use a table of values to graph each equation. state the ... … · use a table of values to graph...

Use a table of values to graph each equation. State the domain and range.

1.y = 2x2 + 4x 6

SOLUTION:

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y 8}.

x y = 2x2 + 4x 6 (x, y)

3 y = 2(3)2 + 4(3) 6 = 0 (3,0)

2 y = 2(2)2 + 4(2) 6 = 6

(2,6)

1 y = 2(1)2 + 4(1) 6 =

8

(1,8)

0 y = 2(0)2 + 4(0) 6 = 6 (0,6)

1 y = 2(1)2 + 4(1) 6 = 0 (1,0)

2 y = 2(2)2 + 4(2) 6 = 10 (2,10)

2.y = x2 + 2x 1

SOLUTION:

Graphtheorderedpairs,andconnectthemtocreateasmoothcurve.Theparabolaextendstoinfinity.


x y = x2 + 2x 1 (x, y)

3 y = (3)2 + 2(3) 1 =

2

(3,2)

2 y = (2)2 + 2(2) 1 = 1

(2,1)

1 y = (1)2 + 2(1) 1 =

2

(1,2)

0 y = (0)2 + 2(0) 1 = 1

(0,1)

1 y = (1)2 + 2(1) 1 = 2 (1,2)

2 y = (2)2 + 2(2) 1 = 7 (2,7)

3.y = x2 6x 3

SOLUTION:



x y = x2 6x 3 (x, y)

1 y = (1)2 6(1) 3 = 4 (1,4)

0 y = (0)2 6(0) 3=3 (0,3)

1 y = (1)2 6(1) 3 = 8 (1,8)

2 y = (2)2 6(2) 3 = 11 (2,11)

3 y = (3)2 6(3) 3=12 (3,12)

4 y = (4)2 6(4) 3=11 (4,11)

5 y = (5)2 6(5) 3 = 8 (5,8)

6 y = (6)2 6(6) 3 = 3 (6,3)

7 y = (7)2 6(7) 3 = 4 (7,4)

4.y = 3x2 6x 5

SOLUTION:



x y = 3x2 6x 5 (x, y)

2 y = 3(2)2 6(2) 5 = 19

(2,19)

1 y = 3(1)2 6(1) 5 = 4 (1,4)

0 y = 3(0)2 6(0) 5 = 5 (0,5)

1 y = 3(1)2 6(1) 5=8 (1,8)

2 y = 3(2)2 6(2) 5=5 (2,5)

3 y = 3(3)2 6(3) 5 = 4 (3,4)

Find the vertex, the equation of the axis of symmetry, and the y-intercept of each graph.

5.

SOLUTION:Find the vertex. Because the parabola opens down, the vertex is located at the maximum point of the parabola. It is located at (1, 5). Find the axis of symmetry. The axis of symmetry is the line that goes through the vertex and divides the parabola into congruent halves. It is located at x = 1. Find the y-intercept. The y-intercept is the point where the graph intersects the y-axis. It is located at (0, 3), so the y-intercept is 3.

6.

SOLUTION:Find the vertex. Because the parabola opens up, the vertex is located at the minimum point of the parabola. It is located at (2, 3). Find the axis of symmetry. The axis of symmetry is the line that goes through the vertex and divides the parabola into congruent halves. It is located at x = 2. Find the y-intercept. The y-intercept is the point where the graph intersects the y-axis. It is located at (0, 1), so the y-intercept is 1.

7.


8.

SOLUTION:Find the vertex. Because the parabola opens down, the vertex is located at the maximum point of the parabola. It is located at (0, 5). Find the axis of symmetry. The axis of symmetry is the line that goes through the vertex and divides the parabola into congruent halves. It is located at x = 0. Find the y-intercept. The y-intercept is the point where the graph intersects the y-axis.Itislocatedat(0,5),sothey-intercept is 5.

Find the vertex, the equation of the graph of the axis of symmetry, and the y-intercept of the graph of eachfunction.

9.y = 3x2 + 6x 1

SOLUTION:Find the vertex.

In the equation y = 3x2 + 6x 1, a = 3, b = 6, and c = 1.

The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 1. Substitute the x-coordinate of the vertex into the original equation to find the value of y .

The vertex is at (1, 2). Find the axis of symmetry. The axis of symmetry is the vertical line that goes through the vertex. It is located at x = 1. Find the y-intercept. The y-intercept always occurs at (0, c). Since c = 1 for this equation, the y-intercept is located at (0, 1).

