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Grade 11 · Functions
Hyperbola & Exponential Graphs (Grade 11)
MARKING GUIDELINE
Marks
39
Duration
1 hour
Questions
3
Name:
Class:
Date:
Mark
/ 39
Instructions and Information
- Answer ALL the questions in this question paper.
- Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
- Show ALL calculations clearly.
- Show all units where applicable.
- Number the answers correctly according to the numbering system used in this question paper.
- A non-programmable calculator may be used, unless stated otherwise.
- Write neatly and legibly.
Question 1
[10 MARKS]Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.
- 1.1The asymptotes of y = 2x − 1 + 3 are …(1)A)x = 1 and y = 3✓B)x = −1 and y = 3C)x = 1 and y = 0D)x = 3 and y = 1Answer: A — The vertical asymptote is where the denominator is zero; the horizontal one is the shift q.
- B — x − 1 = 0 gives x = +1
- C — the + 3 lifts the horizontal asymptote off the x-axis
- D — the two asymptotes have been swapped
- 1.2The y-intercept of y = −2x − 1 + 3 is …(1)A)5✓B)1C)3D)−2Answer: A — At x = 0, y = −2 ÷ (−1) + 3 = 2 + 3 = 5.
- B — took −2 ÷ (−1) as −2 instead of +2
- C — gave only the shift and left out the fraction
- D — gave the value of a instead of evaluating at x = 0
- 1.3The domain of y = 3x + 2 is …(1)A)x ∈ R, x ≠ −2✓B)x ∈ R, x ≠ 2C)x ∈ RD)x > −2Answer: A — x + 2 = 0 at x = −2, where the expression is undefined.
- B — x + 2 = 0 gives x = −2, not +2
- C — x = −2 must be excluded
- D — values below −2 are perfectly allowed
- 1.4The range of y = 4x − 1 − 2 is …(1)A)y ∈ R, y ≠ −2✓B)y ∈ R, y ≠ 1C)y ∈ R, y ≠ 0D)y > −2Answer: A — The curve never reaches its horizontal asymptote, y = −2.
- B — 1 is the VERTICAL asymptote's x-value
- C — the vertical shift moves the excluded value away from 0
- D — the curve takes values both above and below −2
- 1.5The axes of symmetry of y = ax are …(1)A)y = x and y = −x✓B)the x-axis and the y-axisC)y = x onlyD)it has noneAnswer: A — A hyperbola centred at the origin is symmetrical about both diagonals.
- B — the axes are the ASYMPTOTES, which the curve never touches
- C — there are TWO axes of symmetry, one of each gradient
- D — it has two
- 1.6For y = ax − 1 + 3, the axis of symmetry with a POSITIVE gradient is …(1)A)y = x + 2✓B)y = x − 2C)y = −x + 4D)y = xAnswer: A — It runs through the centre (1 ; 3) with gradient 1, and 3 = 1 + c gives c = 2.
- B — substituting the centre gives 3 = 1 + c, so c is +2, not −2
- C — that is the axis with a NEGATIVE gradient
- D — y = x passes through the origin, not through (1 ; 3)
- 1.7The asymptote of y = 3(2x) − 6 is …(1)A)y = −6✓B)y = 0C)y = 3D)x = 0Answer: A — The exponential part shrinks towards 0, leaving y = −6.
- B — that is the asymptote before the vertical shift
- C — 3 is the value of a, not the shift
- D — an exponential graph has no VERTICAL asymptote
- 1.8The x-intercept of y = 3(2x) − 6 is …(1)A)x = 1✓B)x = 2C)x = 6D)x = 0Answer: A — 3(2x) = 6 gives 2x = 2, so x = 1.
- B — solved 2x = 2 and then wrote down the base
- C — divided 6 by 1 instead of by 3
- D — x = 0 gives y = 3 − 6 = −3, not 0
- 1.9The y-intercept of y = 3(2x) − 6 is …(1)A)−3✓B)3C)−6D)0Answer: A — At x = 0, 20 = 1, so y = 3 − 6 = −3.
- B — forgot to subtract the 6
- C — took 20 as 0 instead of 1
- D — that is the y-value at the X-intercept, not the y-intercept
- 1.10As x becomes very large and NEGATIVE, the graph of y = 3(2x) − 6 approaches …(1)A)y = −6 from above✓B)y = −6 from belowC)y = 0D)no limiting value at allAnswer: A — 3(2x) stays POSITIVE however small it becomes, so the curve sits just above −6.
- B — 3(2x) is always positive, so the curve stays ABOVE the asymptote
- C — the −6 shift moves the asymptote off the x-axis
- D — the curve levels off at −6 rather than falling forever
Answer grid — marking guideline
| 1.1 | A | B | C | D |
| 1.2 | A | B | C | D |
| 1.3 | A | B | C | D |
| 1.4 | A | B | C | D |
| 1.5 | A | B | C | D |
| 1.6 | A | B | C | D |
| 1.7 | A | B | C | D |
| 1.8 | A | B | C | D |
| 1.9 | A | B | C | D |
| 1.10 | A | B | C | D |
Question 2
[15 MARKS]Given the hyperbola g(x) = 4x − 2 + 1.
- 2.1Write down the equations of the asymptotes, and explain how each one is read off the equation.(3)Vertical: x = 2 (1)
Horizontal: y = 1 (1)
x = 2 makes the denominator zero, and the + 1 lifts the horizontal asymptote from y = 0 to y = 1 (1) - 2.2Calculate the coordinates of the intercepts with the axes.(5)y-intercept: g(0) = 4−2 + 1 = −1, so (0 ; −1) (2)
x-intercept: 4x − 2 = −1 gives x − 2 = −4 (2)
so (−2 ; 0) (1) - 2.3Write down the equations of the two axes of symmetry of g.(4)Both pass through the centre (2 ; 1) with gradients 1 and −1 (2)
y = x − 1 (1) and y = −x + 3 (1) - 2.4Write down the domain and the range of g, and explain why each one excludes exactly ONE value.(3)Domain: x ∈ ℝ, x ≠ 2 (1)
Range: y ∈ ℝ, y ≠ 1 (1)
The two excluded values are the asymptotes — the curve approaches each one without ever reaching it (1)
Question 3
[14 MARKS]Given the exponential function h(x) = 2(3)x − 6.
- 3.1Write down the equation of the asymptote of h, and the coordinates of its y-intercept.(3)The asymptote is y = −6 (1)
Let x = 0: h(0) = 2(3)0 − 6 = 2 − 6 (1)
= −4, so the y-intercept is (0 ; −4) (1) - 3.2Calculate the coordinates of the intercepts with the axes.(5)y-intercept: h(0) = 2 − 6 = −4, so (0 ; −4) (2)
x-intercept: 2(3)x = 6, so 3x = 3 (2)
so (1 ; 0) (1) - 3.3Write down the range of h, and explain why h has exactly ONE x-intercept.(3)Range: y > −6 (1)
2(3)x is positive for every x, so h never falls to −6, but it grows without bound as x increases (1)
An increasing graph that starts below the x-axis and rises without bound crosses it exactly once (1) - 3.4Determine the equation of the graph obtained when h is shifted 2 units to the LEFT.(3)Replace x by (x + 2) (2)
y = 2(3)x + 2 − 6 (1)
TOTAL: 39 marks
This question paper consists of 3 questions.