Question

Micah is writing a function that models the height a dolphin reaches when it propels itself from underwater to the surface, leaps through the air, and reenters the water. The model is represented by the equation h=-16t+96t-128, where h is the height in feet above the surface of the water and t is the time in seconds. According to Micha's model, how long will the dolphin be above the surface of the water?

Answer 1

Answer:

The dolphin will be above the surface of the water for 2 seconds.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

The height of the dolphin after t seconds is given by:

$h(t) = -16t^2 + 96t - 128$

According to Micha's model, how long will the dolphin be above the surface of the water?

It stays above the surface of the water between the first and the second root. Initially, it is below water, when the first time for which $h(t) = 0$ it crosses the surface upwards, and then the second time for which $h(t) = 0$ it crosses the surface downwards.

We have to find these roots. So

$h(t) = -16t^2 + 96t - 128$

$-16t^2 + 96t - 128 = 0$

Multiplying by -16

$t^2 - 6t + 8 = 0$

$\Delta = (-6)^{2} - 4*1*8 = 36 - 32 = 4$

$t_{1} = \frac{-(-6) + \sqrt{4}}{2} = 4$

$t_{2} = \frac{-(-6) - \sqrt{4}}{2} = 2$

4 - 2 = 2

The dolphin will be above the surface of the water for 2 seconds.