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Understanding a Situation with Quadratic Equations

8.1.1 Understanding a Situation with Quadratic Equations

8.1.1 • Understanding a Situation with Quadratic Equations

Warm Up

A mechanical device is used to launch a potato vertically into the air. The potato is launched from a platform 20 feet above the ground, with an initial vertical velocity of 92 feet per second.

The function h ( t ) = − 16 t 2 + 92 t + 20 h ( t ) = − 16 t 2 + 92 t + 20 models the height of the potato over the ground, in feet, t t seconds after launch.

Here is the graph representing the function.

A graph showing height in feet versus time in seconds; the curve rises, peaks at 4 seconds and 160 feet, then descends, crossing the x axis at 6 seconds.

For each question, explain your reasoning.

1.

What is the height of the potato 1 second after launch?

Compare your answer:

The height of the potato 1 second after launch will be 96 feet. Substituting 1 second into the function h ( t ) h ( t ) yields:

h ( 1 ) = − 16 ( 1 ) 2 + 92 ( 1 ) + 2 = − 16 + 92 + 20 = 96 h ( 1 ) = − 16 ( 1 ) 2 + 92 ( 1 ) + 2 = − 16 + 92 + 20 = 96

2.

Will the potato still be in the air 8 seconds after launch?

Compare your answer:

No, from the graph, we can see the potato will land before 8 seconds after the launch.

3.

Will the potato reach 120 feet? If so, when will it happen?

Compare your answer:

Yes, from the graph, we can see the potato’s height will be 120 feet at two times: approximately 1.5 seconds and 4.3 seconds after launch.

4.

When will the potato hit the ground, or have a height of zero feet?

Compare your answer:

We can see from the graph the potato will hit the ground, or have a height of zero feet, after approximately 6 seconds.