11/10/2004
18:30
Page 626
626
Chapter 10 Quadratic Equations and Inequalities
10-10
EXAMPLE
10
An equation with imaginary solutions Find the complex solutions to x 2 4x 12
0.
Solution
Because the quadratic polynomial cannot be factored, we solve the equation by completing the square.
Calculator Close-Up
The answer key (ANS) can be used to check imaginary answers as shown here.
x2 x2 x2
4x 4x 4x (x
12 4 2)2 x2 x
0 12 12 8 2 2 4
The original equation Subtract 12 from each side. One-half of 4 is 2, and ( 2)2 4.
8 Even-root property i8 2i 2 2i 2 .
Check these values in the original equation. The solution set is 2 Now do Exercises 6776
Warm-Ups
True or false? Explain your answer.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Completing the square means drawing the fourth side. The equation (x 3)2 12 is equivalent to x 3 2 3. Every quadratic equation can be solved by factoring. 4 16 The trinomial x 2 3 x 9 is a perfect square trinomial. Every quadratic equation can be solved by completing the square. To complete the square for 2x 2 6x 4, add 9 to each side. 3 5 (2x 3)(3x 5) 0 is equivalent to x 2 or x 3. 9 In completing the square for x 2 3x 4, add 4 to each side. 8 is equivalent to x 2 2. The equation x 2 All quadratic equations have two distinct complex solutions.
10.1
Exercises
Reading and Writing After reading this section, write out
the answers to these questions. Use complete sentences.
Net Tutor e-Professors
Boost your GRADE at mathzone.com!
Practice Problems Self-Tests Videos
1. What are the three methods discussed in this section for solving a quadratic equation?
dug22241_ch10a.qxd
11/10/2004
18:30
Page 627
10-11
2. Which quadratic equations can be solved by the even-root property?
10.1 Factoring and Completing the Square
627
29. y 2
1 y 4
30. z 2
3 z 2
3. How do you nd the last term for a perfect square trinomial when completing the square?
31. x 2
2 x 3
32. p 2
6 p 5
4. How do you complete the square when the leading coefcient is not 1? Factor each perfect square trinomial. See Example 4. 33. x 2 Solve by factoring. See Example 1. 5. x 2 7. a 2 9. 2x 2 x 2a x 6 15 3 0 0 6. x 2 8. w 2 10. 6x 2 6x 2w x 8 15 15 0 37. z 2 11. y 2 13. a 2 14y 16 49 0 0 12. a 2 14. 4w 2 6a 25 9 0 0 39. t 2 3 t 5 9 100 40. h2 3 h 2 9 16 4 z 7 4 49 6 m 5 9 25 0 35. y 2 5y 25 4 36. w 2 w 1 4 8x 16 34. x 2 10x 25
38. m2
Use the even-root property to solve each equation. See Example 2. 9 16. x 2 15. x 2 81 4 16 2 2 17. x 18. a 32 9
Solve by completing the square. See Examples 57. Use your calculator to check. 41. x 2 42. x
2 2
2x 6x 4x 6x w 3y 5q z h m 4x 6x 8x 10x 4x 2
15 7 70 24 20 10 14 0
0
43. 2x 19. (x 21. (z 3)2 1)2 3 2
2
16 5 7 4
20. (x 22. (a
5)2 2)2 2 3
2
44. 3x 2 4 8 5 9 45. w
2
0 0
46. y 2 47. q 2 48. z 2 49. 2h 2 50. 2m 2
23. w
24. w
3 15 6 8 4 3 1
0 0
Find the perfect square trinomial whose rst two terms are given. See Example 3. 25. x 2 27. x 2 2x 3x 26. m 2 28. w 2 14m 5w
51. x 2 52. x 54. x
2
0 0 0 0
53. x 2
2
55. 4x2
dug22241_ch10a.qxd
11/10/2004
18:30
Page 628
628
Chapter 10 Quadratic Equations and Inequalities
10-12
56. 4x2 57. 2x 2 58. 2x
2
4x
2
0
79. 4x 2 80. 5w 2 81. p y
25 3 1 2 2 3
2
0 0 9 4 4 9 3 1 24 7 9 4 6 12 10 17 7x 0 0 0 0 7 3 x 1 1 4 1 2 0 0 0 0
3x
4
0
5x
1
0 82.
2
Solve each equation by an appropriate method. See Examples 8 and 9. 59. 2x 1 x 1 60. 2x 4 x y1 2 14
83. 5t 2 84. 3v2
4t 4v 2m 6q 2)2 1)2 x x 6x 8x 5
61. w
w1 2
62. y
1
85. m 2 86. q
2
87. (x 63. t t 2 x2 2 4 x 3 x 2t t 3 64. z z 3 5z 3z 1 88. (2x 89. 90. 1 1 0 0 x2 x
2
65.
91. x 2 92. x 2 93. 2x 94. 95. 1 x 1 x 7x
1 66. 2 x
29 1 x 2
Use completing the square to nd the imaginary solutions to each equation. See Example 10. 67. x 2 69. x 2 2x 6x 1 2 5 11 0 0 68. x 2 70. x 2 4x 8x 1 8 5 19 0
96. 0
1
x
71. x2
72. x2
If the solution to an equation is imaginary or irrational, it takes a bit more effort to check. Replace x by each given number to verify each statement. 97. 98. 99. 100. 0 Both 2 Both 1 Both 1 Both 2 3 and 2 3 satisfy x 2 4x 1 0. 2 satisfy x 2 2x 1 0. 2 and 1 i and 1 i satisfy x 2 2x 2 0. 3i and 2 3i satisfy x 2 4x 13 0.
