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

1 Answer
MathFact-orials.blogspot.com
Feb 23, 2017

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]}

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