Question

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

Answer 1

Axis of symmetry x=3

Vertex = (3, 1)

x-intercept = 2 and 4.

Explanation:

`y=-x^2+6x-8`
`=-(x^2-6x)-8`
`=-(x^2-6x+(-6/2)^2)-8+(-1)(-(-6/2)^2)`
`=-(x^2-6x+9)-8+9`
`=-(x-3)^2+1`

`0=x^2-6x+8`
`0=(x-4)(x-2)`
`x=2 or 4`