How can you sketch the set of points in these 2 equations:abs.val(x + 2y) <=(x - y) and (x^2 - y)(x - y^2)=0?

Question

How do you handle absolute value when it comes to graphing and simplifying an equation? And how do you deal with exponents that can't be simplified down very easily? Of course, some calculator manipulation would make these easier to some degree, but I need to understand how simplified these can get before resorting to a calculator.

Answer 1

The second one is fairly easy, if you remember that a product is 0 exactly if one of the factors is 0, so you will get 2 parabolas (one for each factor), one opened upwards, one to the right.

For the first one, You could resort to distinction of cases:

If x+2y&gt;=0 you&#039;ll have x+2y&lt;=x-y or y&lt;=0 (and x&gt;=-2y) which will result in an angle at the origin.

If x+2y&lt;0 you&#039;ll have -x-2y&lt;=x-y or x&gt;=-y/2 (and x&lt;-2y)  which will again result in an angle at the origin. Now the lines x&gt;=-2y and x&lt;-2y will obviously fall together, and you get the angle y&lt;=0 and x&gt;-y/2.

Answer 2

One way to handle absolute value inequalities is to replace them with equations (to find the boundary) and then test around the boundary, like this:

Abs(x+2y) &lt;= x-y  becomes

Abs(x+2y) = x-y, which we solve as...

x+2y = +-(x-y)

That is, x+2y = x-y, or x+2y = y-x.

These simplify to y = 0 and y = -2x.  These two lines split the plane into four chunks: one containing the entire first quadrant and some of the second; one containing the entire third quadrant and some of the fourth; one contained by the second quadrant; one contained by the fourth.  From each region, we pick a point to represent it...I&#039;ll use the four points (+-1, +-1).  Now we test these points against the original inequality:

(1,1) fails, because Abs(1+2) = 3, while 1-1 = 0.  3 &gt; 0.

(1,-1) passes: Abs(1-2) &lt;1-(-1).

(-1,1) fails: Abs(-1+2) &gt; -1-1.

(-1,-1) fails: Abs(-1-2) &gt; -1+1.

So, we keep the region containing (1,-1); that&#039;s the part of the fourth quadrant bounded by the positive x-axis and the lower half of the line y=-2x.

Your second problem is a nod to a special property of zero: if the product of two expressions is zero, then one of the expressions is zero.  In other words, you can just graph x^2-y=0 and x-y^2=0, getting a pair of parabolas.  This pair of parabolas is the graph.  Note: this method works ONLY for zero.