Chapter 4 - Quadratic Functions - 4.2 Graphing Quadratics in Vertex Form - 4.2 Exercises - Page 317: 26

See graph

Given $$\begin{aligned} r(t) &= 2t^2-12. \end{aligned}$$ The vertex of the function is $(h,k)=(0,-12)$. We determine the vertical intercept: $$r(0)=2(0^2)-12=-12.$$ The vertical intercept is $(0,-12)$. We use two pairs of symmetrical points: $(-1,-10)$, $(1,10)$ and $(-3,6)$, $(3,6)$. The graph of the function is shown below.