Mr. Dreyer

Algebra 2 Lv 1

1Warmup

2. Welcome/expectations signatures on desk
3. Got calculators?
4. Write down homework: Read Chap 2.7 (p. 114-116). Problems (p. 117-119): 1-39 odd, 50-53, 56-57.
5. Graph $y=2|x-(-3)|-2$.

Piecewise functions are functions defined differently for different pieces of the domain. For example:

2Evaluating piecewise functions

Basic plan: Look at input (usually $x$), decide which “piece” applies, and evaluate that. For example:

Evaluate $f\left(0\right)$.

We note that $0⩽1$ so we use the first piece:

3Aside: multiple inequalities

A shorthand for

is

4

Graphing piecewise functions

• Graph each piece separately
• Be conscious of open points (not in the function) and closed points (in the function)
• Each $x$ value in the domain should correspond to only one of the pieces

For example, graph

This is called a “step function.” Why?

5Determining piecewise function definition from graph

• Use what you already know to figure out the definition of each piece
• Determine domain of each piece by looking at the $x$ values

Example: consider the displayed function. We see that it looks like two pieces. Use what we already know to determine the functions for each one

• On the left side, we see a line of slope $-1$ and a $y$-intercept of $-2$ so we can use the slope-intercept form to decide that its equation is

  ${\displaystyle y=-x-2}$

• On the right side, we see a line of slope $\frac{1}{2}$ but we do not see a $y$-intercept. We do see some points, like $(2,-2)$. So ywe can use the point-slope form:

  ${\displaystyle y-(-2)=\frac{1}{2}(x-2)}$

  To make this equation into a function, we add $-2$ to both sides to isolate $y$ and then simplify:

  ${\displaystyle \begin{array}{rcl}y& =& \frac{1}{2}(x-2)+(-2)\\ & =& \frac{1}{2}x-3\end{array}}$

• Then we figure out the domains and write it up:

  ${\displaystyle f\left(x\right)=\{\begin{array}{l}-x-2,x<32\\ \frac{1}{2}x-3,2⩽x<8\end{array}.}$