# How do you find the vertex and intercepts for y=x^2+8x-7?

Vertex is at $\left(- 4 , - 23\right)$ Y intercept is at $\left(0 , - 7\right)$ X-intercepts are at $\left(0.796 , 0\right)$ and $\left(- 8.796 , 0\right)$
$y = {x}^{2} + 8 x - 7 = {x}^{2} + 8 x + 16 - 23 = {\left(x + 4\right)}^{2} - 23$ So vertex is at $\left(- 4 , - 23\right)$
y-intercept : putting x=0 in the equation we get $y = - 7$
x-intercept : Putting y=0 we get ${x}^{2} = 8 x - 7 = 0$ solving the quadratic equation we get $x = 0.796$ and $x = - 8.796$ graph{x^2+8x-7 [-80, 80, -40, 40]}[Answer]