Question

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

Answer 1

Vertex is at `(3, -1)` ; y-intercept is at `(0,8)` and x-intercepts are at `(2,0) and (4,0)`

Explanation:

We know the equation of parabola in vertex form is`y=a(x-h)^2+k` where vertex is at `(h,k)`.Here
`y=x^2-6x+8= (x-3)^2-9+8=(x-3)^2-1 :.`Vertex is at `(3, -1)` 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]