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
[tex]h(x)=-3(x-6)^2_{}+242[/tex]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
[tex]\begin{gathered} h(x)=-3(0-6)^2_{}+242 \\ h(x)=-3(36)^{}_{}+242 \\ f(x)=134 \end{gathered}[/tex]the answer is 134 ft
