Chapter 4 - Quadratic Functions - 4.2 Graphing Quadratics in Vertex Form - 4.2 Exercises - Page 317: 25

See graph

Given $$\begin{aligned} h(x) &= x^2+4. \end{aligned}$$ The vertex of the function is $(0,4)$. We determine the vertical intercept: $$h(0)=-0^2+4=4.$$ The vertical intercept is $(0,4)$. We use two pairs of symmetrical points: $(-1,5)$, $(1,5)$ and $(-3,13)$, $(3,13)$. The graph of the function is shown below.