How do you graph #y<x^2-3x#?

1 Answer
Nov 15, 2016

Answer:

graph{y<(x^2-3x) [-10.5, 9.5, -4.8, 5.2]}

Explanation:

First, we know it will look like an upward-facing parabola because of the first term in the function #x^2# with a coefficient of positive one.

Next we can determine that the graph will be shaded everywhere below the parabola, and the parabola will be a dotted line because of the #<# sign.

Now, we find where the dotted upward facing parabola has a vertex and where its intercepts are. I'm going to write this using an equals sign rather than a less than symbol, because we are only looking for where the parabola is right now.

#y=x^2-3x#
#y=(x)(x-3)#
Roots are at #x=0# and #x=3#
#y=x^2-3x#
#y=(x-1.5)^2-2.25#
Vertex is at #(1.5,-2.25)#