Mathbench

IEB practice · Paper I

IEB Functions and graphs practice

8 newly generated questions across all 10 functions and graphs topics the IEB examines in Paper I, 90 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

Download the booklet as PDF · Sit this paper under time · Afrikaans

The parabola

2 questions · 23 marks

110 marks

In the diagram below, the parabola f cuts the x-axis at A(2; 0) and B(6; 0), and the y-axis at C(0; −12).

xyOA(2; 0)B(6; 0)C(0; −12)f
  1. (a)
    Determine the equation of f in the form f(x) = ax^2 + bx + c.
    (3)
  2. (b)
    Determine the coordinates of the turning point of f.
    (2)
  3. (c)
    Write down the range of f.
    (1)
  4. (d)
    For which values of k will f(x) = k have two different real roots?
    (2)
  5. (e)
    The graph of f is moved 3 units to the left and 1 unit up to give g. Write down the coordinates of the turning point of g.
    (2)
Show the answers

(a) Answer: f(x) = −x^2 + 8x − 12

  1. f(x) = a(x − 2)(x − 6)
  2. C: −12 = a(−2)(−6), so a = −1
  3. f(x) = −x^2 + 8x − 12

(b) Answer: (4; 4)

  1. x = (2 + 6)/2 = 4
  2. f(4) = 4

(c) Answer: y ≤ 4

  1. the turning value is 4
  2. y ≤ 4

(d) Answer: k < 4

  1. the line y = k must cut f twice
  2. k < 4

(e) Answer: (1; 5)

  1. (4 − 3; 4 + 1)
  2. (1; 5)
213 marks

In the diagram below, the parabola f with turning point T(−1; 5) and y-intercept (0; 4), and the straight line g, are drawn. The graphs meet at A and B.

xyOT(−1; 5)(0; 4)ABfg
  1. (a)
    Determine the equation of f in the form f(x) = a(x − p)^2 + q.
    (3)
  2. (b)
    g(x) = −3x − 8. Determine the coordinates of A and B, the points where f and g meet.
    (5)
  3. (c)
    For which values of x is f(x) ≥ g(x)?
    (2)
  4. (d)
    Write down the range of f.
    (1)
  5. (e)
    For which values of t will f(x) = t have no real roots?
    (2)
Show the answers

(a) Answer: f(x) = −(x + 1)^2 + 5

  1. f(x) = a(x + 1)^2 + 5
  2. substitute (0; 4): 4 = a(1)^2 + 5
  3. a = −1
  4. f(x) = −(x + 1)^2 + 5

(b) Answer: A(−3; 1) and B(4; −20)

  1. −(x + 1)^2 + 5 = −3x − 8
  2. −x^2 + x + 12 = 0
  3. −(x + 3)(x − 4) = 0
  4. x = −3 or x = 4
  5. A(−3; 1) and B(4; −20)

(c) Answer: −3 ≤ x ≤ 4

  1. read from the graph at A and B
  2. −3 ≤ x ≤ 4

(d) Answer: y ≤ 5

  1. the turning point is a maximum: y = 5

(e) Answer: t > 5

  1. the line y = t must miss the parabola entirely
  2. t > 5

The hyperbola

2 questions · 26 marks

312 marks

Given: f(x) = −8/(x − 3) − 2

xyO
  1. (a)
    Determine f(0).
    (1)
  2. (b)
    Determine the value of x for which f(x) = 0.
    (2)
  3. (c)
    Sketch the graph of f on the axes provided. Label clearly all asymptotes and intercepts with the axes.
    (5)
  4. (d)
    Write down the domain of f.
    (1)
  5. (e)
    For which values of x is f(x) > 0?
    (2)
  6. (f)
    Write down the equation of the graph formed when f is reflected in the x-axis.
    (1)
Show the answers

(a) Answer: f(0) = 2/3

  1. f(0) = −8/(−3) − 2 = 2/3

(b) Answer: x = −1

  1. −8/(x − 3) = 2
  2. −8 = 2(x − 3)
  3. x = −1

(c) Answer: See the sketch below.

xyOx = 3y = −2(0; 2/3)(−1; 0)f
  1. asymptotes: x = 3 and y = −2
  2. y-intercept (0; 2/3); x-intercept (−1; 0)

