# How do you find x-intercepts, coordinates of vertex for parabola y = x^2 - 4x - 12?

Apr 11, 2015

$y = {x}^{2} - 4 x - 12$

Part 1: x-intercepts
The x-intercepts occur at the points on the function where $y = 0$
So, we need to solve
${x}^{2} - 4 x - 12 = 0$
The left side factors fairly easily into:
$\left(x - 6\right) \left(x + 2\right) = 0$
So solution occur when
$x - 6 = 0 \rightarrow x = 6$
and
$x + 2 = 0 \rightarrow x = \left(- 2\right)$
So the x-intercepts are at $\left(0 , 6\right)$ and $\left(0 , - 2\right)$

Part 2: vertex of the parabola
The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to $0$.
The derivative of the given quadratic is
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 4$
By observation, this is equal to $0$ when $x = 2$
When $x = 2$ the original equation becomes
$y = {\left(2\right)}^{2} - 4 \left(2\right) - 12$
$y = - 16$
Therefore the vertex of this parabola is at $\left(2 , - 16\right)$