Chapter 3 - 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 249: 38

$y=2(x+3)^2-10$

Vertex: $(h,k)=(-3,-10)$ Standard form: $y=a(x-h)^2+k$ $y=a[x-(-3)]^2+(-10)$ $y=a(x+3)^2-10$ Now, use the point $(0,8)$ to find $a$: $8=a(0+3)^2-10$ $8+10=a(9)$ $a=\frac{18}{9}=2$ $y=2(x+3)^2-10$