10.y = x2 + 2x + 1


In the equation y = x2 + 2x + 1, a = 1, b = 2, and c = 1.




11.y = x2 4x + 5


In the equation y = x2 4x + 5, a = 1, b = 4, and c = 5.




12.y = 4x2 8x + 9


In the equation y = 4x2 8x + 9, a = 4, b = 8, and c = 9.




Consider each function. a.Determinewhetherthefunctionhasmaximum or minimum value.b.Statethemaximumorminimumvalue. c.Whatarethedomainandrangeofthefunction?

13.y = x2 + 4x 3

SOLUTION:

a. For y = x2 + 4x 3, a = 1, b = 4, and c = 3. Because a is negative, the graph opens downward, so the

function has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y-coordinate.

The maximum value is 1. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {f (x)|f(x)1}.

14.y = x2 2x + 2

SOLUTION:

a. For y = x2 2x + 2, a = 1, b = 2, and c = 2. Because a is negative, the graph opens downward, so the




The maximum value is 3. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {f (x) | f(x)3}.

15.y = 3x2 + 6x + 3

SOLUTION:

a. For y = 3x2 + 6x + 3, a = 3, b = 6, and c = 3. Because a is negative, the graph opens downward, so the function

has a maximum value.




16.y = 2x2 + 8x 6

SOLUTION:

a. For y = 2x2 + 8x 6, a = 2, b = 8, and c = 6. Because a is negative, the graph opens downward, so the





Graph each function.

17.f (x) = 3x2 + 6x + 3

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For f (x) = 3x2 + 6x + 3, a = 3, b = 6, and c = 3.

Step 2Findthevertex,anddeterminewhetheritisamaximumorminimum. The x-coordinate of the vertex is x = 1. Substitute the x-coordinate of the vertex into the original equation to find the value of y .

The vertex lies at (1, 6). Because a is negative, the graph opens down, and the vertex is a maximum. Step 3Findthey-intercept. Use the original equation, and substitute 0 for x.

The y-intercept is (0, 3). Step 4The axis of symmetry divides the parabola into two equal parts. So if there is a point on one side, there is a corresponding point on the other side that is the same distance from the axis of symmetry and has the same y-value.

Step 5Connectthepointswithasmoothcurve.

x 1 0 1 2 3 y 6 3 6 3 6

18.f (x) = 2x2 + 4x + 1

SOLUTION:




The y-intercept is (0, 1). Step 4 The axis of symmetry divides the parabola into two equal parts. So if there is a point on one side, there is a corresponding point on the other side that is the same distance from the axis of symmetry and has the same y-value.


x 1 0 1 2 3 y 5 1 3 1 5

19.f (x) = 2x2 8x 4

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For f (x) = 2x2 8x 4, a = 2, b = 8, and c = 4.


The vertex lies at (2, 12). Because a is positive, the graph opens up, and the vertex is a minimum. Step 3Findthey-intercept. Use the original equation, and substitute 0 for x.



x 0 1 2 3 4 y 4 10 12 10 4

20.f (x) = 3x2 6x 1

SOLUTION:






x 1 0 1 2 3 y 8 1 4 1 8

21.CCSS REASONING A juggler is tossing a ball into the air. The height of the ball in feet can be modeled by the

equation y = 16x2 + 16x + 5, where y represents the height of the ball at x seconds.

a. Graph this equation. b. At what height is the ball thrown? c. What is the maximum height of the ball?

SOLUTION:

a. Step 1 Findtheequationoftheaxisofsymmetry.For y = 16x2 + 16x + 5, a = 16, b = 16, and c = 5.

Step 2Findthevertex,anddeterminewhetheritisamaximumorminimum.

The x-coordinate of the vertex is x = . Substitute the x-coordinate of the vertex into the original equation to find

the value of y .

The vertex lies at . Because a is negative, the graph opens down, and the vertex is a maximum.

