Use a table of values to graph each equation. State …...Use a table of values to graph each equation. State the domain and range. y = 2 x2 + 4 x í 6 62/87,21 Graph the ordered pairs,

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:



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.

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, −12). 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, –4), so the y-intercept is –4.

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. It is located at (0, 5), so the y-intercept is 5.

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

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.



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).

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

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 x-coordinate of the vertex is x = –1. Substitute this value into the function to find the y-coordinate.

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 Find the equation of the axis of symmetry. For f (x) = −3x2 + 6x + 3, a = –3, b = 6, and c = 3.

Step 2 Find the vertex, and determine whether it is a maximum or minimum. 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 3 Find the y-intercept. Use the original equation, and substitute 0 for x.

The y-intercept is (0, 3). 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.

Step 5 Connect the points with a smooth curve.

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:

Step 1 Find the equation of the axis of symmetry. 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 3 Find the y-intercept. Use the original equation, and substitute 0 for x.

The y-intercept is (0, –4). 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 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 Find the equation of the axis of symmetry. For y = −16x2 + 16x + 5, a = –16, b = 16, and c = 5.

Step 2 Find the vertex, and determine whether it is a maximum or minimum.

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 3 Find the y-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:


The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 2}.

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.


29.


30.

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, 4), so the y-intercept is 4.

31.

SOLUTION: Find the vertex. Because the parabola opens up, the vertex is located at the minimum point of the parabola. It is located at (1, 1). 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, 4), so the y-intercept is 4.

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. It is located at (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


In the equation y = x2 + 8x + 10, a = 1, b = 8, and c = 10.


The x-coordinate of the vertex is x = –4. Substitute the x-coordinate of the vertex into the original equation to find the value of y .

The vertex is at (–4, –6). Find the axis of symmetry. The axis of symmetry is the vertical line that goes through the vertex. It is located at x = –4. Find the y-intercept. The y-intercept always occurs at (0, c). Since c = 10 for this equation, the y-intercept is located at (0, 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


In the equation y = −3x2 − 6x + 7, a = –3, b = –6, and c = 7.



The vertex is at (–1, 10). 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 = 7 for this equation, the y-intercept is located at (0, 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





The vertex is at (2, –14). Find the axis of symmetry. The axis of symmetry is the vertical line that goes through the vertex. It is located at x = 2. Find the y-intercept. The y-intercept always occurs at (0, c). Since c = 14 for this equation, the y-intercept is located at (0, 14).

40. y = 2x2 − 12x + 6






41. y = −3x2 + 6x − 18





The vertex is at (1, –15). 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 = –18 for this equation, the y-intercept is located at (0,–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.



The minimum value is 4. 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) ≥ 4}.

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 Find the equation of the axis of symmetry. For y = −3x2 + 6x − 4, a = –3, b = 6, and c = –4.


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



x –1 0 1 2 3 y –13 –4 –1 –4 –13

53. y = −2x2 − 4x − 3

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = −2x2 − 4x − 3, a = –2, b = –4, and c = –3.

Step 2 Find the vertex, and determine whether it is a maximum or minimum. 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, –1). Because a is negative, the graph opens down, and the vertex is a maximum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



x –3 –2 –1 0 1 y –9 –3 –1 –3 –9

54. y = −2x2 − 8x + 2

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = −2x2 − 8x + 2, a = –2, b = –8, and c = 2.


The vertex lies at (–2, 10). Because a is negative, the graph opens down, and the vertex is a maximum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



x –4 –3 –2 –1 0 y 2 8 10 8 2

55. y = x2 + 6x − 6

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = x2 + 6x − 6, a = 1, b = 6, and c = –6.


The vertex lies at (–3, –15). Because a is positive, the graph opens up, and the vertex is a minimum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



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

56. y = x2 − 2x + 2

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = x2 − 2x + 2, a = 1, b = –2, and c = 2.


The vertex lies at (1, 1). Because a is positive, the graph opens up, and the vertex is a minimum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



x –1 0 1 2 3 y 5 2 1 2 5

57. y = 3x2 − 12x + 5

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = 3x2 − 12x + 5, a = 3, b = –12, and c = 5.