73. x2 75. 5z2
12 4z
0 1 0
74.
3x2
21 3w
0 2
76. 2w2
Solve each problem. 101. Approach speed. The formula 1211.1L CA2S is used to determine the approach speed for landing an aircraft, where L is the gross weight of the aircraft in pounds, C is the coefcient of lift, S is the surface area of the wings in square feet (ft2), and A is approach speed in feet per second. Find A for the Piper Cheyenne, which has a gross weight of 8700 lb, a coefcient of lift of 2.81, and wing surface area of 200 ft2.
Find all real or imaginary solutions to each equation. Use the method of your choice. 77. x2 78. w
2
121 225
dug22241_ch10a.qxd
11/10/2004
18:30
Page 629
10-13
102. Time to swing. The period T (time in seconds for one complete cycle) of a simple pendulum is related to the length L 2 (in feet) of the pendulum by the formula 8T 2 L . If a child is on a swing with a 10-foot chain, then how long does it take to complete one cycle of the swing? 103. Time for a swim. Tropical Pools gures that its monthly revenue in dollars on the sale of x aboveground pools is given by R 1500x 3x2, where x is less than 25. What number of pools sold would provide a revenue of $17,568? 104. Pole vaulting. In 1981 Vladimir Poliakov (USSR) set a world record of 19 ft 3 in. for the pole vault 4 (www.polevault.com). To reach that height, Poliakov obtained a speed of approximately 36 feet per second on the runway. The formula h 16t 2 36t gives his height t seconds after leaving the ground. a) Use the formula to nd the exact values of t for which his height was 18 feet. b) Use the accompanying graph to estimate the value of t for which he was at his maximum height. c) Approximately how long was he in the air?
10.2 The Quadratic Formula
629
Getting More Involved
105. Discussion Which of the following equations is not a quadratic equation? Explain your answer. a) x 2 c) 4x 5 106. Exploration Solve x 2 4x k 0 for k 0, 4, 5, and 10. a) When does the equation have only one solution? b) For what values of k are the solutions real? c) For what values of k are the solutions imaginary? 107. Cooperative learning Write a quadratic equation of each of the following types, then trade your equations with those of a classmate. Solve the equations and verify that they are of the required types. a) a single rational solution b) two rational solutions c) two irrational solutions d) two imaginary solutions 108. Exploration In the next section we will solve ax 2 bx c 0 for x by completing the square. Try it now without looking ahead. 5x 0 1 0 b) 3x 2 1 d) 0.009x 2 0 0
25 Height (ft) 20 15 10 5 0 0 1 2 Time (sec)
Graphing Calculator Exercises
For each equation, nd approximate solutions rounded to two decimal places. 109. 110. 111. 112. x 2 7.3x 12.5 0 x 20 1.2x 2 2x 3 20 x x 2 1.3x 22.3 x 2
Figure for Exercise 104
10.2
In this Section
Developing the Formula Using the Formula Number of Solutions Applications
The Quadratic Formula
Completing the square from Section 10.1 can be used to solve any quadratic equation. Here we apply this method to the general quadratic equation to get a formula for the solutions to any quadratic equation.
Developing the Formula
Start with the general form of the quadratic equation, ax 2 bx c 0.
dug22241_ch10a.qxd
11/10/2004
18:30
Page 630
630
Chapter 10 Quadratic Equations and Inequalities
10-14
Assume a is positive for now, and divide each side by a: ax2 x2 bx a b x a x2 One-half of b is
a b , 2a
c c a b x a
0 a 0 c a
b2 : 4a2 Subtract c from each side. a
and
b 2a
squared is x2 b x a
b2 4a2
c a
b2 4a2
Factor the left-hand side and get a common denominator for the right-hand side: x x b 2a b 2a x
2
b2 4a2 b2
4ac 4a2
c(4a) a(4a)
4ac 4a2
2
4ac 4a2 b2 4ac 4a2 b2 4ac 2a b2 2a 4a2 4ac
Even-root property
b 2a x b 2a b
Because a
0,
4a2
2a.
x
We assumed a was positive so that 4a2 2a, and we get x
2a would be correct. If a is negative, then b2
b 2a
4ac . 2a
However, the negative sign can be omitted in 2a because of the symbol preceding it. For example, the results of 5 ( 3) and 5 3 are the same. So when a is negative, we get the same formula as when a is positive. It is called the quadratic formula.
The Quadratic Formula The solution to ax 2 bx
c x
0, with a b b2 2a
0, is given by the formula 4ac .
dug22241_ch10a.qxd
11/10/2004
18:30
Page 631
10-15
10.2 The Quadratic Formula
631
Using the Formula
The quadratic formula solves any quadratic equation. Simply identify a, b, and c and insert those numbers into the formula. Note that if b is positive then b (the opposite of b) is a negative number. If b is negative, then b is a positive number.
EXAMPLE
1
Two rational solutions Solve x 2 2 x 15 0 using the quadratic formula.
Solution
To use the formula, we rst identify the values of a, b, and c: 1x 2
a
2x
b
15
c
0
The coefcient of x 2 is 1, so a 1. The coefcient of 2x is 2, so b 2. The constant term is 15, so c 15. Substitute these values into the quadratic formula: x 2 2 2 2 2 2 2
6
22 4(1)( 15) 2(1) 4 64 8 2 2 8 60
Calculator Close-Up
Note that the two solutions to x2 2x 15 0 correspond to the two x-intercepts for the graph of y x
2
2x
10
15.
