# How do you graph the inequality y> -2x^2-4x+3?

See below:

#### Explanation:

Let's first see that we're going to have a parabola to draw.

The $- 2 {x}^{2}$ tells us that the parabola will look like a $\cap$

The y-intercept is:

$y = - 2 {\left(0\right)}^{2} - 4 \left(0\right) + 3 = 3 \implies \left(0 , 3\right)$

The vertex is at:

$x = - \frac{b}{2 a} = - \frac{- 4}{2 \left(- 2\right)} = - \frac{- 4}{- 4} = - 1$

$y = - 2 {\left(- 1\right)}^{2} - 4 \left(- 1\right) + 3 = - 2 \left(1\right) + 4 + 3 = 5$

and so $\left(- 1 , 5\right)$

We can now draw a graph:

graph{-2x^2-4x+3 [-11.25, 11.25, -4.05, 7.2]}

We want to now work with the inequality. We want values where $y > - 2 {x}^{2} - 4 x + 3$. These will either be inside the parabola or outside. Let's see which.

Let's look at the origin, $\left(0 , 0\right)$. Is this a point that satisfies the inequality?

$0 > - 2 {\left(0\right)}^{2} - 4 \left(0\right) + 3$

$0 > 3 \textcolor{w h i t e}{000} \textcolor{red}{\text{No}}$

Which means the valid solution set is outside the parabola (but doesn't include the line defining the parabola itself, and so we use a dotted line. Had the question been $\ge$, then it'd be a solid line and the line defining the parabola would be a part of the solution set):

graph{y> -2x^2-4x+3 [-11.25, 11.25, -4.05, 7.2]}