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4 Quadratic Equations: Getting Information from a Model

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 ✓





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





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 .





NOW TRY EXERCISE 17







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







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



454



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).



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