8
x Check 3 and
2 2
8
3
or
x
5 5, 3 .
5 in the original equation. The solution set is
20
Now do Exercises 712
CAUTION To identify a, b, and c for the quadratic formula, the equation must be in
the standard form ax 2 bx c must rst rewrite the equation.
0. If it is not in that form, then you
EXAMPLE
2
One rational solution Solve 4x2 12x 9 by using the quadratic formula.
Solution
Rewrite the equation in the form ax 2 4x 2 bx 12 x c 9 0 before identifying a, b, and c: 0
dug22241_ch10a.qxd
11/10/2004
18:30
Page 632
632
Chapter 10 Quadratic Equations and Inequalities
10-16
Calculator Close-Up
Note that the single solution to 4x
2
In this form we get a x 12 12 12 8 12 8 3 2 0
4, b ( 12)2 2(4) 144 8
12, and c 4(4)(9)
9.
Because b 12, b 12.
12x
9
0
corresponds to the single x-intercept for the graph of y 4x
2
144
12x
9.
10
2 2
4
Check 3 in the original equation. The solution set is
2
3 2
.
Now do Exercises 1318
Because the solutions to the equations in Examples 1 and 2 were rational numbers, these equations could have been solved by factoring. In Example 3 the solutions are irrational.
EXAMPLE
3
Two irrational solutions Solve 2x2 6x 3 0.
Solution
Let a 2, b 6, and c 3 in the quadratic formula: x 6 6 (6)2 4(2)(3) 2(2) 36 24 4 12 4 23 4 2( 3 3) 22 3 3 2 Check these values in the original equation. The solution set is
3 2 3
Calculator Close-Up
The two irrational solutions to 2x2 6x 3 0
6 6
correspond to the two x-intercepts for the graph of y 2x2
5
6x
3.
5
5
.
3
Now do Exercises 1924
dug22241_ch10a.qxd
11/10/2004
18:30
Page 633
10-17
10.2 The Quadratic Formula
633
EXAMPLE
4
Two imaginary solutions, no real solutions Find the complex solutions to x2 x 5 0.
Solution Calculator Close-Up
Because x2 x 5 0 has no real solutions, the graph of y x2 x 5
Let a
1, b
1, and c
5 in the quadratic formula: x 1 1 (1)2 4(1)(5) 2(1) 19
has no x-intercepts.
10
2 1 i 19 2 Check these values in the original equation. The solution set is are no real solutions to the equation.
1 i 19 2
. There
6 2
6
Now do Exercises 2530 You have learned to solve quadratic equations by four different methods: the even-root property, factoring, completing the square, and the quadratic formula. The even-root property and factoring are limited to certain special equations, but you should use those methods when possible. Any quadratic equation can be solved by completing the square or using the quadratic formula. Because the quadratic formula is usually faster, it is used more often than completing the square. However, completing the square is an important skill to learn. It will be used in the study of conic sections later in this text.
Methods for Solving ax 2
Method property Even-root Factoring Quadratic formula Completing the square Comments Use when b
bx
0.
c
0
Examples (x 2)2 x2 x2 (x x2 x x2 x2 5x 2)(x 5x 5 3 8 8 6 3) 0 25 2 0 7 2 9 0 0 4(3)
Use when the polynomial can be factored. Solves any quadratic equation Solves any quadratic equation, but quadratic formula is faster
6x 7 6x 9 (x 3)2
Number of Solutions
The quadratic equations in Examples 1 and 3 had two real solutions each. In each of those examples the value of b2 4ac was positive. In Example 2 the quadratic equation had only one solution because the value of b2 4ac was zero. In Example 4
dug22241_ch10a.qxd
11/10/2004
18:30
Page 634
634
Chapter 10 Quadratic Equations and Inequalities
10-18
the quadratic equation had no real solutions because b2 4ac was negative. Because b2 4ac determines the kind and number of solutions to a quadratic equation, it is called the discriminant. Number of Solutions to a Quadratic Equation The quadratic equation ax2 bx c 0 with a 0 has two real solutions if b2 4ac 0, one real solution if b2 4ac 0, and no real solutions (two imaginary solutions) if b2 4ac
0.
EXAMPLE
5
Using the discriminant Use the discriminant to determine the number of real solutions to each quadratic equation.
a) x2 b) x2 c) 4x2
3x 5 0 3x 9 12x 9 0
Solution
a) For x 2
3x b2
5
0, use a 4 ac ( 3)2
1, b
3, and c 9 20
5 in b2 29
4ac:
4(1)( 5)
Because the discriminant is positive, there are two real solutions to this quadratic equation. b) Rewrite x2 3x 9 as x 2 3x 9 0. Then use a 1, b 3, and c 9 in b2 4ac: b2 4ac ( 3)2 4(1)(9) 9 36 27
Because the discriminant is negative, the equation has no real solutions. It has two imaginary solutions. c) For 4x2 12x 9 0, use a 4, b 12, and c 9 in b2 4ac: b2 4ac ( 12)2 4(4)(9) 144 144 0
Because the discriminant is zero, there is only one real solution to this quadratic equation. Now do Exercises 3146
Applications
With the quadratic formula we can easily solve problems whose solutions are irrational numbers. When the solutions are irrational numbers, we usually use a calculator to nd rational approximations and to check.
dug22241_ch10a.qxd
11/10/2004
18:30
Page 635
10-19
10.2 The Quadratic Formula
635
EXAMPLE
6
Area of a tabletop The area of a rectangular tabletop is 6 square feet. If the width is 2 feet shorter than the length, then what are the dimensions?
