## Intermediate Algebra (6th Edition)

$f(x)=5(x+3)^{2}+6$
The vertex-form of a quadratic equation is $y=a(x-h)^{2}+k$, with a vertex at the opposite of $h$ and $k$, or $(h, k)$. The original equation in this problem is $f(x)=5x^{2}$; written in vertex-form it is $f(x)=5(x-0)^{2}+0$ and has a vertex of $(0, 0)$. To translate this parabola so that the vertex is at $(-3, 6)$ we need to rewrite the equation as $f(x)=5(x-(-3))^{2}+6$ or $f(x)=5(x+3)^{2}+6$.