Question

How do find the vertex and axis of symmetry for a quadratic equation `y=-x^2+2x-5`?

Answer 1

vertex at `(1,-4)`
axis of symmetry: `x=1`

Explanation:

By rewriting `y =-x^2+2x-5` into vertex form:
`color(white)("XXXX")` `y=m(x-a)^2+b` (which has a vertex at `(a,b)`)
we can easily extract the vertex.

`y=-x^2+2x-5`
`color(white)("XXXX")`extract `m`
`y = (-1)(x^2-2x) -5`
`color(white)("XXXX")`complete the square
`y=(-1)(x^2-2x+1) -5 +1`
`color(white)("XXXX")`simplify into vertex form
`y=(-1)(x-1)^2+ (-4)`

with a vertex at `(1,-4)`

Since this is a parabola in standard position (vertical axis)
the axis of symmetry runs though the x coordinate of the vertex.