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

$V \left(0 , - 8\right)$
As you know from linear graphs the constant at the end (in this case -8) controls the y-intercept. By default on $y = {x}^{2}$ The vertex is $\left(0 , 0\right)$ so by moving the vertex down 8 units the new vertex coordinates are $\left(0 , - 8\right)$.