Question

How do you find the vertex of a parabola `f(x) = x^2 - 2x - 3`?

Answer 1

The vertex of `f(x)` is `-4` when `x=1` graph{x^2-2x-3 [-8, 12, -8.68, 1.32]}

Explanation:

Let `a,b,c`, 3 numbers with `a!=0`

Let `p` a parabolic function such as `p(x) = a*x^2 + b*x + c`

A parabola always admit a minimum or a maximum (= his vertex).

We have a formula to find easily the abscissa of a vertex of a parabola :

Abscissa of vertex of `p(x) = -b/(2a)`



Then, the vertex of `f(x)` is when `(-(-2))/2=1`

And `f(1) = 1 - 2 - 3 = -4`


Therefore the vertex of `f(x)` is `-4` when `x=1`

Because `a>0` here, the vertex is a minimum.