Solution
x ft x 2 ft
Let x be the length and x 2 be the width, as shown in Fig. 10.1. Because the area is 6 square feet and A LW, we can write the equation x(x or x2 2x 6 0. Because this equation cannot be factored, we use the quadratic formula with a 1, b 2, and c 6: 2) 6
Figure 10.1
x
2 2
( 2)2 4(1)( 6) 2(1) 28 2 27 1 2 2
7
Because 1 7 is a negative number, it cannot be the length of a tabletop. If 72 7 1. Checking the product of x1 7, then x 2 1 7 1 and 7 1, we get
(
7
1)(
7
1)
7
1
6.
The exact length is 7 1 feet, and the width is 7 1 feet. Using a calculator, we nd that the approximate length is 3.65 feet and the approximate width is 1.65 feet. Now do Exercises 7594
Warm-Ups
True or false? Explain your answer.
1. Completing the square is used to develop the quadratic formula. 7. 2. For the equation 3x 2 4x 7, we have a 3, b 4, and c 3. If dx 2 4. 5. 6. 7. 8. ex f 0 and d 0, then x
e e2 2d 4df
.
The quadratic formula will not work on the equation x2 3 0. If a 2, b 3, and c 4, then b2 4ac 41. If the discriminant is zero, then there are no imaginary solutions. If b2 4ac 0, then ax2 bx c 0 has two real solutions. 1, b 2, and c To solve 2x x2 0 by the quadratic formula, let a
0.
9. Two numbers that have a sum of 6 can be represented by x and x 6. 10. Some quadratic equations have one real and one imaginary solution.
dug22241_ch10a.qxd
11/10/2004
18:30
Page 636
636
Chapter 10 Quadratic Equations and Inequalities
10-20
10.2
Exercises
Solve each equation by using the quadratic formula. See Example 3.
Net Tutor e-Professors
Boost your GRADE at mathzone.com!
Practice Problems Self-Tests Videos
19. v 2 21. x2
8v 5x
6 1
0 0
20. p 2 22. x2
6p 3x
4 5
0 0
Reading and Writing After reading this section, write out
the answers to these questions. Use complete sentences. 1. What is the quadratic formula used for? Solve each equation by using the quadratic formula. See Example 4. 25. 2t 2 27. 2x 2 6t 3x 5 6 0 26. 2y 2 28. 3x 2 1 2x 2y 5 0 23. 2t 2 6t 1 0 24. 3z 2 8z 2 0
2. When do you use the even-root property to solve a quadratic equation? 3. When do you use factoring to solve a quadratic equation?
4. When do you use the quadratic formula to solve a quadratic equation?
29.
1 2
x2
13
5x
30.
1 4
x2
17 4
2x
5. What is the discriminant? 6. How many solutions are there to any quadratic equation in the complex number system?
Find b 2 4ac and the number of real solutions to each equation. See Example 5. 31. x 2 6x 2 0 33. 2x 2 5x 6 0 35. 4m 2 25 1 2 y 1 4 20m 0 32. x 2 6x 9 0 34. x 2 3x 4 0 36. v 2 3v 5 1 1 1 38. w 2 w 2 3 4
Solve each equation by using the quadratic formula. See Example 1. 7. x 2 5x 6 0 8. x 2 7x 12 0
37. y 2
0
9. y 2 y 6 11. 6z 2 7z
3
0
10. m2 2m 8 12. 8q2 2q 1
39. 0 41. 9
3t 2 24z
2
5t 7 x 0
6 16z
2
0 0
40. 9m 2 42. 12 44. 46. 6x
2
16 7x 5 7x
24m x2 0 0 0
43. 5x 45. x 2 Solve each equation by using the quadratic formula. See Example 2. 13. 4x 15.
2
3x 2
Solve each equation by the method of your choice. 47. 0 0 49. 12 y 4 12 x 3 y 1 48. 12 x 2 42 w 9 x 1
4x 6x
1 1
0 0
14. 4x 16.
2
12x 24x
9
9x 2
9x 2
16
1 x 2
1 3
50.
1
5 w 3
17. 9
24x
16x 2
0
18. 4
20x
25x 2
dug22241_ch10a.qxd
11/10/2004
18:30
Page 637
10-21
51. 3y2 2y 4 0 52. 2y2 3y 6 0
10.2 The Quadratic Formula
637
79. Bulletin board. The length of a bulletin board is one foot more than the width. The diagonal has a length of 3 feet (ft). Find the length and width of the bulletin board.
53.
w w 9(3x 4 12 x 3 20 x2 8)(x 3(2y 8(y 2 w 5)2
w 3
54.
y 3y 4 y 1)2 12 x 4 6 x 10)(x 7z 12(z 1
2 4 80. Diagonal brace. The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures 6 m. Find the width and height.
55.
1 0 8 x 4) 5) 1) 42
56. 58.
