# How do you find the vertex and the intercepts for f(x)=x^2-6x-6?

Oct 31, 2017

vertex: $\left(3 , - 15\right)$
y-intercept: $\left(- 6\right)$
x-intercepts: $\left(3 + \sqrt{15}\right)$ and $\left(3 - \sqrt{15}\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {x}^{2} - 6 x - 6$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We can convert this into the general vertex form ${\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$
by "completing the square" methodology:

${x}^{2} - 6 x - 6$
$\textcolor{w h i t e}{\text{XXX}} = {x}^{2} - 6 x \textcolor{m a \ge n t a}{+ {3}^{2}} - 6 \textcolor{m a \ge n t a}{- {3}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} = {\left(x - \textcolor{red}{3}\right)}^{2} + \textcolor{b l u e}{\left(- 15\right)}$
which is the vertex form with vertex at $\left(\textcolor{red}{3} , \textcolor{b l u e}{- 15}\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The y-intercept (or equivalently the $f \left(x\right)$-intercept)
is the value of $y$ (or equivalently of $f \left(x\right)$) when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} f \left(0\right) = {0}^{2} - 6 \cdot 0 - 6 = - 6$
So the y (or $f \left(x\right)$) intercept is at $y \left(= f \left(x\right)\right) = - 6$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The x-intercepts are the values of $x$ fro which $f \left(x\right)$ (or equivalently, $y$) are equal to 0.
So we want the values of $x$ for which
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 6 x - 6 = 0$
Since there is not obvious factoring of the left side,
we will use the quadratic formula to find the "roots" of this expression.
$\textcolor{w h i t e}{\text{XXX}}$For the general form $a {x}^{2} + b x + c = 0$
$\textcolor{w h i t e}{\text{XXX}}$the quadratic formula tells us that the roots are
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
In this case $a = 1 , b = - 6 , \mathmr{and} c = - 6$
so we have
$\textcolor{w h i t e}{\text{XXX}} x = \frac{6 \pm \sqrt{{\left(- 6\right)}^{2} - 4 \cdot 1 \cdot \left(- 6\right)}}{2 \cdot 1}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{6 \pm \sqrt{36 + 24}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{6 \pm \sqrt{60}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{6 \pm 2 \sqrt{15}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = 3 \pm \sqrt{15}$

So the $x$-intercepts are at $3 + \sqrt{15}$ and $3 - \sqrt{15}$

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =