Question

Which is the vertex of `x^2+10x=-17`?

Answer 1

`(-5,-8)`

Explanation:

`x^2+10x=-17`
`0=-x^2-10x-17`
`0=-[x^2+10x+17]`
`0=-[(x+5)^2-8]`
`0=-(x+5)^2+8`

`x^2+10x+17=0`
`(x+5)^2-8=0`

Vertex is at `x = -5`. It is unclear whether the coefficient of the highest degree is positive or negative. If the parabola is negative, then the vertex is at `(-5,8)`. If the parabola is positive, the vertex is at `(-5,-8)`