(d) Answer: x ∈ ℝ, x ≠ 3

  1. f is undefined only at x = 3

(e) Answer: −1 < x < 3

  1. read from the graph: the x-intercept is x = −1 and the asymptote x = 3 is never included
  2. −1 < x < 3

(f) Answer: y = 8/(x − 3) + 2

  1. y = −f(x) = 8/(x − 3) + 2
414 marks

In the diagram below, the graph of f(x) = a/(x + p) + q is drawn with its asymptotes. A(2; 3) lies on f.

xyOx = −1y = 2A(2; 3)f
  1. (a)
    Write down the values of p and q.
    (2)
  2. (b)
    Determine the value of a.
    (2)
  3. (c)
    Determine the equations of the axes of symmetry of f.
    (2)
  4. (d)
    The axis of symmetry y = x + 3 meets the graph of f at B and D. Determine the coordinates of B and D, leaving your answers in surd form where necessary.
    (3)
  5. (e)
    Write down the domain of f.
    (1)
  6. (f)
    For which values of x is f(x) ≥ 0?
    (2)
  7. (g)
    The graph of f is shifted 1 unit to the left and 2 units down to form g. Write down the equation of g.
    (2)
Show the answers

(a) Answer: p = 1 and q = 2

  1. the asymptotes are x = −1 and y = 2
  2. x + p = 0 at x = −1, so p = 1; q = 2

(b) Answer: a = 3

  1. 3 = a/(2 + 1) + 2
  2. a = (3 − 2)(2 + 1) = 3

(c) Answer: y = x + 3 and y = −x + 1

  1. both pass through the point where the asymptotes meet, (−1; 2)
  2. y = (x + 1) + 2 and y = −(x + 1) + 2

(d) Answer: (−1 − √3; 2 − √3) and (−1 + √3; 2 + √3)

  1. 3/(x + 1) + 2 = x + 3
  2. (x + 1)^2 = 3
  3. x = −1 ± √3
  4. (−1 − √3; 2 − √3) and (−1 + √3; 2 + √3)

(e) Answer: x ∈ ℝ, x ≠ −1

  1. f is undefined only at x = −1

(f) Answer: x ≤ −5/2 or x > −1

  1. read from the graph: the x-intercept is x = −5/2 and the asymptote x = −1 is never included
  2. x ≤ −5/2 or x > −1

(g) Answer: g(x) = 3/(x + 2) + 0

  1. g(x) = f(x + 1) − 2
  2. g(x) = 3/(x + 2) + 0

Exponential equations and exponent laws

2 questions · 24 marks

511 marks

In the diagram below, the graph of h(x) = 3^(x + p) + q is drawn. The line y = −7 is an asymptote of h, and A(1; 2) lies on h.

xyOy = −7A(1; 2)h
  1. (a)
    Write down the range of h.
    (1)
  2. (b)
    Given that h(x) = 3^(x + p) + q, determine the values of p and q.
    (3)
  3. (c)
    Determine the x-intercept of h, correct to two decimal places.
    (3)
  4. (d)
    The graph of h is reflected in the y-axis to give k. Write down the equation of k.
    (2)
  5. (e)
    For which values of x is h(x) > 0?
    (2)
Show the answers

(a) Answer: y > −7

  1. the asymptote is y = −7 and h lies above it

(b) Answer: p = 1 and q = −7

  1. q = −7, from the asymptote
  2. 2 = 3^(1 + p) − 7
  3. 3^(1 + p) = 9 = 3^2
  4. p = 1

(c) Answer: (0.77; 0)

  1. 3^(x + 1) − 7 = 0
  2. 3^(x + 1) = 7
  3. x = log_3 7 − 1 = 0.77

(d) Answer: k(x) = 3^(−x + 1) − 7

  1. a reflection in the y-axis replaces x by −x
  2. k(x) = 3^(−x + 1) − 7

(e) Answer: x > 0.77

  1. h is increasing and cuts the x-axis at x = 0.77
  2. x > 0.77
613 marks

In the diagram below, f(x) = b^x and g, the reflection of f in the line y = x, are drawn. P(2; 1/4) lies on f.

