Chapter 4 - Quadratic Functions - 4.2 Graphing Quadratics in Vertex Form - 4.2 Exercises - Page 317: 21

See graph

Given $$\begin{aligned} f(x) &= (x+4)^2+ 2. \end{aligned}$$ The vertex of the function is $(h,k)=(-4,2)$. We determine the vertical intercept: $$f(0)=(0+4)^2+2=18.$$ The vertical intercept is $(0,18)$. We use two symmetrical points: $(1,27)$ and $(-9,27)$. The graph of the function is shown below.