25(2x 9 49 2 34
0 0
57. 25
6m
x
59. 1 61. (x
60.
x2 62. (x
2) 4 1)
20
x
2
Figure for Exercise 80
63. y
64. z
Use the quadratic formula and a calculator to solve each equation. Round answers to three decimal places and check your answers. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. x 2 3.2x 5.7 0 x 2 7.15x 3.24 0 x 2 7.4x 13.69 0 1.44x 2 5.52x 5.29 0 1.85x 2 6.72x 3.6 0 3.67x 2 4.35x 2.13 0 3x 2 14,379x 243 0 x 2 12,347x 6741 0 x 2 0.00075x 0.0062 0 4.3x 2 9.86x 3.75 0
81. Area of a rectangle. The length of a rectangle is 4 ft longer than the width, and its area is 10 square feet (ft2). Find the length and width. 82. Diagonal of a square. The diagonal of a square is 2 m longer than a side. Find the length of a side. If an object is given an initial velocity of v0 feet per second from a height of s0 feet, then its height S after t seconds is given by the formula S 16t 2 v0 t s0 . 83. Projected pine cone. If a pine cone is projected upward at a velocity of 16 ft/sec from the top of a 96-foot pine tree, then how long does it take to reach the earth? 84. Falling pine cone. If a pine cone falls from the top of a 96-foot pine tree, then how long does it take to reach the earth? 85. Tossing a ball. A ball is tossed into the air at 10 ft/sec from a height of 5 feet. How long does it take to reach the earth?
Find the exact solution(s) to each problem. If the solution(s) are irrational, then also nd approximate solution(s) to the nearest tenth. See Example 6. 75. Missing numbers. Find two positive real numbers that differ by 1 and have a product of 16.
86. Time in the air. A ball is tossed into the air from a height of 12 feet at 16 ft/sec. How long does it take to reach the earth? 87. Penny tossing. If a penny is thrown downward at 30 ft/sec from the bridge at Royal Gorge, Colorado, how long does it take to reach the Arkansas River 1000 ft below? 88. Foul ball. Suppose Charlie OBrian hits a baseball straight upward at 150 ft/sec from a height of 5 ft. a) Use the formula to determine how long it takes the ball to return to the earth. b) Use the graph on the next page to estimate the maximum height reached by the ball.
76. Missing numbers. Find two positive real numbers that differ by 2 and have a product of 10. 77. More missing numbers. Find two real numbers that have a sum of 6 and a product of 4. 78. More missing numbers. Find two real numbers that have a sum of 8 and a product of 2.
dug22241_ch10a.qxd
11/10/2004
18:30
Page 638
638
Chapter 10 Quadratic Equations and Inequalities
10-22
93. Farmers Delight. The manager of Farmers Delight bought a load of watermelons for $750 and priced the watermelons so that he would make a prot of $2 on each melon. When all but 100 of the melons had been sold, he broke even. How many did he buy originally? 94. Traveling club. The members of a traveling club plan to share equally the cost of a $150,000 motorhome. If they can nd 10 more people to join and share the cost, then the cost per person will decrease by $1250. How many members are there originally in the club?
400 Height (ft) 300 200 100 0 0 2 4 6 8 10 Time (sec)
Figure for Exercise 88
Getting More Involved
Solve each problem. 89. Kitchen countertop. A 30 in. by 40 in. countertop for a work island is to be covered with green ceramic tiles, except for a border of uniform width as shown in the gure. If the area covered by the green tiles is 704 square inches (in.2), then how wide is the border? 95. Discussion Find the solutions to 6x 2 5x 4 0. Is the sum of your solutions equal to b? Explain why the sum a of the solutions to any quadratic equation is b. a (Hint: Use the quadratic formula.) 96. Discussion
30 in. 40 in.
x
Use the result of Exercise 95 to check whether 2, 1 33 is the solution set to 9x 2 3x 2 0. If this solution set is not correct, then what is the correct solution set? 97. Discussion What is the product of the two solutions to 6x2 5x 4 0? Explain why the product of the solutions to any quadratic equation is c.
a
Figure for Exercise 89
98. Discussion Use the result of the previous exercise to check whether 9, 2 is the solution set to 2x 2 13x 18
2
90. Recovering an investment. The manager at Cream of the Crop bought a load of watermelons for $200. She priced the melons so that she would make $1.50 prot on each melon. When all but 30 had been sold, the manager had recovered her initial investment. How many did she buy originally? 91. Baby shower. A group of ofce workers plans to share equally the $100 cost of giving a baby shower for a coworker. If they can get six more people to share the cost then the cost per person will decrease by $15. How many people are in the original group? 92. Sharing cost. The members of a ying club plan to share equally the cost of a $200,000 airplane. The members want to nd ve more people to join the club so that the cost per person will decrease by $2000. How many members are currently in the club?
0.
If this solution set is not correct, then what is the correct solution set?
Graphing Calculator Exercises
Determine the number of real solutions to each equation by examining the calculator graph of y ax2 bx c. Use the discriminant to check your conclusions. 99. 100. 101. 102. 103. 104. x 2 6.33x 3.7 0 1.8x 2 2.4x 895 0 4x 2 67.1x 344 0 2x 2 403 0 x 2 30x 226 0 16x 2 648x 6562 0
dug22241_ch10a.qxd
11/10/2004
18:30
Page 639
10-23
10.3 Graphing Parabolas
639
10.3
In this Section
Finding Ordered Pairs Graphing Parabolas The Vertex and Intercepts Applications
Graphing Parabolas
The graph of any equation of the form y mx b is a straight line. In this section we will see that all equations of the form y ax2 bx c (with a 0) have graphs that are in the shape of a parabola.
