# What is the vertex form of y=x^2+8x-1?

The vertex form of $y = {x}^{2} + 8 x - 1$ is y=(x+4)^2-17.
First find $- \frac{b}{2} = - 4$, so -4 will be added to x inside the parentheses.
Next, find $c - {b}^{2}$ to find the value you add at the end.
$y = {\left(x - \frac{b}{2}\right)}^{2} + c - {b}^{2}$
$y = {\left(x + 4\right)}^{2} - 17$