Chapter 3 - Quadratic Functions - 3.2 The Vertex of a Parabola - Exercises and Problems for Section 3.2 - Exercises and Problems - Page 128: 23

See graph

Not that $m(n+3)+7=\frac{1}{2}(n+3)^2+7=\left(\frac{1}{2} n^2+3 n+\frac{9}{2}\right)+7=\frac{1}{2} n^2+3 n+\frac{23}{2}$. To sketch, shift the graph of $m(n)=\frac{1}{2} n^2$ by 3 units to the left and 7 units up, as in the figure.