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

1 Answer
Gió
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=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]}

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