Chapter 3 - Quadratic Functions - 3.1 Introduction to the Family of Quadratic Functions - Exercises and Problems for Section 3.1 - Exercises and Problems - Page 121: 25

$ y=\frac{7}{4}(x+2)^2 $

There is one zero at $x=-2$, so by symmetry the vertex is $(-2,0)$. We have $y=a(x+2)^2$. Solving for $a$, we have $$ \begin{aligned} a(0+2)^2 & =7 \\ 4 a & =7 \\ a & =\frac{7}{4} \end{aligned} $$ Hence $$ y=\frac{7}{4}(x+2)^2 $$