Problem 4
Answer:
[tex]f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}[/tex]
------------------
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.
---
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 [tex]f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}[/tex]
This is only graphed when [tex]x \ge 1[/tex] (note the closed or filled in circle for the endpoint of this portion).
===================================================
Problem 5
Answer:
[tex]f(x) = \frac{1}{2}|x+3|[/tex] is the absolute value function
while this is the piecewise function
[tex]f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\[/tex]
------------------
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|
------
Break that down into a piecewise function
when x < -3, then y = -(1/2)(x+3)
when [tex]x \ge -3[/tex], 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 [tex]x \ge 0[/tex]
So that is how we get [tex]f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\[/tex]as the piecewise function.
