# How to graph a parabola y=x^2-4x-5?

May 25, 2015

You can start by finding the x-intercepts, if there are any. That is when y = 0.

$0 = {x}^{2} - 4 x - 5$
$0 = \left(x - 5\right) \left(x + 1\right)$

So you have one intercept at $x = 5$ and one at $x = - 1$. Or, at $\text{(5, 0)}$ and $\text{(-1, 0)}$.

Then, you can find where the vertex is (where the graph turns around). This can be found using:

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

$\frac{- \left(- 4\right)}{2 \cdot 1} = \frac{4}{2} = 2$

Then you can plug it into the original equation and solve for the y-coordinate.

Plugging it in:
${x}^{2} - 4 x - 5 = y$
${2}^{2} - 4 \cdot 2 - 5 = 4 - 8 - 5 = - 9$

So I would expect a vertex at $\text{(2, -9)}$.

If you want to go even further, you can solve to see if there are any y-intercepts (at $x = 0$).

$y = {0}^{2} - 4 \cdot 0 - 5 = - 5$

So there is also a y-intercept at $\text{(0, -5)}$.

Here's the graph:
graph{x^2 - 4x - 5 [-20.07, 20.06, -10.03, 10.04]}

You can click on the graphed curve to locate the intercepts and vertex.