SOLUTION: What is the area of the region enclosed by the graphs of y = 2│x − 3│ − 2 and y = 4 − 2│x − 2│?

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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│?

Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!

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The area is made up of a triangle on top, a triangle on the bottom, and a parallelogram in the middle.
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+A=At1%2BAt2%2BAp
+A=%281%2F2%29bh%2B%281%2F2%29bh%2Bbh
+A=%281%2F2%29%282%29%282%29%2B%281%2F2%29%282%29%282%29%2B%282%29%282%29
+A=2%2B2%2B4
+A=8+

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
What is the area of the region enclosed by the graphs of y = 2│x − 3│ − 2
and y = 4 − 2│x − 2│?

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:



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%28x-3%29-2 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%28x-3%29-2
y=2abs%283-3%29-2
y=2abs%280%29-2
y=2%280%29-2
y=0-2
y=-2

So the vertex of y=2abs%28x-3%29-2 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%28x-2%29
y=4-2abs%282-2%29
y=4-2abs%280%29
y=4-2%280%29
y=4-0
y=4

So the vertex of y=2abs%28x-3%29-2 is (2,4). 

Now we find the other two vertices of the figure

We solve the system

system%28y=2abs%28x-3%29-2%2C+y=4-2abs%28x-2%29%29

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

2abs%28x-3%29-2=4-2abs%28x-2%29

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

abs%28x-3%29-1=2-abs%28x-2%29

Add 1 to both sides:

abs%28x-3%29=3-abs%28x-2%29

There are four cases to consider:

1.  x+-+3+%3E+0 and x-2%3E0
which is the same as 
    x+%3E+3 and x%3E2
which is the same as
    x%3E3

abs%28x-3%29=3-abs%28x-2%29
becomes
x-3=3-%28x-2%29
x-3=3-x%2B2
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+%3E+0 and x-2%3C0
which is the same as 
    x+%3E+3 and x%3C2

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

3.  x+-+3+%3C+0 and x-2%3E0
which is the same as 
    x+%3C+3 and x%3E2
which is the same as
    2%3Cx%3C3

abs%28x-3%29=3-abs%28x-2%29
becomes
-%28x-3%29=3-%28x-2%29
-x%2B3=3-%28x-2%29
-x%2B3=3-x%2B2
3=5
Thats a contradiction too. So we ignore this case

4.  x+-+3+%3C+0 and x-2%3C0
which is the same as 
    x+%3C+3 and x%3C2
which is the same as
    x%3C2

abs%28x-3%29=3-abs%28x-2%29
becomes
-%28x-3%29=3-%28-%28x-2%29%29
-x%2B3=3-%28-x%2B2%29
-x%2B3=3%2Bx-2
-x%2B3=3%2Bx-2
-x%2B3=1%2Bx
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:



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%22=%221%2F2abs%28matrix%285%2C2%2C1%2C2%2C3%2C-2%2C4%2C0%2C2%2C4%2C1%2C2%29%29

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%22=%221%2F2abs%28matrix%285%2C2%2C1%2C2%2C3%2C-2%2C4%2C0%2C2%2C4%2C1%2C2%29%29%22=%221%2F2[]%22=%221%2F2%28-2%2B0%2B16%2B4-6%2B8%2B0-4%29%22=%221%2F2%2816%29%22=%228  

Edwin