The vertex lies at (2, −7). Because a is positive, the graph opens up, and the vertex is a minimum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



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 the Casper

Marina. The distance traveled by Miranda over time can be modeled by the equation d = −16t2 + 66t, where d is the number of feet she travels in t 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 Find the equation of the axis of symmetry. d = −16t2 + 66t, a = –16 and b = 66.

Step 2 Find the vertex, and determine whether it is a maximum or minimum. 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 3 Find the d-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

GRAPHING CALCULATOR Graph each equation. Use the 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

cursor to the minimum point.

[-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

move the cursor to the minimum point.

[-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

move the cursor to the maximum point.

[-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 and move the cursor to the maximum point.

[-5, 5] scl: 1 by [-20, 2] scl: 2 The vertex is at (0.857, –4.857).

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.

SOLUTION:

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.

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.


7.

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, −12). 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, –4), so the y-intercept is –4.

8.


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

9. y = −3x2 + 6x − 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






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

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





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






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 Find the equation of the axis of symmetry. For y = −16x2 + 16x + 5, a = –16, b = 16, and c = 5.

Step 2 Find the vertex, and determine whether it is a maximum or minimum.

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 3 Find the y-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.

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, 4), so the y-intercept is 4.

31.

SOLUTION: Find the vertex. Because the parabola opens up, the vertex is located at the minimum point of the parabola. It is located at (1, 1). 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, 4), so the y-intercept is 4.

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. It is located at (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


In the equation y = x2 + 8x + 10, a = 1, b = 8, and c = 10.




35. y = 2x2 + 12x + 10






36. y = −3x2 − 6x + 7


In the equation y = −3x2 − 6x + 7, a = –3, b = –6, and c = 7.



The vertex is at (–1, 10). 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 = 7 for this equation, the y-intercept is located at (0, 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





The vertex is at (1, –15). 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 = –18 for this equation, the y-intercept is located at (0,–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.



The minimum value is 4. 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) ≥ 4}.

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 Find the equation of the axis of symmetry. For y = −3x2 + 6x − 4, a = –3, b = 6, and c = –4.


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



x –1 0 1 2 3 y –13 –4 –1 –4 –13

53. y = −2x2 − 4x − 3

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = −2x2 − 4x − 3, a = –2, b = –4, and c = –3.


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



x –3 –2 –1 0 1 y –9 –3 –1 –3 –9

54. y = −2x2 − 8x + 2

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = −2x2 − 8x + 2, a = –2, b = –8, and c = 2.


The vertex lies at (–2, 10). Because a is negative, the graph opens down, and the vertex is a maximum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



x –4 –3 –2 –1 0 y 2 8 10 8 2

55. y = x2 + 6x − 6

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = x2 + 6x − 6, a = 1, b = 6, and c = –6.


The vertex lies at (–3, –15). Because a is positive, the graph opens up, and the vertex is a minimum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



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

56. y = x2 − 2x + 2

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = x2 − 2x + 2, a = 1, b = –2, and c = 2.


The vertex lies at (1, 1). Because a is positive, the graph opens up, and the vertex is a minimum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



x –1 0 1 2 3 y 5 2 1 2 5

57. y = 3x2 − 12x + 5

SOLUTION:

Step 1 Find the equation of the axis of symmetry. For y = 3x2 − 12x + 5, a = 3, b = –12, and c = 5.


The vertex lies at (2, −7). Because a is positive, the graph opens up, and the vertex is a minimum. Step 3 Find the y-intercept. Use the original equation, and substitute 0 for x.



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 the Casper

Marina. The distance traveled by Miranda over time can be modeled by the equation d = −16t2 + 66t, where d is the number of feet she travels in t 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 Find the equation of the axis of symmetry. d = −16t2 + 66t, a = –16 and b = 66.

Step 2 Find the vertex, and determine whether it is a maximum or minimum. 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 3 Find the d-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

GRAPHING CALCULATOR Graph each equation. Use the 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

cursor to the minimum point.

[-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

move the cursor to the minimum point.

[-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

move the cursor to the maximum point.

[-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 and move the cursor to the maximum point.

[-5, 5] scl: 1 by [-20, 2] scl: 2 The vertex is at (0.857, –4.857).

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.

SOLUTION:

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.

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

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Use a table of values to graph each equation. State …...Use a table of values to graph each equation. State the domain and range. y = 2 x2 + 4 x í 6 62/87,21 Graph the ordered pairs,