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

May 18, 2017

#### Explanation:

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

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 $y < {x}^{2} - 3 x$, points on parabola do not satisfy this and hence the parabola is not included.

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

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

Hence graph of $y < {x}^{2} - 3 x$ 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 $y \le {x}^{2} - 3 x$, we would have kept it complete i.e. not dotted.