Question

What is the the vertex of `y = x^2 - x - 6`?

Answer 1

`(1/2, -13/2)`

Explanation:

Vertex of a parabola in the form `ax^2+bx+c` is given by:
`x=-b/(2a)`
Note this is only gives the `x`-coordinate; we will have to evaluate this value to get the `y`-coordinate.

Our parabola `x^2-x-6` has `a=1`, `b=-1`, and `c=-6`. Using the vertex formula above, we see:
`x=-(-1)/(2(1))=1/2`

Evaluating `y` at this value:
`y=(1/2)^2-(1/2)-6=1/4-1/2-6=-13/2`

Therefore our vertex occurs at the point `(1/2, -13/2)`.