# How do you graph y>=-2x^2-x+3?

Jul 20, 2017

I would graph the equation $y = - 2 {x}^{2} - x + 3$. This divides the plane into two regions. Test points in each region to see if the region contains solutions to the inequality.

#### Explanation:

Graph $y = - 2 {x}^{2} - x + 3$.

The graph is a parabola.
We could graph this by finding the vertex and so on, but I find it simpler to just find the $x$ intercepts

$- 2 {x}^{2} - x + 3 - - \left(2 {x}^{2} + x - 3\right) = - \left(2 x \textcolor{w h i t e}{\text{XXX")(xcolor(white)"XXX}}\right)$

$= - \left(2 x + 3\right) \left(x - 1\right)$

So the $x$ intercepts are $- \frac{3}{2}$ and $1$.

The $y$ intercept is $3$, so the graph of the equation is

graph{ -2x^2-x+3 [-8.62, 7.186, -3.96, 3.94]}

Testing $\left(0 , 0\right)$ we see that $0 \ge - 2 {\left(0\right)}^{2} - \left(0\right) + 3$ is false so there are no solutions in the region containing the origin.

Testing $\left(0 , 5\right)$ (remember the $y$ intercept is $3$) or $\left(1 , 3\right)$ or $\left(5 , 0\right)$ or some other point outside the region containing $\left(0 , 0\right)$, we learn the the region not containing $\left(0 , 0\right)$ contains the solutions

OR We can reason the the values of $y$ greater than -2x^2-x+3 are above the curve.

graph{y >= -2x^2-x+3 [-8.62, 7.186, -3.96, 3.94]} .