## Algebra 1: Common Core (15th Edition)

$y = -11.125$
The vertex form of a quadratic function is $y = a(x-h)^2+k$, where $(h,k)$ is the vertex. Since the coefficient of $x^2$ is $2$, $a$ must be equal to 2. $y = 2(x-h)^2 + k$ If we divide the right side of the function by $2$, we will get $x^2 + 2.5x-4$. Try to make this a perfect square. $(x^2+2.5x + 1.25^2) - 1.25^2 -4$ $(x+1.25)^2 - 1.5625 - 4$ $(x+1.25)^2 - 5.5625$ Multiply this by $2$ to get the right side of the function. $y = 2(x+1.25)^2 -11.125$ The coordinates of the vertex are $(-1.25,\boxed{-11.125})$