Mr. Dreyer

Algebra 2 Lv 1

1Warmup

1. Write down homework: Read chapter review (p. 130-132, except for 2.5). Go over any content you are not comfortable with. Do the chapter test on p.133, problems 1-26. (will count as homework, not as a test).
2. Write numbers of homework problems you had difficulty with at the top of the board; if you can solve them, go up and do so (and get class participation points)!

2Test grading

Every question (or lettered sub-question) was worth 5 points except question 10, which was worth 10 points. Because this test was harder than I had expected, I will apply the following piecewise function to the raw score you see on your paper:

So, for example, if you scored 83 it would count as 83+10=93. If you scored 93 it would count as ?.

3Things to think about

• Functions must have only one $f\left(x\right)$ value for every $x$. It is fine to have more than one $x$ value mapping onto the same $f\left(x\right)$ value. At the most extreme case, the function defined in problem 2, $f\left(x\right)=3$, has the same $f\left(x\right)$ value for every $x$ and it's still a function.
• $f\left(x\right)$ does not in any way mean to multiply $f$ times $x$. It means to apply the function $f$ to $x$. I know it looks like multiplication but it's not.

• Remember the distributive law!

  ${\displaystyle a(b+c)=ab+ac}$

  Do not do something like this:

  ${\displaystyle \begin{array}{rcl}y-5& =& 2(x-3)\\ y& =& 2x-3+5???\end{array}}$
• To find a linear function that goes through a point, plug in the $x$ value and solve for $y$, or use the point-slope form.

• Point-slope form is very useful. For example:

  • Problem: Find a point on the graph of this equation:

    ${\displaystyle y-7=3(x-5)}$

    Answer: $(5,7)$

  • Problem: Find an equation with a slope of 3 that goes through the point $(5,7)$.

    Answer:

    ${\displaystyle y-7=3(x-5)}$
• To evaluate something means to find its value.

4Questions?