Question

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

Answer 1

Vertex is at `(1,5)` with y-intercept is at `(0,6)` and having no x-intercept.

Explanation:

`y= x^2-2x+6 or y = (x-1)^2+-1+6 or y = (x-1)^2+5` Comparing with standard equation of parabola `y=a(x-h)^2+k` we find vertex as `(h,k) or 1,5`

To find y intercept putting `x=0` in the equation we get `y=6`

To find x intercept putting `y=0` in the equation we get `x^2-2x+6=0 or (x-1)^2= -5 or (x-1)= +-sqrt(-5) or x=1+-sqrt(5) i :. x` has complex roots and there will be no x-intercept. graph{x^2-2x+6 [-45, 45, -22.5, 22.5]}[Ans]