Question

How do you find the vertex of `g(x)=-x^2+6x-8`?

Answer 1

Rearrange the equation into Vertex Form (or use the shortcut to find axis of symmetry and substitute x) .

Explanation:

To get the vertex for an equation your best bet is to convert the form into vertex form,which is `y=a(x-h)+k`. In vertex form `(h,k)` is the vertex.

The work to get an answer:
`y=-x^2+6x-8`
`y+8=-x^2+6x`
`y+8=-(x^2-6x)`
`y+8+9=-(x-3)^2`
`y=-(x-3)^2-17`

So the vertex is `(3,-17)`

PS; the shortcut is when the equation is in the form `y=ax^2+bx+c` is to make it `y=a(x-(-b/(2a)))^2+c/a` and just use the needed into to substitute it into it so `((-b/(2a)), c/a)` is your answer.