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Grade 11 · Functions and Graphs
Hyperbola & Exponential Graphs (Grade 11)
MEMORANDUM
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Mark
  / 23

Answer all questions. Show all your working — marks are awarded for method as well as the final answer. Teacher copy: accept any correct equivalent method.

  1. 1
    Write down the asymptotes of y = 4/(x − 1) + 2.
    (2)
    x = 1y = 2xy(−1 ; 0)(0 ; −2)−6−4−22468−6−4−22468
    x = 1 and y = 2✓✓ (2)
  2. 2
    Determine the y-intercept of y = 4/(x − 1) + 2.
    (2)
    4/−1 + 2 = −2✓✓ (2)
  3. 3
    Determine the x-intercept of y = 4/(x − 1) + 2.
    (3)
    4 = −2(x − 1) → x = −1✓✓✓ (3)
  4. 4
    Write down the domain and range of y = 4/(x − 1) + 2.
    (2)
    Domain x ∈ ℝ, x ≠ 1; range y ∈ ℝ, y ≠ 2✓✓ (2)
  5. 5
    Write down the equation of the asymptote of y = 2x − 1.
    (1)
    y = −1xyy = 2x − 1(0 ; 0)−4−3−2−11234−4−2246810
    y = −1✓ (1)
  6. 6
    Determine the y-intercept of y = 2x − 1.
    (2)
    2⁰ − 1 = 0✓✓ (2)
  7. 7
    Write down the range of y = 2x − 1.
    (1)
    y > −1✓ (1)
  8. 8
    Describe the transformation from y = 2x to y = 2x − 1.
    (2)
    A translation of 1 unit downwards✓✓ (2)
  9. 9
    For y = 2x − 3 + 1, write down the equation of the asymptote and describe the transformation from y = 2x.
    (3)
    Asymptote y = 1 ✓; the graph is shifted 3 units RIGHT ✓ and 1 unit UP ✓(3)
  10. 10
    Determine the y-intercept of y = 4 · 3x + 1 − 5.
    (3)
    x = 0: y = 4 × 31 − 5 ✓ = 12 − 5 ✓ = 7 ✓(3)
  11. 11
    Explain why the value of p does NOT change the horizontal asymptote of an exponential graph.
    (2)
    p is a horizontal shift ✓; the asymptote is a horizontal line whose height is set by q only, so sliding the graph sideways leaves it at y = q ✓(2)
TOTAL: 23 marks
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