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

1 Answer
May 18, 2017

Answer:

Please see below.

Explanation:

Let us draw a graph of #y=x^2-3x#, by choosing a few values of #x# say #{-6,-3,0,1.5,3,6,9}# and putting these values we get corresponding values of #y# which are #{54,18,0,-2.25,0,18,54}# and then drawing the points i.e. #(-6,54),(-3,18),(0,0),(1,-2.25),(3,0),(6,18),(9,54)#

and joining them, we get a graph as follows
graph{x^2-3x [-18.42, 21.58, -4.32, 15.68]}

Observe that it divides the entire Cartesian plane in three parts #-# one on the line, which is a parabola, #-# two inside the parabola and #-# three outside the parabola.

As we have #y < x^2-3x#, points on parabola do not satisfy this and hence the parabola is not included.

Now select a point inside parabola say #(1,1)# and it is apparent that while #y=1#, #x^2-3x=-2# and hence #y > x^2-3x#. Therefore points inside the parabola also do not satsfy the equality.

Selecting a point outside parabola say #(-1,0)#, we have that #y=0# and #x^2-3x=1+3=4# and we have #y < x^2-3x# and hence area outside parrabola satisfies the given inequality.

Hence graph of #y < x^2-3x# is as follows:

graph{y < x^2-3x [-18.42, 21.58, -4.32, 15.68]}

Note #-# The line parabola has been kept dotted as points on the line are not included. Had it been #y <= x^2-3x#, we would have kept it complete i.e. not dotted.