# How do you find the vertex of y = x^2 - 4x - 12?

May 31, 2016

It is in $\left(2 , - 16\right)$.

#### Explanation:

The vertex is the minimum or the maximum of a parabola. Then it is enough to search the zero of the derivative. We want to solve

$\frac{d}{\mathrm{dx}} \left({x}^{2} - 4 x - 12\right) = 0$
that is
$2 x - 4 = 0$
and
$x = 2$.
To find the $y$ it is enough to substitute in the original equation

$y = {2}^{2} - 4 \cdot 2 - 12$
$y = - 16$.

The vertex is then in the point $\left(2 , - 16\right)$ and we can verify it looking at the plot.

graph{x^2-4x-12 [-9.255, 10.735, -17.9, -7.89]}

May 31, 2016

Vertex at $\left(2 , - 16\right)$

#### Explanation:

The general vertex form for a parabola is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Converting the given equation $y = {x}^{2} - 4 x - 12$ into vertex form:

(extracting the $\textcolor{g r e e n}{m}$ and completing the square:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{1} \left({x}^{2} - 4 x + 4\right) - 12 - 4$

(into vertex form)
$\textcolor{w h i t e}{\text{XXX")y=color(green)(m)(x-color(red)(2))^2+color(blue)(} \left(- 16\right)}$
with vertex at (color(red)(2),color(blue)(""(-16)))

For verification purposes, here is the graph of the given equation:
graph{x^2-4x-12 [-13.07, 18.96, -16.62, -0.59]}