Respuesta :

check the picture below, that one use feet, but is the same thing, in this case is just centimeters.

a)

so the parabola highest point is at the vertex's y-coordinate, which one will that be?

[tex]\bf h=-4t^2+60\implies h=-4t^2+0t+60 \\\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ \begin{array}{lcccl} h = & -4t^2& +0t& +60\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(~~~~~~,~~60-\cfrac{0^2}{4(-4)} \right)\implies \left(~~~~~~,~~60-0 \right)\implies (~~~~,~60)[/tex]



b)

well, when t = 3, you'll just get h = -4(3)² + 60.

and surely you know what that is.



c)

recall that the x-intercepts occur when y = 0, in this case, when h = 0,

 [tex]\bf h=-4t^2+60\implies \stackrel{h}{0}=-4(t^2-15)\implies 0=t^2-15 \\\\\\ 0=\stackrel{\textit{difference of squares}}{t^2-(\sqrt{15})^2}\implies 0=(t-\sqrt{15})(t+\sqrt{15})\\\\ -------------------------------\\\\ 0=t-\sqrt{15}\implies \sqrt{15}=t\\\\ -------------------------------\\\\ 0=t+\sqrt{15}\implies -\sqrt{15}=t[/tex]
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