Question

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

Answer 1

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

Consider the general form of your equation given by:
`y=ax^2+bx+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=-b/(2a)` and `y_v=-Delta/(4a)`
Where `Delta=b^2-4ac`
giving: `x_v=-4 and y_v=3`
2) the y-axis intercept:
obtained setting `x=0`.
giving: `x=0 and y=-13`
3) The x-axis intercept(s):
obtained setting `y=0` and solving the corresponding second degree equation:
i.e.: `-x^2-8x-13=0`
giving: `x_1=-5.7 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]}