Question

Find the vertex of this parabola y=-x^2+8x-22

Answer 1

Vertex of parabola is y=
`-x^(2) +8x-22` at (4,6)

Explanation:

The given equation of parabola is y=
`-x^(2) +8x-22`

Simplifying the equation,

y=
`-x^(2) +8x-22`

y=
`(-1)(x^(2)-8x-+22)`

y=
`(-1)(x^(2) -8x + 16-16+22)`

y=
`(-1)[(x^(2) -8x + 16)-(16-22)]`

y=
`(-1)[(x-4)^(2)+(+6)]`

y=
`(-1)(x-4)^(2)+(+6)(-1)`

y=
`(-1)(x-4)^(2)+(-6)`

The general equation of parabola is y = y=
`a(x+h)^(2)+k`

Where, (h,k) is vertex of parabola.

On comparing the equations

we get,

Vertex of parabola is y=
`-x^(2) +8x-22` at (4,-6)