xyOy = xP(2; 1/4)ABfg
  1. (a)
    Determine the value of b.
    (2)
  2. (b)
    Write down the equation of g in the form y = …
    (2)
  3. (c)
    Write down the domain of g.
    (1)
  4. (d)
    For which values of x is g(x) > −1?
    (2)
  5. (e)
    The line x = 2 meets f at A and g at B. Determine the length of AB.
    (3)
  6. (f)
    h(x) = f(x) − 3. Write down the equation of the asymptote of h and determine the x-intercept of h, correct to two decimal places.
    (3)
Show the answers

(a) Answer: b = 1/2

  1. P lies on f: b^2 = 1/4
  2. b = 1/2

(b) Answer: y = log_(1/2) x

  1. g is the reflection of f in y = x: x = (1/2)^y
  2. y = log_(1/2) x

(c) Answer: x > 0

  1. the domain of g is the range of f

(d) Answer: 0 < x < 2

  1. g(x) = −1 when x = (1/2)^(−1) = 2
  2. g is decreasing (0 < b < 1), so the inequality sign is reversed
  3. 0 < x < 2

(e) Answer: AB = 5/4 units

  1. A(2; 1/4): f(2) = (1/2)^2 = 1/4
  2. B(2; −1): g(2) = log_(1/2) 2 = −1
  3. AB = 1/4 − (−1) = 5/4

(f) Answer: y = −3; x-intercept (−1.58; 0)

  1. f's asymptote y = 0 moves down 3: y = −3
  2. (1/2)^x − 3 = 0, so (1/2)^x = 3
  3. x = log_(1/2) 3 = −1.58

Fitting a graph to something real

2 questions · 17 marks

78 marks

A ball is thrown from the top of a building 4 m high. It reaches its greatest height of 7 m at a horizontal distance of 6 m from the building. In the diagram the foot of the building is at the origin and the ground lies along the x-axis.

xyO(0; 4)M(6; 7)L
  1. (a)
    Determine the equation of the path of the ball in the form y = a(x − p)^2 + q.
    (3)
  2. (b)
    Calculate the height of the ball 11 m horizontally from the building.
    (2)
  3. (c)
    Calculate, correct to two decimal places, how far from the building the ball lands at L.
    (3)
Show the answers

(a) Answer: y = −1/12(x − 6)^2 + 7

  1. y = a(x − 6)^2 + 7
  2. at (0; 4): 4 = a(−6)^2 + 7
  3. a = −1/12
  4. y = −1/12(x − 6)^2 + 7

(b) Answer: 59/12 m

  1. y = −1/12(11 − 6)^2 + 7
  2. y = 59/12 m

(c) Answer: 15.17 m

  1. −1/12(x − 6)^2 + 7 = 0
  2. (x − 6)^2 = 84
  3. x = 6 + √(84)
  4. x = 15.17 m
89 marks

A cable hangs between two poles, each 9 m high and 16 m apart. Its lowest point is 2 m above the ground, halfway between the poles. In the diagram the foot of the left pole is at the origin and the ground lies along the x-axis.

xyO(0; 9)(16; 9)E(8; 2)
  1. (a)
    Determine the equation of the cable in the form y = a(x − p)^2 + q.
    (3)
  2. (b)
    Calculate the height of the cable 9 m horizontally from the left pole.
    (2)
  3. (c)
    Determine, correct to two decimal places, the distances from the left pole at which the cable is 6 m above the ground.
    (4)
Show the answers

(a) Answer: y = 7/64(x − 8)^2 + 2

  1. y = a(x − 8)^2 + 2
  2. at (0; 9): 9 = a(−8)^2 + 2
  3. a = 7/64
  4. y = 7/64(x − 8)^2 + 2

(b) Answer: 135/64 m

  1. y = 7/64(9 − 8)^2 + 2
  2. y = 135/64 m

(c) Answer: 1.95 m and 14.05 m

  1. 7/64(x − 8)^2 + 2 = 6
  2. (x − 8)^2 = 256/7
  3. x = 8 ± √(256/7)
  4. x = 1.95 m or x = 14.05 m

Download this booklet

The same questions as a booklet to print, with the memorandum as a separate file so you can work without the answers in front of you. In English and Afrikaans.

Every question on this page was generated for this site by a computer engine, and its answer was confirmed by an independent method at the moment the question was made.

All IEB practice · Sit this paper under time · Afrikaans