# How do you graph y=-x^2-8x-13?

##### 1 Answer
Feb 5, 2015

This is the equation of a PARABOLA (basically in the shape of "U").

Consider the general form of your equation given by:
$y = a {x}^{2} + b x + c$
in your case:
$a = - 1$
$b = - 8$
$c = - 13$

With this in mind you can start:

You have several interesting points to consider in order to plot your graph.

1) The vertex: this is the lowest point reached by your parabola.
The coordinates of this point are given by:
${x}_{v} = - \frac{b}{2 a}$ and ${y}_{v} = - \frac{\Delta}{4 a}$
Where $\Delta = {b}^{2} - 4 a c$
giving: ${x}_{v} = - 4 \mathmr{and} {y}_{v} = 3$
2) the y-axis intercept:
obtained setting $x = 0$.
giving: $x = 0 \mathmr{and} y = - 13$
3) The x-axis intercept(s):
obtained setting $y = 0$ and solving the corresponding second degree equation:
i.e.: $- {x}^{2} - 8 x - 13 = 0$
giving: ${x}_{1} = - 5.7 \mathmr{and} {x}_{2} = - 2.3$
Your Parabola has $a < 0$ so that is downward oriented.
Finally:
graph{-x^2-8x-13 [-10, 10, -5, 5]}