# How do you find the vertex and the intercepts for f(x)=x^2 - 12x -2?

Nov 20, 2017

Vertex: $\left(6 , - 38\right)$

$x$-intercepts are:

$\left(6 + \sqrt{38} , 0\right)$ and $\left(6 - \sqrt{38} , 0\right)$

$y$-intercept is:

$\left(0 , - 2\right)$

#### Explanation:

From this form of the parabola equation, i.e. $a {x}^{2} + b x + c$, the $x$-coordinate of the vertex is:

$x = \frac{- b}{2 a}$

Here, $a = 1$ and $b = - 12$, and $c$ is $- 2$ therefore:

$x = \frac{- \left(- 12\right)}{2 \left(1\right)} = \frac{12}{2} = 6$

Now we plug that into the equation to find $y$:

$y = {6}^{2} - 12 \left(6\right) - 2 = 36 - 72 - 2 = 36 - 74 = - 38$

Vertex: $\left(6 , - 38\right)$

To find the $x$-intercepts, you have to set $y$ (the equation) equal to $0$ and solve for $x$

${x}^{2} - 12 x - 2 = 0$

We cannot factor this. Let's use the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

x=(-(-12)+-sqrt((-12)^2-4(1)(-2)))/(2(1)

$x = \frac{12 \pm \sqrt{144 + 8}}{2} = \frac{12 \pm \sqrt{152}}{2} = \frac{12 \pm \sqrt{{2}^{2} \cdot 38}}{2} = \frac{12 \pm 2 \sqrt{38}}{2} = \frac{2 \left(6 \pm \sqrt{38}\right)}{2} = \frac{\cancel{2} \left(6 \pm \sqrt{38}\right)}{\cancel{2}}$

$x = 6 \pm \sqrt{38}$

Therefore, the $x$-intercepts are:

$\left(6 + \sqrt{38} , 0\right)$ and $\left(6 - \sqrt{38} , 0\right)$

To find the $y$-intercept, you have to plug in $0$ for $x$:

$y = {0}^{2} - 12 \left(0\right) - 2 = - 2$

Therefore, the $y$-intercept is:

$\left(0 , - 2\right)$