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. a) the factored expression is - 16t(t - 6). (factor completely.) b) - 16(2)^2+96(2)=□

Explanation:

Step1: Identify the original expression

The original height - time function is \(h(t)=-16t^{2}+96t\).

Step2: Factor out the common factor

We can factor out \(- 16t\) from the expression \(-16t^{2}+96t\). So, \(h(t)=-16t(t - 6)\).

Step3: Evaluate the original and factored - form at \(t = 2\)

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

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

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

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

\]

Step4: Calculate \(-16(2)^{2}+96(2)\)

\[

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

\]

Answer:

128