Function Questions And Answers Mathematics Grade 12 Pdf Download

Function Questions and Answers Mathematics Grade 12. In mathematics, a function from a set X to a set Y assigns to each element of X exactly one element of Y. The set X is called the domain of the function and the set Y is called the codomain of the function. Functions were originally the idealization of how a varying quantity depends on another quantity

FUNCTIONS QUESTIONS AND ANSWERS GRADE 12

Activity 1

If h(x) = (½)^xdetermine the value of h(–4). (3)
If the function g(x) = – x² –3x, find g(x + h) (2)
If f(x) = 4x + 1, determine the value of:
3.1 f(x + a)
3.2 f(x) + a
3.3 af(x) (3)
If g(x) = 2x², determine the value of:
4.1 g(–x)
4.2 –g(x) (2)
[10]

Solutions

READ >> Matric (Grade 12) Mid-Year May June Exams Timetable for 2024 download

h(x) = (½)^x
∴ h(–4) = (½)^-4 (2^–1)^–4 = 2⁴ = 16
So when x = –4, y = 16 and the point (–4; 16) lies on the graph of the function 3h. (3)
g(x) = –x² – 3x
∴g(x + h) = –(x + h)² –3(x + h) wherever there is an x, replace it with (x + h)
= –(x² + 2xh + h²) – 3x – 3h
= –x² – 2xh – h² – 3x – 3h
This means that when x = x + h, y = –x² – 2xh – h² – 3x – 3h (2)
3.1 f(x) = 4x + 1
f(x + a) = 4(x + a)+1
= 4x + 4a + 1
3.2 f(x) = 4x + 1
f(x) + a = 4x + 1 + a
3.3 f(x) = 4x + 1
af(x) = a(4x + 1)
= 4ax + a
(3)
4.1 g(x) = 2x²g(–x) = 2(–x)²
= 2x²
4.2 g(x) = 2x²
–g(x) = –2x²
(2)
[10]

In each example, there is only one possible y-value for each x-value, so f(x); h(x) and g(x) are functions.

Activity 2
The sketch represents the graph of the parabola given by f(x) = 2 – x – x².
Points A, B and C are the intercepts on the axes and D is the turning point of the graph.
$15$
1.1 Determine the co-ordinates of A, B and C. (4)
1.2 Determine the co-ordinates of the turning point D. (3)
1.3 Write down the equation of the axes of symmetry of f(x– 5). (1)
1.4 Determine the values of x for which – f(x) > 0. (2)
[10]

Solutions
1.1B(0; 2)
2 – x – x² = 0
x² + x – 2 = 0
(x – 1)(x + 2) = 0
x = 1 or x = –2
A(–2; 0) and C(1; 0) 3 (4)

1.2 x = –b
2a
= – (–1)
2(–1)
= -½
f (-½) = 2 – (-½) – (-½)²
=⁹/₄ = 2¼
D (-½;⁹/₄)
1.3 x = 9 or x = 4½ (1)
1.4 x ≤ –2 or x ≥ 1 (2)
[10]

Activity 3
The sketch represents the graph of the parabola given by f(x) = ax² + bx + c and the straight line defined by g(x) = mx + c
Points A, B ,C and D are the intercepts on the axes. E is the point of intersection of the two graphs.
$17$
2.1 Write down the co-ordinates of point D if D is the image of B after B has been translated two units to the right. (1)
2.2 Determine the equation of g. (3)
2.3 Determine the equation of the function f in the form f(x) = ax² + bx + c. (4)
2.4 Determine the coordinates of E. (4)
2.5 Write down the values of x for which f(x) ≥ g(x). (2)
[14]

Solutions
2.1 D(5; 0) 3 (1)
2.2 g(x) = mx + 3
0 = m(5) + 3 or m_g = 3 – 0 = – 3
0 – 5 5
m = ^{– 3}/₅
g(x) = ^{– 3}/₅x + 3 (3)
2.3 f(x) = a(x + 1)(x – 3)
3 = a(0 + 1)(0 – 3)
a = 1
f(x) = –(x + 1)(x – 3)
f(x) = –x² + 2x + 3 (4)

2.4 ^{– 3}/₅x + 3 = –x² + 2x + 3
x²^{– 13}/₅ x = 0
$18$
2.5 0 ≤ x ≤ ¹³/₅ (2)
[14]

Activity 4

1. Sketch the graph of y = ¹/_x by plotting points.
Describe the main features of the graph. (4)
Solution
a = 1
a > 0, the graph lies in the 1st and 3rd quadrant