Step 3Findthey-intercept. Use the original equation, and substitute 0 for x.



b. The ball is thrown when time equals 0, or at the y-intercept. So, the ball was 5 feet from the ground when it was thrown. c. The maximum height of the ball occurs at the vertex. So, the ball reaches a maximum height of 9 feet.

x 0 1

y 5 9 5


22.y = x2 + 4x + 6

SOLUTION:



x y = x2 + 4x + 6 (x, y)

4 y = (4)2 + 4(4) + 6 = 6 (4,6)

3 y = (3)2 + 4(3) + 6 = 3 (3,3)

2 y = (2)2 + 4(2) + 6 = 2 (2,2)

1 y = (1)2 + 4(1) + 6 = 3 (1,3)

0 y = (0)2 + 4(0) + 6 = 6 (0,6)

23.y = 2x2 + 4x + 7

SOLUTION:



x y = 2x2 + 4x + 7 (x, y)

3 y = 2(3)2 + 4(3) + 7 = 13 (3, 13)

2 y = 2(2)2 + 4(2) + 7 = 7 (2,7)

1 y = 2(1)2 + 4(1) + 7 = 5 (1,5)

0 y = 2(0)2 + 4(0) + 7 = 7 (0,7)

1 y = 2(1)2 + 4(1) + 7 = 13 (1,13)

24.y = 2x2 8x 5

SOLUTION:



x y = 2x2 8x 5 (x, y)

4 y = 2(4)2 8(4) 5 = 5 (4, 5)

3 y = 2(3)2 8(3) 5 = 11 (3,11)

2 y = 2(2)2 8(2) 5 = 13 (2,13)

1 y = 2(1)2 8(1) 5 = 11 (1,11)

0 y = 2(0)2 8(0) 5 = 5 (0,5)

25.y = 3x2 + 12x + 5

SOLUTION:


The domain of the equation is all real numbers. The range of the equation is all real numbers greater than or equal to

the minimum value, or {y |y 7}.

x y = 3x2 + 12x + 5 (x, y)

0 y = 3(0)2 +12(0) + 5 = 5 (0, 5)

1 y = 3(1)2 +12(1) + 5 = 4 (1,4)

2 y = 3(2)2 +12(2) + 5 = 7 (2,7)

3 y = 3(3)2 +12(3) + 5 = 4 (3,4)

4 y = 3(4)2 +12(4) + 5 = 5 (4, 5)

26.y = 3x2 6x 2

SOLUTION:




x y = 3x2 6x 2 (x, y)

3 y = 3(3)2 6(3) 2 = 7 (3, 7)

2 y = 3(2)2 6(2) 2 = 2 (2,2)

1 y = 3(1)2 6(1) 2 = 5 (1,5)

0 y = 3(0)2 6(0) 2 = 2 (0,2)

1 y = 3(1)2 6(1) 2 = 7 (1,7)

27.y = x2 2x 1

SOLUTION:




x y = x2 2x 1 (x, y)

3 y = (3)2 2(3) 1 = 2 (3, 2)

2 y = (2)2 2(2) 1 = 1 (2,1)

1 y = (1)2 2(1) 1 = 2 (1,2)

0 y = (0)2 2(0) 1 = 1 (0,1)

1 y = (1)2 2(1) 1 = 2 (1,2)


28.

SOLUTION:Find the vertex. Because the parabola opens up, the vertex is located at the minimum point of the parabola. It is located at (3, 6). Find the axis of symmetry. The axis of symmetry is the line that goes through the vertex and divides the parabola into congruent halves. It is located at x = 3. Find the y-intercept. The y-intercept is the point where the graph intersects the y-axis.Itislocatedat(0,3),sothey-intercept is 3.

29.


30.


31.


32.

SOLUTION:Find the vertex. Because the parabola opens up, the vertex is located at the minimum point of the parabola. It is located at (0, 4). Find the axis of symmetry. The axis of symmetry is the line that goes through the vertex and divides the parabola into congruent halves. It is located at x = 0. Find the y-intercept. The y-intercept is the point where the graph intersects the y-axis.Itislocatedat(0,4), so the y-intercept is 4.

33.


Find the vertex, the equation of the axis of symmetry, and the y-intercept of each function.

34.y = x2 + 8x + 10






35.y = 2x2 + 12x + 10


In the equation y = 2x2 + 12x + 10, a = 2, b = 12, and c = 10.




36.y = 3x2 6x + 7






37.y = x2 6x 5


In the equation y = x2 6x 5, a = 1, b = 6, and c = 5.



The vertex is at (3, 4). Find the axis of symmetry. The axis of symmetry is the vertical line that goes through the vertex. It is located at x = 3. Find the y-intercept. The y-intercept always occurs at (0, c). Since c = 5 for this equation, the y-intercept is located at (0,5).

38.y = 5x2 + 20x + 10






39.y = 7x2 28x + 14






40.y = 2x2 12x + 6






41.y = 3x2 + 6x 18






42.y = x2 + 10x 13


In the equation y = x2 + 10x 13, a = 1, b = 10, and c = 13.




