# Sketch the region: y ≤ x^2 and 4 ≥ y ≥ 0 and y ≥ 2x-4

1. Sep 12, 2015

### HermitOfThebes

1. The problem statement, all variables and given/known data

Sketch the region defined by: y ≤ x^2 and 4 ≥ y ≥ 0 and y ≥ 2x − 4. Evaluate the area defined by the above inequalities.
2. Relevant equations

3. The attempt at a solution
I have already plotted the positive part and found the area. However, the negative part seems to be infinite. I am required to find the area, surely it's not infinite then.
The area under x^2 in the negative part, above 0, below 4 and above the line is an infinite rectangle starting from x=-2 and stretching to negative infinity. What am I doing wrong?
Here's a diagram: http://www.quickmath.com/webMathema...=y>+2x-4&v7=x&v8=y&v9=-10&v10=10&v11=-6&v12=6

2. Sep 12, 2015

### Staff: Mentor

There is no "negative" part. The region of interest here is bounded on the left by $y = x^2$, on the right by y = 2x - 4, above by the line y = 4, and below by the line y = 0. The entire region lies within the first quadrant.

3. Sep 12, 2015

### Ray Vickson

I agree with the OP: the region sought lies below the parabola y = x^2, above the line y = 0, below the line y = 4 and above the line y = 2x - 4. The part of the region in {x < 0} is, indeed, unbounded: except for a parabolic northeast corner, it is a strip extending left to $x \to -\infty$. Perhaps there were some restrictions on $x$ that the OP omitted.

4. Sep 12, 2015

### pasmith

I suspect that whoever set the question either didn't realize that the unbounded part exists, or assumed that anyone reading the question would interpret it as referring only to the bounded part. In either case the question is poorly designed. It happens.

You cannot possibly be penalized if you both give (with working) the correct area for the bounded part and point out the existence of the unbounded part.

5. Sep 12, 2015

### Staff: Mentor

I don't see how. Here is the region as I see it.

The author of the question could have been clearer in the description.

Last edited: Sep 12, 2015
6. Sep 12, 2015

### Ray Vickson

The OP did not include the restriction x >= 0; in fact, he used a package to produce a plot having both {x>0} and {x<0} regions. There are (x,y) points with x < 0 that satisfy all the listed inequalities, such as (x,y) = (-3,1). It is a feasible point because 1 < (-3)^2, 1 > 0, 1 < 4 and 1 > - 4 + 2(-3).