# How do you graph the inequality y<=x^2-4?

Sep 5, 2017

See below

#### Explanation:

First consider the limiting case: $y = {x}^{2} - 4$

This is a parabola with a absolute minimum value
(since the coefficient of ${x}^{2} > 0$)

Since there is no term in $x$, ${y}_{\min} = y \left(o\right) = - 4$

Considering $y = \left(x + 2\right) \left(x - 2\right)$

$y$ will vave zeros at $x = - 2 \mathmr{and} x = 2$

From the results above, it is possible to graph $y$ in the limiting case.

Now, turning to the inequality: $y \le {x}^{2} - 4$

This may be represented graphically as the entire area of the $x y -$plane that is below and on the limiting case graph. The area that satisifies the inequality is shown shaded below extended to $\left(- \infty , + \infty\right)$ on both axes.

graph{y<=x^2-4 [-10, 10, -5, 5]}