Consider each function. a. Determine whether the function has a maximum or minimum value.b. State the maximum or minimum value. c. What are the domain and range of the function?

43.y = 2x2 8x + 1

SOLUTION:

a. For y = 2x2 8x + 1, a = 2, b = 8, and c = 1. Because a is negative, the graph opens downward, so the





44.y = x2 + 4x 5

SOLUTION:

a. For y = x2 + 4x 5, a = 1, b = 4, and c = 5. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .


The minimum value is 9. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {f (x)|f (x)9}.

45.y = 3x2 + 18x 21

SOLUTION:

a. For y = 3x2 + 18x 21, a = 3, b = 18, and c = 21. Because a is positive, the graph opens upward, so the function

has a minimum value.




46.y = 2x2 16x + 18

SOLUTION:






47.y = x2 14x 16

SOLUTION:

a. For y = x2 14x 16, a = 1, b = 14, and c = 16. Because a is negative, the graph opens downward, so the





48.y = 4x2 + 40x + 44

SOLUTION:

a. For y = 4x2 + 40x + 44, a = 4, b = 40, and c = 44. Because a is positive, the graph opens upward, so the function





49.y = x2 6x 5

SOLUTION:






50.y = 2x2 + 4x + 6

SOLUTION:

a. For y = 2x2 + 4x + 6, a = 2, b = 4, and c = 6. Because a is positive, the graph opens upward, so the function has a

minimum value.




51.y = 3x2 12x 9

SOLUTION:

a. For y = 3x2 12x 9, a = 3, b = 12, and c = 9. Because a is negative, the graph opens downward, so the






52.y = 3x2 + 6x 4

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = 3x2 + 6x 4, a = 3, b = 6, and c = 4.





x 1 0 1 2 3 y 13 4 1 4 13

53.y = 2x2 4x 3

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = 2x2 4x 3, a = 2, b = 4, and c = 3.





x 3 2 1 0 1 y 9 3 1 3 9

54.y = 2x2 8x + 2

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = 2x2 8x + 2, a = 2, b = 8, and c = 2.





x 4 3 2 1 0 y 2 8 10 8 2

55.y = x2 + 6x 6

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = x2 + 6x 6, a = 1, b = 6, and c = 6.





x 5 4 3 2 1 y 11 14 15 14 11

56.y = x2 2x + 2

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = x2 2x + 2, a = 1, b = 2, and c = 2.





x 1 0 1 2 3 y 5 2 1 2 5

57.y = 3x2 12x + 5

SOLUTION:






x 0 1 2 3 4 y 5 4 7 4 5

58.BOATING Miranda has her boat docked on the west side of Casper Point. She is boating over to Casper Marina,

which is located directly east of where her boat is docked. The equation d = 16t2 + 66t models the distance she travels north of her starting point, where d is the number of feet and t the time traveled in minutes.

a. Graph this equation. b. What is the maximum number of feet north that she traveled? c. How long did it take her to reach Casper Marina?

SOLUTION:

a. Step 1 Findtheequationoftheaxisofsymmetry.d = 16t2 + 66t, a = 16 and b = 66.

Step 2Findthevertex,anddeterminewhetheritisamaximumorminimum. The t-coordinate of the vertex is t = 2. Substitute the t-coordinate of the vertex into the original equation to find the value of N.

The vertex lies at about (2, 68). Because a is negative, the graph opens down, and the vertex is a maximum. Step 3Findthed-intercept. Use the original equation, and substitute 0 for t.

The d-intercept is (0, 0). Step 4 The axis of symmetry divides the parabola into two equal parts. So if there is a point on one side, there is a corresponding point on the other side that is the same distance from the axis of symmetry and has the same d-value.


b. The maximum distance Miranda traveled north is the y-coordinate of the vertex, or about 68 feet. c. It takes the boat 2 minutes to reach the vertex, and due to symmetry it will take the boat another 2 minutes to reach Casper Marina. So, it will take Miranda 2 + 2 or about 4 minutes to reach Casper Marina.

t 1 2 3 d 50 68 50

GRAPHINGCALCULATORGrapheachequation.Usethe TRACE feature to find the vertex on the graph. Round to the nearest thousandth if necessary.

59.y = 4x2 + 10x + 6

SOLUTION:

Let = 4x2 + 10x + 6. Use the WINDOW option to adjust the viewing window. Use the TRACE option to move to

cursortotheminimumpoint.

[-5, 5] scl: 1 by [-5, 5] scl: 1 The vertex is at (1.25, 0.25).

