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An Olympic pole vaulter's height can be modeled by the equation f ( x ) = −16 x 2 + 32 x , where x is the time in seconds after the vaulter leaves the ground. Find the vaulter's maximum height and the time it takes the vaulter to reach this height. Then find how long the vaulter is in the air. At 1 second, the vaulter has reached a maximum height of 48 feet. The vaulter is in the air for 2 seconds. At 2 seconds, the vaulter has reached a maximum height of 16 feet. The vaulter is in the air for 4 seconds. At 2 seconds, the vaulter has reached a maximum height of 32 feet. The vaulter is in the air for 4 seconds. At 1 second, the vaulter has reached a maximum height of 16 feet. The vaulter is in the air for 2 seconds. f ( x ) = −16 x 2 + 32 x Step 1 Find the axis of symmetry. The axis of symmetry is x = 1 . Step 2 Find the vertex. y = −16( 1 ) 2 + 32( 1 ) y = −16 + 32
y = 16 The vertex is (1, 16) . The vaulter's maximum height is 16 ft and it takes 1 s to reach this height. The vaulter takes the same time to fall back to the ground. Therefore the vaulter is in the air for 2 s. 12 / 12 Question 2 : 11 pts Graph y = −3 x 2 − 3 x + 1. Graph y = −3 x 2 − 3 x + 1.
y = −3 x 2 − 3 x + 1
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Step 1 Find the axis of symmetry. The axis of symmetry is x = −1/2 . Step 2 Find the vertex. y = −3( −1/2 ) 2 − 3( −1/2 ) + 1 y = −3/4 + 6/4 + 4/4 y = 7/4 The vertex is (−1/2, 7/4) . Step 3 Find the y -intercept. y = −3(0) 2 − 3(0) + 1 y = 1 The y -intercept is 1 ; the graph passes through (0, 1 ).
Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y -intercept. y = −3(0 . 5) 2 − 3(0 . 5) + 1 = −1 . 25 y = −3(1) 2 − 3(1) + 1 = −5 Two other points are ( 0 . 5, −1 . 25 ) and ( 1, −5 ). Step 5 Graph the axis of symmetry , the vertex , the point containing the y -intercept , and two other points . Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
11 / 11 Question 3 : 11 pts Identify the graph of the quadratic function y = x 2 + 6 x + 7. Identify the graph of the quadratic function y = x 2 + 6 x + 7.
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y = x 2 + 6 x + 7 Step 1 Find the axis of symmetry. The axis of symmetry is x = −3 .
Step 2 Find the vertex. y = ( −3 ) 2 + 6( −3 ) + 7 y = 9 − 18 + 7 y = −2 The vertex is (−3, −2) . Step 3 Find the y -intercept. y = 0 2 + 6(0) + 7 y = 7 The y -intercept is 7 ; the graph passes through (0, 7 ). Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y -intercept. y = (−2) 2 + 6(−2) + 7 = −1 y = (−1) 2 + 6(−1) + 7 = 2
Two other points are ( −2, −1 ) and ( −1, 2 ). Step 5 Graph the axis of symmetry , the vertex , the point containing the y -intercept , and two other points . Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. 11 / 11 Question 4 : 11 pts Skip to question text. The height in feet of a toy rocket is modeled by the function y = −16 t 2 + 128 t , where t is the time in seconds after it is launched. Find the maximum height of the rocket and the time it takes the rocket to reach this height. Then find how long the rocket will be in air.
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256 ft; 4 s; 8 s 356 ft; 6 s; 12 s 312 ft; 5 s; 10 s 200 ft; 2 s; 4 s y = −16 t 2 + 128 t Step 1 Find the axis of symmetry. The axis of symmetry is x = 4 . Step 2 Find the vertex. y = −16( 4 ) 2 + 128( 4 ) y = −256 + 512 y = 256 The vertex is (4, 256) . The rocket's maximum height is 256 ft and it takes 4 s to reach this height.
The rocket takes the same time to fall back to the ground. Therefore it is in the air for 8 s. 11 / 11 Question 5 : 11 pts Graph y = − x 2 + 4 x − 3 . Graph y = − x 2 + 4 x − 3 .
y = − x 2 + 4 x − 3
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Step 1 Find the axis of symmetry. The axis of symmetry is x = 2 . Step 2 Find the vertex. y = −( 2 ) 2 + 4( 2 ) − 3 y = −4 + 8 − 3 y = 1 The vertex is (2, 1) . Step 3 Find the y -intercept. y = 0 2 + 4(0) − 3 y = −3 The y -intercept is −3 ; the graph passes through (0, −3 ).
Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y -intercept. y = −(−1) 2 + 4(−1) − 3 = −8 y = −(1) 2 + 4(1) − 3 = 0 Two other points are ( −1, −8 ) and ( 1, 0 ). Step 5 Graph the axis of symmetry , the vertex , the point containing the y -intercept , and two other points . Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
11 / 11 Question 6 : 11 pts Graph y = 3 x 2 + 4 x − 1 . Graph y = 3 x 2 + 4 x − 1 .
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y = 3 x 2 + 4 x − 1 Step 1 Find the axis of symmetry. The axis of symmetry is x = −2/3 .
Step 2 Find the vertex. y = 3( −2/3 ) 2 + 4( −2/3 ) − 1 y = 4/3 − 8/3 − 3/3 y = −7/3 The vertex is (−2/3, −7/3) . Step 3 Find the y -intercept. y = 3(0) 2 + 4(0) − 1 y = −1 The y -intercept is −1 ; the graph passes through (0, −1 ). Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y -intercept. y = 3(0 . 5) 2 + 4(0 . 5) − 1 = 1 . 75 y = 3(1) 2 + 4(1) − 1 = 6
Two other points are ( 0 . 5, 1 . 75 ) and ( 1, 6 ). Step 5 Graph the axis of symmetry , the vertex , the point containing the y -intercept , and two other points . Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. 0 / 11 Question 7 : 11 pts Skip to question text.
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The height in feet of a baseball can be modeled by the function y = −16 t 2 + 64 t , where t is the time in seconds after the ball is hit. Find the baseball's maximum height and the time it takes to reach this height. Then find how long the baseball is in the air. 140 ft; 3 s; 6 s 64 ft; 2 s; 4 s 164 ft; 4 s; 8 s 100 ft; 1 s; 2 s y = −16 t 2 + 64 t Step 1 Find the axis of symmetry. The axis of symmetry is x = 2 . Step 2 Find the vertex. y = −16( 2 ) 2 + 64( 2 ) y = −64 + 128 y = 64
The vertex is (2, 64) . The baseball's maximum height is 64 ft and it takes 2 s to reach this height. The baseball takes the same time to fall back to the ground. Therefore it is in the air for 4 s. 11 / 11 Question 8 : 11 pts Graph y = 2 x 2 + 3 x −1 . Graph y = 2 x 2 + 3 x −1 .
y = 2 x 2 + 3 x −1
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Step 1 Find the axis of symmetry. The axis of symmetry is x = −3/4 . Step 2 Find the vertex. y = 2( −3/4 ) 2 + 3( −3/4 ) − 1 y = 9/8 − 18/8 − 8/8 y = −17/8 The vertex is (−3/4, −17/8) . Step 3 Find the y -intercept. y = 2(0) 2 + 3(0) − 1 y = −1 The y -intercept is −1 ; the graph passes through (0, −1 ).
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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y -intercept. y = 2(0 . 5) 2 + 3(0 . 5) − 1 = 1 y = 2(1) 2 + 3(1) − 1 = 4 Two other points are ( 0 . 5, 1 ) and ( 1, 4 ). Step 5 Graph the axis of symmetry , the vertex , the point containing the y -intercept , and two other points . Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
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11 / 11 Question 9 : 11 pts Graph y = − x 2 − 4 x − 3 . Graph y = − x 2 − 4 x − 3 .
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y = − x 2 − 4 x − 3 Step 1 Find the axis of symmetry. The axis of symmetry is x = −2 .
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Step 2 Find the vertex. y = −( −2 ) 2 − 4( −2 ) − 3 y = −4 + 8 − 3 y = 1 The vertex is (−2, 1) . Step 3 Find the y -intercept. y = 0 2 − 4(0) − 3 y = −3 The y -intercept is −3 ; the graph passes through (0, −3 ). Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y -intercept. y = −(−1) 2 − 4(−1) − 3 = 0 y = −(1) 2 − 4(1) − 3 = −8
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Two other points are ( −1, 0 ) and ( 1, −8 ). Step 5 Graph the axis of symmetry , the vertex , the point containing the y -intercept , and two other points . Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
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