Problem 4
<h3>Answer:</h3>

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Work Shown:
The left line goes through (-2,0) and (0,4)
The slope of this line is
m = (y2-y1)/(x2-x1)
m = (4-0)/(0-(-2))
m = (4-0)/(0+2)
m = 4/2
m = 2
The y intercept is b = 4
Since m = 2 and b = 4, this means y = mx+b turns into y = 2x+4. This portion is only done when x < 1. Note the open circle at the endpoint of this portion. So we do not include x = 1 as part of this piece.
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The line on the right side goes through (1,-2) and (2,-1)
Slope
m = (y2-y1)/(x2-x1)
m = (-1-(-2))/(2-1)
m = (-1+2)/(2-1)
m = 1/1
m = 1
The y intercept is b = -3. You can see this if you extend the line until it crosses the y axis.
Alternatively, plug in (x,y) = (1,-2) and m = 1 into y = mx+b to find that b = -3
So y = mx+b turns into y = 1x+(-3) or just y = x-3
We combine both parts to end up with 
This is only graphed when  (note the closed or filled in circle for the endpoint of this portion).
 (note the closed or filled in circle for the endpoint of this portion).
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Problem 5
Answer: 
<h3>

 is the absolute value function</h3><h3>while this is the piecewise function</h3>

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Work Shown:
y = |x| .... parent function
y = |x+3| ... shift 3 units to the left
y = (1/2)*|x+3| .... vertically compress by factor of 1/2
f(x) = (1/2)*|x+3|
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Break that down into a piecewise function
when x < -3, then y = -(1/2)(x+3)
when  , then y = (1/2)(x+3)
, then y = (1/2)(x+3)
I'm using the rule that y = |x| turns into y = -x when x < 0 and y = x when  
 
So that is how we get  as the piecewise function.
as the piecewise function.