Question

A rock is thrown from the side of a ledge. The height (in feet), x seconds after the toss is modeled by: h(x) = –3(x-6)2 +242 How high was the platform at the time of the launch?

Answer 1

A rock is thrown from the side of a ledge. The height (in feet), x seconds after the toss is modeled by: h(x) = –3(x-6)2 +242 How high was the platform at the time of the launch?​

we have the equation

`h(x)=-3(x-6)^2_{}+242`

The high of the plattform a t the time of the lauch is for x=0

so

you need to calculate the y-intercept

therefore

For x=0

substitute

`\begin{gathered} h(x)=-3(0-6)^2_{}+242 \\ h(x)=-3(36)^{}_{}+242 \\ f(x)=134 \end{gathered}`

the answer is 134 ft