# What is the the vertex of y = (x-3)(x+5) -x+12?

$\left(- \frac{1}{2} , - \frac{13}{4}\right)$
The equation x-alpha)^2 = $4 a \left(y - \beta\right)$ represents the parabola with vertex at $\left(\alpha , \beta\right)$
The focus is at $\left(\alpha , \beta + a\right)$.
Our equation is equivalent to ${\left(x + \frac{1}{2}\right)}^{2}$ = $4 \left(\frac{1}{4}\right) \left(y + \frac{13}{4}\right)$
Vertex is $\left(- \frac{1}{2} , - \frac{13}{4}\right)$
Focus is $\left(- \frac{1}{2} , - 3\right)$ .