Chapter 3 - Quadratic Functions - 3.2 The Vertex of a Parabola - Exercises and Problems for Section 3.2 - Exercises and Problems - Page 128: 32

$ y=-\frac{2}{49}\cdot(x-4)^2+2 $

We have $(h, k)=(4,2)$, and $y=a(x-4)^2+2$. Solve for $a$ $$ \begin{aligned} a(11-4)^2+2 & =0 \\ 49 a & =-2 \\ a & =-\frac{2}{49} \end{aligned} $$ Hence $$ y=-\frac{2}{49}\cdot(x-4)^2+2 $$