# How do you graph the inequality y<=-x^2-3x+10?

May 19, 2017

Shade under the parabola.

#### Explanation:

First, graph the parabola $y = - {x}^{2} - 3 x + 10$. If you complete the square, you can rewrite this in vertex form giving you
$y = - \left({x}^{2} + 3 x + \frac{9}{4} - \frac{9}{4} - 10\right)$
$y = - \left({\left(x + \frac{3}{2}\right)}^{2} - \frac{49}{4}\right)$
$y = - {\left(x + \frac{3}{2}\right)}^{2} + \frac{49}{4}$

So the parabola opens downward and has a vertex at
$\left(- \frac{3}{2} , \frac{49}{4}\right) = \left(- \frac{1}{5} , 12.25\right)$

The $y$ intercept occurs when $x = 0$, giving $\left(0 , 10\right)$

The $x$ intercept occurs when $y = 0$, giving $\left(2 , 0\right)$ and $\left(- 5 , 0\right)$

Finally, if your $y$ values are less than the parabola, simply shade underneath the parabola. The graph would look like this: