# How do you find the vertex and the intercepts for  y = x^2-6x+8?

Vertex is at $\left(3 , - 1\right)$ ; y-intercept is at $\left(0 , 8\right)$ and x-intercepts are at $\left(2 , 0\right) \mathmr{and} \left(4 , 0\right)$
We know the equation of parabola in vertex form is$y = a {\left(x - h\right)}^{2} + k$ where vertex is at $\left(h , k\right)$.Here
$y = {x}^{2} - 6 x + 8 = {\left(x - 3\right)}^{2} - 9 + 8 = {\left(x - 3\right)}^{2} - 1 \therefore$Vertex is at $\left(3 , - 1\right)$ we find y-intercept by putting x=0 in the equation. So $y = 0 - 0 + 8 = 8$ and x-intercept by putting y=0 in the equation. So x^2-6x+8=0 or (x-4)(x-2)=0 or x=4 ; x=2 graph{x^2-6x+8 [-20, 20, -10, 10]}[Ans]