60.y = 8x2 8x + 8

SOLUTION:

Let Y1= 8x2 8x + 8. Use the WINDOW option to adjust the viewing window. Select the TRACE option and

movethecursortotheminimumpoint.

[-5, 5] scl: 1 by [-2, 18] scl: 2 The vertex is at (0.5, 6).

61.y = 5x2 3x 8

SOLUTION:

Let Y1 = 5x2 3x 8 Use the WINDOW option to adjust the viewing window. Select the TRACE option and

movethecursortothemaximumpoint.

[-5, 5] scl: 1 b [-20, 2] scl: 2 The vertex is at (0.3, 7.55).

62.y = 7x2 + 12x 10

SOLUTION:

Let Y1 = 7x2 + 12x 10 . Use the WINDOW option to adjust the viewing window. Select the TRACE option andmovethecursortothemaximumpoint.


63.GOLF The average amateur golfer can hit the ball with an initial upward velocity of 31.3 meters per second. The

height can be modeled by the equation h = 4.9t2 + 31.3t, where h is the height of the ball, in feet, after t seconds.

a. Graph this equation. b. At what height is the ball hit? c. What is the maximum height of the ball? d. How long did it take for the ball to hit the ground? e. State a reasonable range and domain for this situation.

eSolutions Manual - Powered by Cognero Page 1

9-1 Graphing Quadratic Functions


1.y = 2x2 + 4x 6

SOLUTION:



x y = 2x2 + 4x 6 (x, y)

3 y = 2(3)2 + 4(3) 6 = 0 (3,0)

2 y = 2(2)2 + 4(2) 6 = 6

(2,6)

1 y = 2(1)2 + 4(1) 6 =

8

(1,8)

0 y = 2(0)2 + 4(0) 6 = 6 (0,6)

1 y = 2(1)2 + 4(1) 6 = 0 (1,0)

2 y = 2(2)2 + 4(2) 6 = 10 (2,10)

2.y = x2 + 2x 1

SOLUTION:



x y = x2 + 2x 1 (x, y)

3 y = (3)2 + 2(3) 1 =

2

(3,2)

2 y = (2)2 + 2(2) 1 = 1

(2,1)

1 y = (1)2 + 2(1) 1 =

2

(1,2)

0 y = (0)2 + 2(0) 1 = 1

(0,1)

1 y = (1)2 + 2(1) 1 = 2 (1,2)

2 y = (2)2 + 2(2) 1 = 7 (2,7)

3.y = x2 6x 3

SOLUTION:



x y = x2 6x 3 (x, y)

1 y = (1)2 6(1) 3 = 4 (1,4)

0 y = (0)2 6(0) 3=3 (0,3)

1 y = (1)2 6(1) 3 = 8 (1,8)

2 y = (2)2 6(2) 3 = 11 (2,11)

3 y = (3)2 6(3) 3=12 (3,12)

4 y = (4)2 6(4) 3=11 (4,11)

5 y = (5)2 6(5) 3 = 8 (5,8)

6 y = (6)2 6(6) 3 = 3 (6,3)

7 y = (7)2 6(7) 3 = 4 (7,4)

4.y = 3x2 6x 5

SOLUTION:



x y = 3x2 6x 5 (x, y)

2 y = 3(2)2 6(2) 5 = 19

(2,19)

1 y = 3(1)2 6(1) 5 = 4 (1,4)

0 y = 3(0)2 6(0) 5 = 5 (0,5)

1 y = 3(1)2 6(1) 5=8 (1,8)

2 y = 3(2)2 6(2) 5=5 (2,5)

3 y = 3(3)2 6(3) 5 = 4 (3,4)


5.


6.


7.


8.



9.y = 3x2 + 6x 1






10.y = x2 + 2x + 1






11.y = x2 4x + 5






12.y = 4x2 8x + 9







13.y = x2 + 4x 3

SOLUTION:






14.y = x2 2x + 2

SOLUTION:






15.y = 3x2 + 6x + 3

SOLUTION:






16.y = 2x2 + 8x 6

SOLUTION:







17.f (x) = 3x2 + 6x + 3

SOLUTION:






x 1 0 1 2 3 y 6 3 6 3 6

18.f (x) = 2x2 + 4x + 1

SOLUTION:






x 1 0 1 2 3 y 5 1 3 1 5

19.f (x) = 2x2 8x 4

SOLUTION:






x 0 1 2 3 4 y 4 10 12 10 4

20.f (x) = 3x2 6x 1

SOLUTION:






x 1 0 1 2 3 y 8 1 4 1 8

21.CCSS REASONING A juggler is tossing a ball into the air. The height of the ball in feet can be modeled by the

equation y = 16x2 + 16x + 5, where y represents the height of the ball at x seconds.

a. Graph this equation. b. At what height is the ball thrown? c. What is the maximum height of the ball?

