## Algebra and Trigonometry 10th Edition

$y=2(x+3)^2-8$
Vertex: $(h,k)=(-3,-8)$ Standard form: $y=a(x-h)^2+k$ $y=a[x-(-3)]^2+(-8)$ $y=a(x+3)^2-8$ Now, use the point $(-6,10)$ to find $a$: $10=a(-6+3)^2-8$ $10+8=a(9)$ $a=\frac{18}{9}=2$ $y=2(x+3)^2-8$