Chapter 4 - Quadratic Functions - 4.2 Graphing Quadratics in Vertex Form - 4.2 Exercises - Page 317: 22

See graph

Given $$\begin{aligned} w(t) &= (t+3)^2-8. \end{aligned}$$ The vertex of the function is $(h,k)=(-3,-8)$. We determine the vertical intercept: $$w(0)=(0+3)^2-8=1.$$ The vertical intercept is $(0,1)$. We use two symmetrical points: $(-7,8)$ and $(1,8)$. The graph of the function is shown below.