Quadratic inequalities (1 Viewer)

[planino]

When representing a quadratic inequality, say (x-2)(x-5) > 0 on the Cartesian plane, do you shade the areas above y=0 and outside the parabola, or do you simply emboss the part of actual parabola itself that's above y=0 with open circles at the x-intercepts? Or does either way not matter?
Similarly for (x-2)(x-5) < 0, do you shade inside of the parabola under y=0, or do you emboss the part parabola itself that's below y=0 and draw open circles at the x-intercepts? Or does it not matter which way I choose?

 

[skillstriker]

i usually just shade inside the parabola

 

[Carrotsticks]

Embossing the curve is the way to go here since we are working in 1 dimension.

 

[RivalryofTroll]

When representing a quadratic inequality, say (x-2)(x-5) > 0 on the Cartesian plane, do you shade the areas above y=0 and outside the parabola, or do you simply emboss the part of actual parabola itself that's above y=0 with open circles at the x-intercepts? Or does either way not matter?
Similarly for (x-2)(x-5) < 0, do you shade inside of the parabola under y=0, or do you emboss the part parabola itself that's below y=0 and draw open circles at the x-intercepts? Or does it not matter which way I choose?

Like this?

 

[Siddy123]

View attachment 24946

Like this?

i see you used latex for your solution :L

 

[planino]

Ok then shading it is, thanks guys! RivalryOfTroll did you use paint?

 

[Drongoski]

Shading is news to me.

 

[D94]

I've rarely seen a question like that asking you to shade a region with only 1 variable in a quadratic. Anyway, Wolfram it, and you'll see they concur with Carrotsticks. A more common question would ask you to find the solutions of the inequality.

 

[Carrotsticks]

Oh dear! I just noticed that it was single variable.

At the time of posting, I read it as (x-2)(x-5) > Y, so putting in some value of Y would result in a region.

Apologies for that.

BUT you could argue the following:

y=(x-2)(x-5)

So (x-2)(x-5) > 0 would be the equivalent of y > 0 bounded by the curve, which would result in the region.