SOLUTION:

a. Step 1 Findtheequationoftheaxisofsymmetry.For y = 16x2 + 16x + 5, a = 16, b = 16, and c = 5.

Step 2Findthevertex,anddeterminewhetheritisamaximumorminimum.

The x-coordinate of the vertex is x = . Substitute the x-coordinate of the vertex into the original equation to find

the value of y .

The vertex lies at . Because a is negative, the graph opens down, and the vertex is a maximum.

Step 3Findthey-intercept. Use the original equation, and substitute 0 for x.



b. The ball is thrown when time equals 0, or at the y-intercept. So, the ball was 5 feet from the ground when it was thrown. c. The maximum height of the ball occurs at the vertex. So, the ball reaches a maximum height of 9 feet.

x 0 1

y 5 9 5


22.y = x2 + 4x + 6

SOLUTION:



x y = x2 + 4x + 6 (x, y)

4 y = (4)2 + 4(4) + 6 = 6 (4,6)

3 y = (3)2 + 4(3) + 6 = 3 (3,3)

2 y = (2)2 + 4(2) + 6 = 2 (2,2)

1 y = (1)2 + 4(1) + 6 = 3 (1,3)

0 y = (0)2 + 4(0) + 6 = 6 (0,6)

23.y = 2x2 + 4x + 7

SOLUTION:



x y = 2x2 + 4x + 7 (x, y)

3 y = 2(3)2 + 4(3) + 7 = 13 (3, 13)

2 y = 2(2)2 + 4(2) + 7 = 7 (2,7)

1 y = 2(1)2 + 4(1) + 7 = 5 (1,5)

0 y = 2(0)2 + 4(0) + 7 = 7 (0,7)

1 y = 2(1)2 + 4(1) + 7 = 13 (1,13)

24.y = 2x2 8x 5

SOLUTION:



x y = 2x2 8x 5 (x, y)

4 y = 2(4)2 8(4) 5 = 5 (4, 5)

3 y = 2(3)2 8(3) 5 = 11 (3,11)

2 y = 2(2)2 8(2) 5 = 13 (2,13)

1 y = 2(1)2 8(1) 5 = 11 (1,11)

0 y = 2(0)2 8(0) 5 = 5 (0,5)

25.y = 3x2 + 12x + 5

SOLUTION:




x y = 3x2 + 12x + 5 (x, y)

0 y = 3(0)2 +12(0) + 5 = 5 (0, 5)

1 y = 3(1)2 +12(1) + 5 = 4 (1,4)

2 y = 3(2)2 +12(2) + 5 = 7 (2,7)

3 y = 3(3)2 +12(3) + 5 = 4 (3,4)

4 y = 3(4)2 +12(4) + 5 = 5 (4, 5)

26.y = 3x2 6x 2

SOLUTION:




x y = 3x2 6x 2 (x, y)

3 y = 3(3)2 6(3) 2 = 7 (3, 7)

2 y = 3(2)2 6(2) 2 = 2 (2,2)

1 y = 3(1)2 6(1) 2 = 5 (1,5)

0 y = 3(0)2 6(0) 2 = 2 (0,2)

1 y = 3(1)2 6(1) 2 = 7 (1,7)

27.y = x2 2x 1

SOLUTION:




x y = x2 2x 1 (x, y)

3 y = (3)2 2(3) 1 = 2 (3, 2)

2 y = (2)2 2(2) 1 = 1 (2,1)

1 y = (1)2 2(1) 1 = 2 (1,2)

0 y = (0)2 2(0) 1 = 1 (0,1)

1 y = (1)2 2(1) 1 = 2 (1,2)


28.


29.


30.


31.


32.

SOLUTION:Find the vertex. Because the parabola opens up, the vertex is located at the minimum point of the parabola. It is located at (0, 4). Find the axis of symmetry. The axis of symmetry is the line that goes through the vertex and divides the parabola into congruent halves. It is located at x = 0. Find the y-intercept. The y-intercept is the point where the graph intersects the y-axis.Itislocatedat(0,4), so the y-intercept is 4.

33.