Finding Ordered Pairs
It is straightforward to calculate y when given x for an equation of the form y ax2 bx c. However, if we are given y and want to nd x, then we must use methods for solving quadratic equations.
EXAMPLE
1
Finding ordered pairs Complete each ordered pair so that it satises the given equation.
a) (2, ), ( , 0), y x 2 x 6 b) (0, ), ( , 20), y 16x 2 48x 84
Solution
a) If x 2, then y 22 2 6 4. So the ordered pair is (2, 4). To nd x when y 0, replace y by 0 and solve the resulting quadratic equation: x2 x 6 0 (x 3)(x 2) 0 x30 or x20 x3 or x 2 The ordered pairs are ( 2, 0) and (3, 0). b) If x 0, then y 16 02 48 0 84 84. The ordered pair is (0, 84). To nd x when y 20, replace y by 20 and solve the equation for x: 48x 84 48x 64 2 x 3x 4 (x 4)(x 1) x4 x The ordered pairs are ( 16x2 16x2 20 0 0 0 0 or x1 4 or x 1, 20) and (4, 20).
Subtract 20 from each side. Divide each side by Factor. 16.
0 1
Zero factor property
Now do Exercises 710
Graphing Parabolas
All equations of the form y ax2 bx c with a 0 have graphs that are similar in shape. The graph of any equation of this form is called a parabola. Note that any real number can be used in place of x.
dug22241_ch10a.qxd
11/10/2004
18:30
Page 640
640
Chapter 10 Quadratic Equations and Inequalities
10-24
EXAMPLE
2
The simplest parabola Make a table of ordered pairs that satisfy y of y x2.
x 2 and then sketch the graph
Solution
Make a table of values for x and y :
x y x2 2 4 1 1 0 0 1 1 2 4
Calculator Close-Up
This close-up view of y x2 shows how rounded the curve is at the bottom. When drawing a parabola by hand, be sure to draw it smoothly.
4
Plot the ordered pairs from the table and draw a parabola through the points as shown in Fig. 10.2.
y 8 6 4 2 43 2 1 2 1 2
y 3
x2 4 x
2 1
2
Figure 10.2
Now do Exercises 1720
The parabola in Fig. 10.2 is said to open upward. In the next example we see a parabola that opens downward. If a 0 in the equation y a x 2 bx c, then the parabola opens upward. If a 0, then the parabola opens downward.
EXAMPLE
3
A parabola opening downward Graph y 4 x 2.
y 5 y 4 x2
Solution
We plot enough points to get the correct shape of the graph:
x y 4 x2 2 0 1 3 0 4 1 3 2 0
3 1
3 2 1 1 2 1 3 x
See Fig. 10.3 for the graph.
Figure 10.3
Now do Exercises 2126
dug22241_ch10a.qxd
11/10/2004
18:30
Page 641
10-25
10.3 Graphing Parabolas
641
The Vertex and Intercepts
The lowest point on a parabola that opens upward or the highest point on a parabola that opens downward is called the vertex. The y-coordinate of the vertex is the minimum value of y if the parabola opens upward, and it is the maximum value of y if the parabola opens downward. For y x2 the vertex is (0, 0), and 0 is the minimum value of y. For y 4 x2 the vertex is (0, 4), and 4 is the maximum value of y. If y ax2 bx c has x-intercepts, they can be found by solving ax2 bx c 0 by the quadratic formula. The vertex is midway between the x-intercepts as shown in Fig. 10.4. Note that in the quadratic formula x b2 b b2 2a 4ac ,
b . 2a
4ac is added and subtracted from the numerator of
So
b , 2a
0 is the point
midway between the x-intercepts and the vertex has the same x-coordinate. Even if the b parabola has no x-intercepts, the x-coordinate of the vertex is still .
2a
y
Study Tip
Although you should avoid cramming, there are times when you have no other choice. In this case concentrate on what is in your class notes and the homework assignments. Try to work one or two problems of each type. Instructors often ask some relatively easy questions on a test to see if you have understood the major ideas.
y
ax2
bx
c
b
b 2a
2
4ac
b 2a b Vertex b2 2a
x
4ac
Figure 10.4
Vertex of a Parabola The x-coordinate of the vertex of y
ax2
bx
c is
b , 2a
provided that a
0.
When you graph a parabola, you should always locate the vertex because it is the point at which the graph turns around. With the vertex and several nearby points you can see the correct shape of the parabola. Recall from Section 4.1 how we used function notation for naming and evaluating polynomials. So instead of y x2 we can write f (x) x2, where we read f (x) as f of x. Note that f (x) is simply used as the second coordinate in place of y. So instead of writing y 4 when x 2, we simply write f (2) 4. We will use this function notation in Example 4.
dug22241_ch10a.qxd
11/10/2004
18:31
Page 642
642
Chapter 10 Quadratic Equations and Inequalities
10-26
EXAMPLE
4
Using the vertex in graphing a parabola Find the vertex and graph f (x) x 2 x 2.
Solution
First nd the x-coordinate of the vertex: x Now nd f
y 4 3 1 3 1 1 2 3 4 2 3 x 1 2
b 2a
( 1) 2( 1)
1 2
1 2
: 1 2
19 , 24
f The vertex is
1 2
2
1 2
2
1 4
1 2
2
9 4
f (x)
x2
x
2
. Now nd a few points on either side of the vertex:
x 2 0 1 2 1 2 9 4 0 2 1 0
f (x)
x2
x
2
Sketch a parabola through these points as in Fig. 10.5.
