Question

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

Answer 1

See below:

Explanation:

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

The `-2x^2` tells us that the parabola will look like a `nn`

The y-intercept is:

`y=-2(0)^2-4(0)+3=3=>(0,3)`

The vertex is at:

`x=-b/(2a)=-(-4)/(2(-2))=- (-4)/(-4)=-1`

`y=-2(-1)^2-4(-1)+3=-2(1)+4+3=5`

and so `(-1,5)`

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> -2x^2-4x+3`. These will either be inside the parabola or outside. Let's see which.

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

`0> -2(0)^2-4(0)+3`

`0>3 color(white)(000)color(red)"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 `>=`, 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]}