Question

suppose that an object is thrown upward from ground level with an initial velocity of 96 ft/sec. its height after t seconds is a function h given by h(t)= - 16t^2 + 96t. a) find an equivalent expression for h(t) by factoring out a common factor with a negative coefficient. b) check your factoring by evaluating both expressions for h(t) at t = 2.

Explanation:

Step1: Factor the function

The height - function is \(h(t)=-16t^{2}+96t\). We can factor out \(-16t\) (a common factor with a negative coefficient). So, \(h(t)=-16t(t - 6)\).

Step2: Evaluate the original and factored - form at \(t = 2\)

For the original function \(h(t)=-16t^{2}+96t\), when \(t = 2\), we substitute \(t\) into the function:
\[

$$\begin{align*} h(2)&=-16\times2^{2}+96\times2\\ &=-16\times4 + 192\\ &=-64+192\\ &=128 \end{align*}$$

\]
For the factored - form \(h(t)=-16t(t - 6)\), when \(t = 2\):
\[

$$\begin{align*} h(2)&=-16\times2\times(2 - 6)\\ &=-32\times(-4)\\ &=128 \end{align*}$$

\]

Answer:

a) The factored form of \(h(t)=-16t^{2}+96t\) is \(h(t)=-16t(t - 6)\).
b) When \(t = 2\), both the original function \(h(t)=-16t^{2}+96t\) and the factored function \(h(t)=-16t(t - 6)\) evaluate to \(128\).