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Grade 10 · Functions & Graphs
The Hyperbola & Exponential Graph (Grade 10)
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 = 3x are …(1)A)x = 0 and y = 0✓B)x = 3 and y = 3C)y = 3 onlyD)x = 0 onlyAnswer: A — The curve approaches both axes but never reaches either.
- B — 3 is the value of a, which controls the shape, not the asymptotes
- C — y = 3 would be the asymptote of y = 3/x + 3
- D — there are TWO asymptotes, one vertical and one horizontal
- 1.2The horizontal asymptote of y = 2x + 3 is …(1)A)y = 3✓B)y = 0C)y = 2D)x = 3Answer: A — The + 3 shifts the whole graph up by 3, and the asymptote moves with it.
- B — that is the asymptote BEFORE the vertical shift
- C — 2 is the value of a, which controls the shape, not the shift
- D — the VERTICAL asymptote is still x = 0
- 1.3The graph of y = ax with a < 0 lies in quadrants …(1)A)2 and 4✓B)1 and 3C)1 and 2D)3 and 4Answer: A — With a negative, x and y always have OPPOSITE signs.
- B — that is where y = a/x lies when a is POSITIVE
- C — quadrants 1 and 2 are separated by the y-axis, which the hyperbola never crosses
- D — quadrants 3 and 4 are separated by the y-axis, which the hyperbola never crosses
- 1.4The y-intercept of y = 3 × 2x is …(1)A)3✓B)1C)2D)0Answer: A — At x = 0, 20 = 1, so y = 3 × 1 = 3.
- B — took 20 as 1 but then forgot to multiply by the 3
- C — read off the base instead of working the value out
- D — took 20 as 0 instead of 1
- 1.5The asymptote of y = 2x + 1 is …(1)A)y = 1✓B)y = 0C)y = 2D)x = 0Answer: A — The + 1 lifts the whole curve, and the asymptote lifts with it.
- B — that is the asymptote before the shift
- C — 2 is the base, not the shift
- D — an exponential graph has no VERTICAL asymptote
- 1.6Does y = 2x have an x-intercept?(1)A)no — 2x is never 0✓B)yes, at x = 0C)yes, at x = 1D)yes, at x = 2Answer: A — A positive base raised to any power stays positive, so the curve never reaches the x-axis.
- B — at x = 0 the graph is at y = 1, which is the Y-intercept
- C — at x = 1 the graph is at y = 2
- D — at x = 2 the graph is at y = 4
- 1.7y = 3x is increasing. The graph of y = (13)x is …(1)A)decreasing✓B)also increasingC)a straight lineD)undefined for negative xAnswer: A — A base between 0 and 1 gives a decreasing exponential graph.
- B — a base smaller than 1 makes each step SMALLER, not bigger
- C — an exponential graph is a curve, never a straight line
- D — (1/3)x is defined for every real value of x
- 1.8The domain of y = 4x is …(1)A)x ∈ R, x ≠ 0✓B)x ∈ RC)x > 0D)y ∈ R, y ≠ 0Answer: A — Division by zero is undefined, so x = 0 must be left out.
- B — x = 0 must be excluded — it would divide by zero
- C — negative values of x are perfectly allowed
- D — that is the RANGE, not the domain
- 1.9The point (2 ; 6) lies on y = ax. Then a = …(1)A)12✓B)3C)8D)13Answer: A — 6 = a ÷ 2, so a = 12.
- B — DIVIDED 6 by 2 instead of multiplying
- C — ADDED 6 and 2 instead of multiplying them
- D — inverted the ratio and worked out 2 ÷ 6
- 1.10For y = 5 × 3x, each time x increases by 1 the value of y is …(1)A)multiplied by 3✓B)increased by 3C)multiplied by 5D)increased by 5Answer: A — Each unit step in x multiplies y by the BASE, which is 3.
- B — an exponential graph MULTIPLIES; only a straight line adds a fixed amount
- C — 5 is the starting value, not the growth factor
- D — 5 is the starting value, and the change is a multiplication, not an addition
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
[13 MARKS]Given the function f(x) = 6x − 1 + 2.
- 2.1Write down the equations of the asymptotes of f, and explain how each one is read off the equation.(3)Vertical: x = 1 (1)
Horizontal: y = 2 (1)
x = 1 is the value that makes the denominator zero, and the + 2 lifts the horizontal asymptote from y = 0 to y = 2 (1) - 2.2Calculate the coordinates of the y-intercept of f.(3)Let x = 0: f(0) = 6−1 + 2 (2) = (0 ; −4) (1)
- 2.3Calculate the coordinates of the x-intercept of f.(4)Let y = 0: 6x − 1 = −2 (1)
6 = −2(x − 1) (1)
x − 1 = −3 (1)
(−2 ; 0) (1) - 2.4Write down the domain and the range of f, and explain why each one excludes exactly ONE value.(3)Domain: x ∈ ℝ, x ≠ 1 (1)
Range: y ∈ ℝ, y ≠ 2 (1)
The excluded values are precisely the asymptotes — the graph approaches each one but never reaches it (1)
Question 3
[16 MARKS]Given the function g(x) = 2(3)x − 6.
- 3.1Write down the equation of the asymptote of g, and the coordinates of its y-intercept.(3)The asymptote is y = −6 (1)
For the y-intercept let x = 0: g(0) = 2(3)0 − 6 = 2 − 6 (1)
= −4, so the y-intercept is (0 ; −4) (1) - 3.2Calculate the coordinates of the y-intercept of g.(3)g(0) = 2(1) − 6 (2) = (0 ; −4) (1)
- 3.3Calculate the coordinates of the x-intercept of g.(4)2(3)x = 6 (1)
3x = 3 (2)
(1 ; 0) (1) - 3.4Describe what happens to g as x becomes very small (very negative), and write down the range of g.(3)3x approaches 0 (1), so g approaches −6 without ever reaching it (1)
Range: y > −6 (1) - 3.5Determine the value of g(2), and hence state whether the point (2 ; 10) lies on g.(3)g(2) = 2(9) − 6 = 12 (2)
12 ≠ 10, so (2 ; 10) does NOT lie on g (1)
TOTAL: 39 marks
This question paper consists of 3 questions.