Find the vertex, the equation of the axis of symmetry, and the y-intercept of each function.

34.y = x2 + 8x + 10






35.y = 2x2 + 12x + 10






36.y = 3x2 6x + 7






37.y = x2 6x 5


In the equation y = x2 6x 5, a = 1, b = 6, and c = 5.




38.y = 5x2 + 20x + 10






39.y = 7x2 28x + 14






40.y = 2x2 12x + 6






41.y = 3x2 + 6x 18






42.y = x2 + 10x 13


In the equation y = x2 + 10x 13, a = 1, b = 10, and c = 13.




Consider each function. a. Determine whether the function has a maximum or minimum value.b. State the maximum or minimum value. c. What are the domain and range of the function?

43.y = 2x2 8x + 1

SOLUTION:






44.y = x2 + 4x 5

SOLUTION:

a. For y = x2 + 4x 5, a = 1, b = 4, and c = 5. Because a is positive, the graph opens upward, so the function has a

minimum value.




45.y = 3x2 + 18x 21

SOLUTION:

a. For y = 3x2 + 18x 21, a = 3, b = 18, and c = 21. Because a is positive, the graph opens upward, so the function





46.y = 2x2 16x + 18

SOLUTION:






47.y = x2 14x 16

SOLUTION:






48.y = 4x2 + 40x + 44

SOLUTION:

a. For y = 4x2 + 40x + 44, a = 4, b = 40, and c = 44. Because a is positive, the graph opens upward, so the function





49.y = x2 6x 5

SOLUTION:






50.y = 2x2 + 4x + 6

SOLUTION:

a. For y = 2x2 + 4x + 6, a = 2, b = 4, and c = 6. Because a is positive, the graph opens upward, so the function has a

minimum value.




51.y = 3x2 12x 9

SOLUTION:

a. For y = 3x2 12x 9, a = 3, b = 12, and c = 9. Because a is negative, the graph opens downward, so the






52.y = 3x2 + 6x 4

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = 3x2 + 6x 4, a = 3, b = 6, and c = 4.





x 1 0 1 2 3 y 13 4 1 4 13

53.y = 2x2 4x 3

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = 2x2 4x 3, a = 2, b = 4, and c = 3.





x 3 2 1 0 1 y 9 3 1 3 9

54.y = 2x2 8x + 2

SOLUTION:






x 4 3 2 1 0 y 2 8 10 8 2

55.y = x2 + 6x 6

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = x2 + 6x 6, a = 1, b = 6, and c = 6.





x 5 4 3 2 1 y 11 14 15 14 11

56.y = x2 2x + 2

SOLUTION:

Step 1 Findtheequationoftheaxisofsymmetry.For y = x2 2x + 2, a = 1, b = 2, and c = 2.





x 1 0 1 2 3 y 5 2 1 2 5

57.y = 3x2 12x + 5

SOLUTION:






x 0 1 2 3 4 y 5 4 7 4 5

58.BOATING Miranda has her boat docked on the west side of Casper Point. She is boating over to Casper Marina,

which is located directly east of where her boat is docked. The equation d = 16t2 + 66t models the distance she travels north of her starting point, where d is the number of feet and t the time traveled in minutes.

a. Graph this equation. b. What is the maximum number of feet north that she traveled? c. How long did it take her to reach Casper Marina?

SOLUTION:

a. Step 1 Findtheequationoftheaxisofsymmetry.d = 16t2 + 66t, a = 16 and b = 66.

Step 2Findthevertex,anddeterminewhetheritisamaximumorminimum. The t-coordinate of the vertex is t = 2. Substitute the t-coordinate of the vertex into the original equation to find the value of N.

The vertex lies at about (2, 68). Because a is negative, the graph opens down, and the vertex is a maximum. Step 3Findthed-intercept. Use the original equation, and substitute 0 for t.

The d-intercept is (0, 0). Step 4 The axis of symmetry divides the parabola into two equal parts. So if there is a point on one side, there is a corresponding point on the other side that is the same distance from the axis of symmetry and has the same d-value.


b. The maximum distance Miranda traveled north is the y-coordinate of the vertex, or about 68 feet. c. It takes the boat 2 minutes to reach the vertex, and due to symmetry it will take the boat another 2 minutes to reach Casper Marina. So, it will take Miranda 2 + 2 or about 4 minutes to reach Casper Marina.

t 1 2 3 d 50 68 50

GRAPHINGCALCULATORGrapheachequation.Usethe TRACE feature to find the vertex on the graph. Round to the nearest thousandth if necessary.