Figure 10.5
Now do Exercises 2734
The y-intercept of a parabola is the point that has 0 as the rst coordinate. The x-intercepts are the points that have 0 as their second coordinates.
EXAMPLE
5
Using the intercepts in graphing a parabola Find the vertex and intercepts, and sketch the graph of each parabola.
a) f (x) b) s x 2 2x 8 16t 2 64t
Solution
a) Use x
b 2a
to get x
1 as the x-coordinate of the vertex. If x f (1) 12 2 1 9. 0, then 02 2 0 8. 8 8
1, then
So the vertex is (1,
9). If x f (0)
dug22241_ch10a.qxd
11/10/2004
18:31
Page 643
10-27
O
10.3 Graphing Parabolas
643
The y-intercept is (0,
B (N) N2
8). To nd the x-intercepts, replace f (x) by 0: 0 0 0 4
6 4 2 3 1 2 4 8 10
2N
8
x2 (x
1
2
3
5
N
2x 8 4)(x 2) x4 x
or or
x
2 x
0 2
(1,
9)
Figure 10.6
The x-intercepts are ( 2, 0) and (4, 0). The graph is shown in Fig. 10.6. b) Because s is expressed in terms of t in the equation s 16t2 64t, the independent variable is t and the dependent variable is s. Since we always put the independent variable rst in an ordered pair, the ordered pairs are b to get written in the form (t, s). To nd the vertex use t 2a t 64 2( 16) 16 22 64. 0, then 16 02 0. 64 0 2.
s 60 40 20 (2, 64) s 16t 2 64t
If t
2, then s 64 2
So the vertex is (2, 64). If t s
1
1
2
3
5
6
7
t
So the s-intercept is (0, 0). To nd the t-intercepts, replace s by 0: 16t 2 64t 16t (t 4) 16t t 0 0 0 0
Figure 10.7
or or
t
4 t
0 4
The t-intercepts are (0, 0) and (4, 0). The graph is shown in Fig. 10.7. Now do Exercises 3550
Calculator Close-Up
You can nd the vertex of a parabola with a calculator by using either the maximum or minimum feature. First graph the parabola as shown.
4 4 10 10 10 10
y-coordinate on the graph. Press CALC and choose minimum.
moving the cursor to the point and pressing ENTER. For the right bound choose a point to the right of the vertex. For the guess choose a point close to the vertex.
12
Because this parabola opens upward, the y-coordinate of the vertex is the minimum
The calculator will ask for a left bound, a right bound, and a guess. For the left bound choose a point to the left of the vertex by
12
dug22241_ch10a.qxd
11/10/2004
18:31
Page 644
644
Chapter 10 Quadratic Equations and Inequalities
10-28
Applications
In applications we are often interested in nding the maximum or minimum value of a variable. If the graph of a parabola opens downward, then the maximum value of the dependent variable is the second coordinate of the vertex. If the parabola opens upward, then the minimum value of the dependent variable is the second coordinate of the vertex.
EXAMPLE
6
Finding the maximum height If a projectile is launched with an initial velocity of v0 feet per second from an initial height of s0 feet, then its height s(t) in feet is determined by s(t) 16t2 v0t s0, where t is the time in seconds. If a ball is tossed upward with velocity 64 feet per second from a height of 5 feet, then what is the maximum height reached by the ball?
I
80 60 40
I ( J)
16J 2
64J
5
(2, 69)
Solution
The height s(t) of the ball for any time t is given by s(t) 16 t 2 64t Because the maximum height occurs at the vertex of the parabola, we use t to nd the vertex: t 64 2( 16) 2
b 2a
5.
20
Now use t
0 1 2 3 4 5 6 7
J
2 to nd the second coordinate of the vertex: s (2) 16(2)2 64(2) 5 69
Figure 10.8
The maximum height reached by the ball is 69 feet. See Fig. 10.8. Now do Exercises 5967
Warm-Ups
True or false? Explain your answer.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. The ordered pair ( 2, 1) satises f (x) x 2 5. The y-intercept for g(x) x 2 3x 9 is (9, 0). 5, 0). The x-intercepts for y x 2 5 are ( 5, 0) and ( 2 12 opens upward. The graph of f (x) x The graph of y 4 x 2 opens downward. The vertex of y x 2 2x is ( 1, 1). The parabola y x 2 1 has no x-intercepts. The y-intercept for g(x) a x 2 bx c is (0, c). If w 2v 2 9, then the maximum value of w is 9. If y 3x 2 7x 9, then the maximum value of y occurs when x
7 . 6
dug22241_ch10a.qxd
11/10/2004
18:31
Page 645
10-29
10.3 Graphing Parabolas
645
10.3
Exercises
Graph each parabola. See Examples 2 and 3.
Net Tutor e-Professors
Boost your GRADE at mathzone.com!
