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

Jul 10, 2015

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

#### Explanation:

Let $a , b , c$, 3 numbers with $a \ne 0$

Let $p$ a parabolic function such as $p \left(x\right) = a \cdot {x}^{2} + b \cdot x + c$

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 $p \left(x\right) = - \frac{b}{2 a}$



Then, the vertex of $f \left(x\right)$ is when $\frac{- \left(- 2\right)}{2} = 1$

And $f \left(1\right) = 1 - 2 - 3 = - 4$


Therefore the vertex of $f \left(x\right)$ is $- 4$ when $x = 1$

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