59.y = 4x2 + 10x + 6

SOLUTION:

Let = 4x2 + 10x + 6. Use the WINDOW option to adjust the viewing window. Use the TRACE option to move to

cursortotheminimumpoint.


60.y = 8x2 8x + 8

SOLUTION:

Let Y1= 8x2 8x + 8. Use the WINDOW option to adjust the viewing window. Select the TRACE option and

movethecursortotheminimumpoint.

[-5, 5] scl: 1 by [-2, 18] scl: 2 The vertex is at (0.5, 6).

61.y = 5x2 3x 8

SOLUTION:

Let Y1 = 5x2 3x 8 Use the WINDOW option to adjust the viewing window. Select the TRACE option and

movethecursortothemaximumpoint.

[-5, 5] scl: 1 b [-20, 2] scl: 2 The vertex is at (0.3, 7.55).

62.y = 7x2 + 12x 10

SOLUTION:

Let Y1 = 7x2 + 12x 10 . Use the WINDOW option to adjust the viewing window. Select the TRACE option andmovethecursortothemaximumpoint.


63.GOLF The average amateur golfer can hit the ball with an initial upward velocity of 31.3 meters per second. The

height can be modeled by the equation h = 4.9t2 + 31.3t, where h is the height of the ball, in feet, after t seconds.

a. Graph this equation. b. At what height is the ball hit? c. What is the maximum height of the ball? d. How long did it take for the ball to hit the ground? e. State a reasonable range and domain for this situation.

eSolutions Manual - Powered by Cognero Page 2

9-1 Graphing Quadratic Functions


1.y = 2x2 + 4x 6

SOLUTION:



x y = 2x2 + 4x 6 (x, y)

3 y = 2(3)2 + 4(3) 6 = 0 (3,0)

2 y = 2(2)2 + 4(2) 6 = 6

(2,6)

1 y = 2(1)2 + 4(1) 6 =

8

(1,8)

0 y = 2(0)2 + 4(0) 6 = 6 (0,6)

1 y = 2(1)2 + 4(1) 6 = 0 (1,0)

2 y = 2(2)2 + 4(2) 6 = 10 (2,10)

2.y = x2 + 2x 1

SOLUTION:



x y = x2 + 2x 1 (x, y)

3 y = (3)2 + 2(3) 1 =

2

(3,2)

2 y = (2)2 + 2(2) 1 = 1

(2,1)

1 y = (1)2 + 2(1) 1 =

2

(1,2)

0 y = (0)2 + 2(0) 1 = 1

(0,1)

1 y = (1)2 + 2(1) 1 = 2 (1,2)

2 y = (2)2 + 2(2) 1 = 7 (2,7)

3.y = x2 6x 3

SOLUTION:



x y = x2 6x 3 (x, y)

1 y = (1)2 6(1) 3 = 4 (1,4)

0 y = (0)2 6(0) 3=3 (0,3)

1 y = (1)2 6(1) 3 = 8 (1,8)

2 y = (2)2 6(2) 3 = 11 (2,11)

3 y = (3)2 6(3) 3=12 (3,12)

4 y = (4)2 6(4) 3=11 (4,11)

5 y = (5)2 6(5) 3 = 8 (5,8)

6 y = (6)2 6(6) 3 = 3 (6,3)

7 y = (7)2 6(7) 3 = 4 (7,4)

4.y = 3x2 6x 5

SOLUTION:



x y = 3x2 6x 5 (x, y)

2 y = 3(2)2 6(2) 5 = 19

(2,19)

1 y = 3(1)2 6(1) 5 = 4 (1,4)

0 y = 3(0)2 6(0) 5 = 5 (0,5)

1 y = 3(1)2 6(1) 5=8 (1,8)

2 y = 3(2)2 6(2) 5=5 (2,5)

3 y = 3(3)2 6(3) 5 = 4 (3,4)


5.


6.


7.


8.



9.y = 3x2 + 6x 1






10.y = x2 + 2x + 1






11.y = x2 4x + 5






12.y = 4x2 8x + 9







13.y = x2 + 4x 3

SOLUTION:






14.y = x2 2x + 2

SOLUTION:






15.y = 3x2 + 6x + 3

SOLUTION:






16.y = 2x2 + 8x 6

SOLUTION:







17.f (x) = 3x2 + 6x + 3

SOLUTION:






x 1 0 1 2 3 y 6 3 6 3 6

use a table of values to graph each equation. state the ... … · use a table of values to graph...

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