Practice Problems Self-Tests Videos
17. y
x2
2
Reading and Writing After reading this section, write out
the answers to these questions. Use complete sentences. 1. What equation has a graph called a parabola? 2. When does a parabola open upward and when does a parabola open downward? 18. y x2 4
3. How can you nd the x-intercepts for a parabola? 4. How can you nd the y-intercept? 5. What is the x-coordinate of the vertex of a parabola? 6. How do you nd the y-coordinate of the vertex? 19. y 12 x 2 4
Complete each ordered pair so that it satises the given equation. See Example 1. 7. y 8. y x2 x 12 (3, x 1 (0, ), ( ), ( , 0) 20. y , 3) 12 x 3 6
12 x 2 16x2 x2 4x
9. y 10. y
32x (4, 5 ( 2,
), ( ), (
, 0) , 2)
Determine whether the graph of each parabola opens upward or downward. See Examples 2 and 3. 11. y 13. y 15. y x2 3x2 ( 2x 5 4x 3)2 2 12. y 14. y 16. y 2x2
x2
21. y
2x 2
5
x
3
1
(5
x)2
dug22241_ch10a.qxd
11/10/2004
18:31
Page 646
646
22. y
Chapter 10 Quadratic Equations and Inequalities
10-30
31. f (x) 33. y x2 2x2 x 1 20x 1 32. f (x) 34. y 3x2 3x2 2x 18x 1 7
x2
1
Find all intercepts for the graph of each parabola. See Example 5. 35. f (x) 23. y 12 x 3 5 37. y 39. f (x) x2 16 2x 4x2 x2 8 12x 9 36. f (x) 38. y 40. f (x) x2 x2 x 2x2 9 6 x 3
Find the vertex and intercepts for each parabola. Sketch the graph. See Examples 4 and 5. 41. f (x) 24. y 12 x 2 3 x2 x 2
42. f (x)
x2
2x
3
25. y
(x
2)2
43. g(x)
x2
2x
8
26. y
(x
3)2
44. g(x)
x2
x
6
Find the vertex for the graph of each parabola. See Example 4. 27. f (x) x2 9 29. y x2 4x 1 28. f (x) x2 12 30. y x2 8x 3
dug22241_ch10a.qxd
11/10/2004
18:31
Page 647
10-31
45. y x2 4x 3 50. v u2 8u 9
10.3 Graphing Parabolas
647
46. y
x2
5x
4
Find the maximum or minimum value of y. 51. y 53. y 55. y 57. y x2 3x 2 x2 2x 8 14 3 4x 52. y 54. y 56. y 58. y 33 6 x2 3x 2 x2 5x 2 2x 5 24x
2x2
47. h(x)
x2
Solve each problem. See Example 6. 3x 4 59. Maximum height. If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height in feet is given by s(t) 16t 2 64t t 4. where t is time in seconds. Graph this parabola for 0 What is the maximum height reached by the ball?
48. h(x)
x2
2x
8
49. a
b2
6b
16
60. Maximum height. If a soccer ball is kicked straight up from the ground with an initial velocity of 32 feet per second, then its height above the earth in feet is given by s(t) 16t2 32t where t is time in seconds. Graph this parabola for 0 t 2. What is the maximum height reached by the ball?
dug22241_ch10a.qxd
11/10/2004
18:31
Page 648
648
Chapter 10 Quadratic Equations and Inequalities
10-32
where y is the number of births divided by the number of deaths in the year 1950 x (World Resources Institute, www.wri.org). a) Use the graph to estimate the year in which the stabilization ratio was at its maximum. b) Use the formula to nd the year in which the stabilization ratio was at its maximum. c) What was the maximum stabilization ratio from part (b)? d) What is the signicance of a stabilization ratio of 1?
61. Minimum cost. It costs Acme Manufacturing C dollars per hour to operate its golf ball division. An analyst has determined that C is related to the number of golf balls produced per hour, x, by the equation C 0.009x2 1.8x 100. What number of balls per hour should Acme produce to minimize the cost per hour of manufacturing these golf balls? 62. Maximum prot. A chain store manager has been told by the main ofce that daily prot, P, is related to the number of clerks working that day, x, according to the equation P 25x2 300x. What number of clerks will maximize the prot, and what is the maximum possible prot? 63. Maximum area. Jason plans to fence a rectangular area with 100 meters of fencing. He has written the formula A w (50 w) to express the area in terms of the width w. What is the maximum possible area that he can enclose with his fencing?
y 4 Stabilization ratio (births/deaths) 3 2 1 0
Stabilization ratio for South and Central America
10 20 30 40 Years after 1950
50 x
Figure for Exercise 66
67. Suspension bridge. The cable of the suspension bridge shown in the gure hangs in the shape of a parabola with equation y 0.0375x 2, where x and y are in meters. What is the height of each tower above the roadway? What is the length z for the cable bracing the tower?
Photo for Exercise 63
y 20
64. Minimizing cost. A company uses the formula C(x) 0.02x 2 3.4x 150 to model the unit cost in dollars for producing x stabilizer bars. For what number of bars is the unit cost at its minimum? What is the unit cost at that level of production? 65. Air pollution. The amount of nitrogen dioxide A in parts per million (ppm) that was present in the air in the city of Homer on a certain day in June is modeled by the formula A(t) 2t 2 32t 12,
10
z x 0 5 10 15 25 30 35 40
Figure for Exercise 67
where t is the number of hours after 6:00 A.M. Use this formula to nd the time at which the nitrogen dioxide level was at its maximum. 66. Stabilization ratio. The stabilization ratio (births/deaths) for South and Central America can be modeled by the formula y 0.0012x 2 0.074x 2.69
Getting More Involved
68. Exploration a) Write the equation y 3(x 2)2 6 in the form y ax 2 bx c, and nd the vertex of the parabola b using the formula x .
2a