Honors Algebra 2: Unit 2 Review Worksheet



  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 graph
    33. Sketch the graph of \(y=q(x)\)
    34. Solution graph
  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
    graph
  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, \( x \in \Re, y \in \Re \) graph
    3. \( y = 6 \)
    4. Solution Yes, no, \(x \in \Re, y = 6 \) graph
    5. \( x = -4\)
    6. Solution No, yes \(x = -4, y \in \Re \)graph
    7. \( y = x^2 - 3\)
    8. Solution Yes, no, \( x \in \Re, y \ge -3 \)graph
    9. \(y = 2x^5 + 4 \)
    10. Solution Yes, yes, \( x \in \Re, y \in \Re \)graph
    11. \(y = 2^x - 5\)
    12. Solution Yes, yes, \(x \in \Re, y > -5 \)graph
    13. \(y = \frac{1}{2}\sqrt{x + 2} - 1\)
    14. Solution Yes, yes, \( x \ge -2, y \ge -1\)graph