Question 317209
What is the area of the region enclosed by the graphs of y = 2│x − 3│ − 2 
and y = 4 − 2│x − 2│?
<pre><b>
The other tutor's solution is correct but I think you should
calculate the vertices rather that assume they are exactly as 
they look.

Let's graph it:

{{{drawing(400,3600/11,-3,8,-4,5, graph(400,3600/11,-3,8,-4,5,y=2abs(x-3)-2),
graph(400,3600/11,-3,8,-4,5,y=4-2abs(x-2)) )}}}

The region looks like a parallelogram.

We need to find the corner points, which
are the two vertices and the two points of intersection

To find the vertex of {{{y=2abs(x-3)-2}}} we set the part
in the absolute value = 0

{{{x-3=0}}}

and solve for x

{{{x=3}}}

And we substitute this in

{{{y=2abs(x-3)-2}}}
{{{y=2abs(3-3)-2}}}
{{{y=2abs(0)-2}}}
{{{y=2(0)-2}}}
{{{y=0-2}}}
{{{y=-2}}}

So the vertex of {{{y=2abs(x-3)-2}}} is (3,-2).

To find the vertex of y=4-2abs(x-2)}}} we also set the part
in the absolute value = 0

{{{x-2=0}}}

and solve for x

{{{x=2}}}

And we substitute this in

{{{y=4-2abs(x-2)}}}
{{{y=4-2abs(2-2)}}}
{{{y=4-2abs(0)}}}
{{{y=4-2(0)}}}
{{{y=4-0}}}
{{{y=4}}}

So the vertex of {{{y=2abs(x-3)-2}}} is (2,4). 

Now we find the other two vertices of the figure

We solve the system

{{{system(y=2abs(x-3)-2, y=4-2abs(x-2))}}}

Since the right sides both equal to y, set them equal:

{{{2abs(x-3)-2=4-2abs(x-2)}}}

We can divide every term through by 2 without getting fractions,
so we do so:

{{{abs(x-3)-1=2-abs(x-2)}}}

Add 1 to both sides:

{{{abs(x-3)=3-abs(x-2)}}}

There are four cases to consider:

1.  {{{x - 3 > 0}}} and {{{x-2>0}}}
which is the same as 
    {{{x > 3}}} and {{{x>2}}}
which is the same as
    {{{x>3}}}

{{{abs(x-3)=3-abs(x-2)}}}
becomes
{{{x-3=3-(x-2)}}}
{{{x-3=3-x+2}}}
{{{x-3=5-x}}}
{{{2x=8}}}
{{{x=4}}}

Substitute in

y=2abs(x-3)-2
y=2abs(4-3)-2
y=2abs(1)-2
y=2(1)-2
y=0

So one point of intersection is (4,0)

2.  {{{x - 3 > 0}}} and {{{x-2<0}}}
which is the same as 
    {{{x > 3}}} and {{{x<2}}}

That's a contradiction, so we ignore this case.

3.  {{{x - 3 < 0}}} and {{{x-2>0}}}
which is the same as 
    {{{x < 3}}} and {{{x>2}}}
which is the same as
    {{{2<x<3}}}

{{{abs(x-3)=3-abs(x-2)}}}
becomes
{{{-(x-3)=3-(x-2)}}}
{{{-x+3=3-(x-2)}}}
{{{-x+3=3-x+2}}}
{{{3=5}}}
Thats a contradiction too. So we ignore this case

4.  {{{x - 3 < 0}}} and {{{x-2<0}}}
which is the same as 
    {{{x < 3}}} and {{{x<2}}}
which is the same as
    {{{x<2}}}

{{{abs(x-3)=3-abs(x-2)}}}
becomes
{{{-(x-3)=3-(-(x-2))}}}
{{{-x+3=3-(-x+2)}}}
{{{-x+3=3+x-2}}}
{{{-x+3=3+x-2}}}
{{{-x+3=1+x}}}
{{{2=2x}}}
{{{1=x}}}

Substitute in

y=2abs(x-3)-2
y=2abs(1-3)-2
y=2abs(-2)-2
y=2(2)-2
y=4-2
y=2

So the other point of intersection is (1,2)

So the area we want to find is the area of the polygon whose vertices
are (1,2), (3,-2), (4,0), and (2,4)  figure:

{{{drawing(400,3600/11,-3,8,-4,5, graph(400,3600/11,-3,8,-4,5),
red(line(1,2,2,4)), red(line(4,0,2,4)),red(line(4,0,3,-2)),red(line(3,-2,1,2)), 
locate(1.1,2.2,"(1,2)"),locate(3,-2,"(3,-2)"),locate(4,.5,"(4,0)"),locate(2,4.3,"(2,4)") 

)}}}

Since the polygon is convex (doesn't "sink in"

 anywhere). we use the determinant formula, whose
rows are the coordinates in counter-clockwise order, with the first
row repeated at the bottom:

{{{A}}}{{{"="}}}{{{1/2}}}{{{abs(matrix(5,2,1,2,3,-2,4,0,2,4,1,2))}}}

To expand it add the sum of the products of the diagonals going down
to the right and subtract the sum of the products of the diagonals
going up to the right:

{{{A}}}{{{"="}}}{{{1/2}}}{{{abs(matrix(5,2,1,2,3,-2,4,0,2,4,1,2))}}}{{{"="}}}{{{1/2}}}[{{{(1)(-2)+(3)(0)+(4)(4)+(2)(2)-(3)(2)-(4)(-2)-(2)(0)-(1)(4)}}}]{{{"="}}}{{{1/2}}}{{{(-2+0+16+4-6+8+0-4)}}}{{{"="}}}{{{1/2}}}{{{(16)}}}{{{"="}}}{{{8}}}  

Edwin</pre></b>