# Honors Algebra 2: Unit 2 Review Worksheet

• Class: Honors Algebra 2
• Author: Peter Atlas
• Text: Algebra and Trigonometry: Structure and Method, Brown

1. Given the three functions defined below, answer the following questions:
 $$p(x) = \begin{cases} x + 1, & \text{if } x < -1 \\ 4, & \text{if } -1 < x < 2 \\ -x - 1, & \text{if }x \ge 2 \end{cases}$$ $$q(x) = 3(x - 1)^2$$ $$\displaystyle r(x) = \frac{1}{4 - 2x}$$
1. What is the domain of $$p(x)$$?
2. Solution $$x \ne -1$$
3. What is the range of $$p(x)$$?
4. Solution $$y < 0 \bigvee y = 4$$
5. What is the domain of $$q(x)$$?
6. Solution $$\Re$$
7. What is the range of $$q(x)$$?
8. Solution $$y \ge 0$$
9. What is the domain of $$r(x)$$?
10. Solution $$x \ne 2$$
11. Find $$p(-8)$$
12. Solution -7
13. Find $$p(0)$$
14. Solution 4
15. Find $$p \left( q(-1) \right)$$
16. Solution -13
17. Find and simplify $$\left( r \circ q \right) (x+2)$$
18. Solution $$\displaystyle \frac{1}{-6x^2 - 12x - 2}$$
19. Find and simplify $$\displaystyle \frac{q}{r}(x)$$ and state its domain.
20. Solution $$- 6x^3 + 24x^2 - 30x + 12.$$ Domain: $$x \ne 2.$$
21. Find and simplify $$\displaystyle 2q(x) + \frac{1}{r(x)} - 4$$
22. Solution $$6x^2 - 14x + 6$$
23. What is the maximum value of $$p(x)$$?
24. Solution 4
25. What is the minimum value of $$p(x)$$?
26. Solution It doesn't have one
27. What is the maximum value of $$q(x)$$?
28. Solution It doesn't have one
29. What is the minimum value of $$q(x)$$?
30. Solution 0
31. Sketch the graph of $$y = p(x)$$
32. Solution
33. Sketch the graph of $$y=q(x)$$
34. Solution
2. Rewrite the following as piecewise functions:
1. $$y = |x|$$
2. Solution $$y = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if }x < 0 \end{cases}$$
3. $$y = 2 \left|x - 4 \right| + 1$$
4. Solution $$y = \begin{cases} 2x - 7, & \text{if } x \ge 4 \\ -2x + 9, & \text{if }x < 4 \end{cases}$$
5. $$\displaystyle y=-3 \left| 2x+16 \right| - 7$$
6. Solution $$y = \begin{cases} -6x - 55, & \text{if } x \ge -8 \\ 6x + 41, & \text{if }x < -8 \end{cases}$$
3. 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.
4. Solution $$y = -2(x - 4)^2 + 3$$ would be a parabola that is flipped over the x-axis, stretched vertically by a factor of 2, moved up 3, and moved right 4. Point A would become (4, 3), and Point B would become (5, 1).
5. 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?
6. Solution This would be a V, squished vertically by a factor of 3, flipped over the y axis (which doesn't make it look different) and moved to the right by 6. The vertex becomes (6, 0), and the point (1, 1) becomes $$\displaystyle \left( 5, \frac{1}{3} \right)$$
7. Using 2 rules, give the piecewise definition of the function whose graph is below
8. Solution $$\displaystyle f(x) = \begin{cases} |x + 1| - 2, & \text{if } -2 \le x \lt 3 \\ -\frac{2}{3}\left(x - 6 \right) + 1, & \text{if } x \ge 3 \\ \end{cases}$$
9. Given $$\displaystyle f(x) = \frac{2}{3}x+\frac{4}{5},$$ find $$\displaystyle f^{-1}(x)$$ in standard form.
10. Solution $$15x - 10y = 12$$
11. Of the following relations, indicate
1. whether it is a function
2. whether its inverse is a function
3. the domain
4. the range
5. sketch the graph of the relation
1. $$y = 2x - 3$$
2. Solution Yes, yes, $$x \in \Re, y \in \Re$$
3. $$y = 6$$
4. Solution Yes, no, $$x \in \Re, y = 6$$
5. $$x = -4$$
6. Solution No, yes $$x = -4, y \in \Re$$
7. $$y = x^2 - 3$$
8. Solution Yes, no, $$x \in \Re, y \ge -3$$
9. $$y = 2x^5 + 4$$
10. Solution Yes, yes, $$x \in \Re, y \in \Re$$
11. $$y = 2^x - 5$$
12. Solution Yes, yes, $$x \in \Re, y > -5$$
13. $$y = \frac{1}{2}\sqrt{x + 2} - 1$$
14. Solution Yes, yes, $$x \ge -2, y \ge -1$$