- Class: Honors Algebra 2

- Author: Peter Atlas

- Text:
__Algebra and Trigonometry: Structure and Method__, Brown

- Given the three functions defined below, answer the following questions:

\( p(x) = \begin{cases} x + 1, & \text{if } x < -1 \\ 4, & \text{if } -1 < x < 2 \\ -x - 1, & \text{if }x \ge 2 \end{cases} \) \( q(x) = 3(x - 1)^2\) \( \displaystyle r(x) = \frac{1}{4 - 2x}\) - What is the domain of \(p(x)\)?
- What is the range of \(p(x)\)?
- What is the domain of \(q(x)\)?
- What is the range of \(q(x)\)?
- What is the domain of \(r(x)\)?
- Find \(p(-8)\)
- Find \(p(0)\)
- Find \(p \left( q(-1) \right) \)
- Find and simplify \( \left( r \circ q \right) (x+2) \)
- Find and simplify \( \displaystyle \frac{q}{r}(x)\) and state its domain.
- Find and simplify \( \displaystyle 2q(x) + \frac{1}{r(x)} - 4\)
- What is the maximum value of \(p(x)\)?
- What is the minimum value of \(p(x)\)?
- What is the maximum value of \(q(x)\)?
- What is the minimum value of \(q(x)\)?
- Sketch the graph of \(y = p(x)\)
- Sketch the graph of \(y=q(x)\)

## Solution

\( x \ne -1\)## Solution

\( y < 0 \bigvee y = 4\)## Solution

\( \Re \)## Solution

\( y \ge 0 \)## Solution

\( x \ne 2\)## Solution

-7## Solution

4## Solution

-13## Solution

\( \displaystyle \frac{1}{-6x^2 - 12x - 2}\)## Solution

\( - 6x^3 + 24x^2 - 30x + 12.\) Domain: \( x \ne 2. \)## Solution

\(6x^2 - 14x + 6\)## Solution

4## Solution

It doesn't have one## Solution

It doesn't have one## Solution

0## Solution

## Solution

- Rewrite the following as piecewise functions:
- \(y = |x| \)
- \(y = 2 \left|x - 4 \right| + 1\)
- \( \displaystyle y=-3 \left| 2x+16 \right| - 7\)

## Solution

\(y = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if }x < 0 \end{cases}\)## Solution

\(y = \begin{cases} 2x - 7, & \text{if } x \ge 4 \\ -2x + 9, & \text{if }x < 4 \end{cases}\)## Solution

\(y = \begin{cases} -6x - 55, & \text{if } x \ge -8 \\ 6x + 41, & \text{if }x < -8 \end{cases}\) - The graph of \(y = x^2\) contains the points A(0, 0) and B(1, 1). Describe the graph of \(y = -2(x - 4)^2 + 3\), and give the coordinates of the points that correspond to A and B on the transformed graph.
- Describe the graph of \( \displaystyle y = \frac{|6 - x|}{3}\). A(0, 0) and B(1, 1) are on the graph of \(y = |x|\). What are the coordinates of the corresponding points on the transformed graph?
- Using 2 rules, give the piecewise definition of the function whose graph is below

- Given \( \displaystyle f(x) = \frac{2}{3}x+\frac{4}{5},\) find \( \displaystyle f^{-1}(x)\) in standard form.
- Of the following relations, indicate
- whether it is a function
- whether its inverse is a function
- the domain
- the range
- sketch the graph of the relation

- \(y = 2x - 3\)
- \( y = 6 \)
- \( x = -4\)
- \( y = x^2 - 3\)
- \(y = 2x^5 + 4 \)
- \(y = 2^x - 5\)
- \(y = \frac{1}{2}\sqrt{x + 2} - 1\)

## Solution

Yes, yes, \( x \in \Re, y \in \Re \)## Solution

Yes, no, \(x \in \Re, y = 6 \)## Solution

No, yes \(x = -4, y \in \Re \)## Solution

Yes, no, \( x \in \Re, y \ge -3 \)## Solution

Yes, yes, \( x \in \Re, y \in \Re \)## Solution

Yes, yes, \(x \in \Re, y > -5 \)## Solution

Yes, yes, \( x \ge -2, y \ge -1\)