Quadratic problem - verifying answer approach

[tathagata7]

Greetings:

I would like to verify my work answer for the below SSAT question.

Thank you


Robert throws a rock into the air. The rock’s height, y, in feet with respect to time, x, in seconds, can be modeled by the function y=−2x^2 +4x +16 How many seconds does it take the rock to hit the ground after Robert threw it?

Test Answer 4

My work:

y=−2x^2 +4x +16

-2(x^2 -2x -8) = 0 (is it correct to divide everything by -2)

(x-4) (x+2) = 0

(4, -2)

Since answer has to be positive it’s 4

 

[Steven G]

Greetings:

I would like to verify my work answer for the below SSAT question.

Thank you


Robert throws a rock into the air. The rock’s height, y, in feet with respect to time, x, in seconds, can be modeled by the function y=−2x^2 +4x +16 How many seconds does it take the rock to hit the ground after Robert threw it?

Test Answer 4

My work:

y=−2x^2 +4x +16

-2(x^2 -2x -8) = 0 (is it correct to divide everything by -2)

(x-4) (x+2) = 0

(4, -2)

Since answer has to be positive it’s 4

Yes, your work is correct. Of course -2 is multiplied by something ad the result is 0 then that something must equal 0.

 

[tathagata7]

Yes, your work is correct. Of course -2 is multiplied by something ad the result is 0 then that something must equal 0.

Thank you so much for verifying. If you have a moment, I just posted a challenging triangle perimeter problem in the Geometry section.

 

[hoosie]

Quadratic Problem - Projectile Motion

Robert throws a rock into the air. The rock’s height, y, in feet with respect to time, x, in seconds, can be modeled by the function y=−2x^2 +4x +16 How many seconds does it take the rock to hit the ground after Robert threw it?


Good question. We have to assume here the rock is projected from the ground. The graph of y=−2x^2 +4x +16 is shown below:

Your working was correct to find the graph's x-intercepts:


−2x^2 +4x +16 = 0
=> -2(x^2 -2x -8) = 0
=> x^2 -2x -8 = 0
=> (x - 4)(x + 2) = 0
=> x = 4 or x = -2

As you can see from the graph the stone (starting at x = -2) takes 2 sec to reach a height of 16 ft (y-intercept: when x = 0, y = 16), a further 1 sec to reach maximum height of 18m (when x = 1, y = -2(1)^2 + 4(1) + 16 => y = 18) and then another 3 sec to reach the ground again (when x = 4, y = 0).
This means the rock is in the air for 6 sec.

Note: If the rock was projected from a point 16 ft above the ground (when x = 0 on the graph) it would take 4 sec to hit the ground (1 sec to reach max height, further 3 sec to reach ground).

Robert is clearly not 16 ft tall so the answer in this case cannot be 4sec. We have to assume Robert is kneeling down and flipping the rock upwards from ground level. If Robert was standing on a platform 13 ft high and throwing the rock from a height of 3 ft above the platform then it would take 4 sec for the rock to hit the ground.

Note: The best way to understand the what a negative time value of -2 means in the original question is to think of the stop watch not starting until the rock reaches a height of 16ft (at x = 0). At x = -2 the rock is at ground level 2 sec before the stop watch starts. If the stop watch starts from the moment the stone leaves the ground then the parabolic graph would pass through the origin.

By completing the square:
y = -2(x^2 - 2x - 8)
= -2{(x^2 - 2x + 1) - 1 - 8}
= -2(x - 1)^2 + 18
Max TP occurs at (1, 18)

To make this graph pass through the origin we need to shift it 2 units to the right. The equation becomes y = -2(x - 3)^2 + 18
with max TP occurs at (3, 18).

If the rock follows the path shown by the graph y = -2x^2 + 12x it is in the air for 6 sec. It takes 3 sec to reach max height. In this case the stop watch starts from the moment the rock is thrown.

 

[tathagata7]

Robert throws a rock into the air. The rock’s height, y, in feet with respect to time, x, in seconds, can be modeled by the function y=−2x^2 +4x +16 How many seconds does it take the rock to hit the ground after Robert threw it?


Good question. We have to assume here the rock is projected from the ground. The graph of y=−2x^2 +4x +16 is shown below:

Your working was correct to find the graph's x-intercepts:


−2x^2 +4x +16 = 0
=> -2(x^2 -2x -8) = 0
=> x^2 -2x -8 = 0
=> (x - 4)(x + 2) = 0
=> x = 4 or x = -2

As you can see from the graph the stone (starting at x = -2) takes 2 sec to reach a height of 16 ft (y-intercept: when x = 0, y = 16), a further 1 sec to reach maximum height of 18m (when x = 1, y = -2(1)^2 + 4(1) + 16 => y = 18) and then another 3 sec to reach the ground again (when x = 4, y = 0).
This means the rock is in the air for 6 sec.

Note: If the rock was projected from a point 16 ft above the ground (when x = 0 on the graph) it would take 4 sec to hit the ground (1 sec to reach max height, further 3 sec to reach ground).

Robert is clearly not 16 ft tall so the answer in this case cannot be 4sec. We have to assume Robert is kneeling down and flipping the rock upwards from ground level. If Robert was standing on a platform 13 ft high and throwing the rock from a height of 3 ft above the platform then it would take 4 sec for the rock to hit the ground.

Note: The best way to understand the what a negative time value of -2 means in the original question is to think of the stop watch not starting until the rock reaches a height of 16ft (at x = 0). At x = -2 the rock is at ground level 2 sec before the stop watch starts. If the stop watch starts from the moment the stone leaves the ground then the parabolic graph would pass through the origin.

By completing the square:
y = -2(x^2 - 2x - 8)
= -2{(x^2 - 2x + 1) - 1 - 8}
= -2(x - 1)^2 + 18
Max TP occurs at (1, 18)

To make this graph pass through the origin we need to shift it 2 units to the right. The equation becomes y = -2(x - 3)^2 + 18
with max TP occurs at (3, 18).

If the rock follows the path shown by the graph y = -2x^2 + 12x it is in the air for 6 sec. It takes 3 sec to reach max height. In this case the stop watch starts from the moment the rock is thrown.

Thanks for the additional explanation. Can you resend your graph as image files? It's not visible in the post. Thank you