Chapter 3 - Review Exercises - Page 300: 15

$y=-\frac{1}{2}(x-4)^2+1$

Vertex: $(h,k)=(4,1)$ Standard form: $y=a(x-h)^2+k$ $y=a(x-4)^2+1$ Now, use the point $(2,-1)$ to find $a$: $-1=a(2-4)^2+1$ $-1-1=a(4)$ $a=\frac{-2}{4}=-\frac{1}{2}$ $y=-\frac{1}{2}(x-4)^2+1$