-3	– 2	– 1	– ½	0	½	1	2	3
^-1/₃	– ½	-1	-2	undefined	2	1	– ½	¹/₃

$20$

Domain: x ∈ R; x ≠ 0
Range: y ∈ R; y ≠ 0
Asymptotes: x = 0 and y = 0
Lines of symmetry y = x and y = –x (4)

2. Sketch the graph of y = ^–4/_x by plotting the points. Describe the main features of the graphs. (4)
Solution
a = –4
a < 0, the graph lies in the 2nd and 4th quadrant

-4	-2	-1	0	1	2	4
1	2	4	undefined	-4	-2	-1

$21$

Domain: x ∈ R; x ≠ 0
Range: y ∈ R; y ≠ 0
Asymptotes: x = 0 and y = 0
Lines of symmetry y = x and y = –x (4)

[8]

Activity 5

1. Consider the function y = ¹/_x– 2
1.1 Determine :

the equations of the asymptotes
the coordinates of the x–intercepts

1.2 Sketch the graph
1.3 Write down:

the domain and range
the lines of symmetry
y = x + c and y = –x + c
(10)

Solutions
1.1

The horizontal asymptote is y = –2 since the graph moved 2 units down and the vertical asymptote is x = 0 denominator cannot equal to zero.
For x – intercepts let y = 0
0 = 1 – 2
0 = ¹/_x – 2x (multiplying by LCD which is x)
2x = 1
x = ½
( ½;0)

2. Consider the function f(x) = ^–4/_x+ 1
2.1 Determine:

the equations of the asymptotes
the coordinates of the x–intercepts

2.2 Sketch the graph
2.3 Write down the domain and range
2.4 If the graph of f is reflected by the line having the equation y = –x + c, the new graph coincides with the graph of f(x).
Determine the value of c.
(9)

Solutions
2.1

The horizontal asymptote is y = 1 since the graph moved
1 units up and the vertical asymptote is x = 0 denominator cannot equal to zero.
For x–intercepts let y = 0
0 = ^-4/_x + 1
0 = –4 + x (m3ultiplying by LCD which is x
x = 4
(4; 0)

1.2

x	-4	-2	-1	0	1	2	4
y	-2¼	-2½	-3	undefined	-1	-1½	-1³/₄

$22$
1.3

Domain: x ∈ R; x ≠ 0
Range: y ∈ R; y ≠ 2
y = x and y = –x
translation 2 units down therefore
y = x – 2 and y = –x – 2
∴ c = –2
Or substitute (0; 2) point of intersection of the two asymptotes in
y = x + c or y = –x + c
And calculate the value of c
[10]

Compare this graph with the one in activity 4 (a)

2.2

x	-4	-2	-1	0	1	2	4
y	2	2	5	undefined	-3	-1	0

$23$
2.3 Domain: x ∈ R; y ≠ 0
Range: y ∈ R; y ≠ 1
2.4 The asymptotes are
x = 0 and y = 1
y = –x + c
1 = –(0) + c
1 = c
lines are y = –x + 1 and y = 3x + 1
[9]
Compare this graph with the one in activity 4 (b)

Activity 6

Consider the function f(x) = 2 + 1
x – 3
1. Write down the equations of the asymptotes of f (2)
2. Calculate the coordinates of the x and y-intercepts of f (4)
3. Write the domain and range (2)
4. Sketch the graph of f clearly showing ALL asymptotes and intercepts with the axes. (3)
Consider the function f(x) = 3 – 2
x – 1
1. Write down the equation of the asymptotes. (2)
2. Calculate the coordinates of the intercepts of the graph of f with the axes. (3)
3. Sketch the graph of f clearly showing the intercepts with the axes and the asymptotes. (3)
4. Write down the range of y = –f(x). (1)
5. Describe, in words, the transformation of f to g if g(x) = -3 – 2 (2)
  x + 1
  [22]

Solution

1. x = 3 and y = 1 (2)
2. f(x) = 2   + 1
  x – 3
  y – intercept y = 2   + 1  =  1
  0 – 3 3 3
  (0;¹/₃)
  x – intercept 0 = 2   + 1
  x – 3
  0 = 2 + 1(x – 3)
  0 = 2 + x – 3
  x = 1 ∴(1; 0) (4)
3. Domain: x ∈ R ; x ≠ 3
  Range: y ∈ R ; y ≠ 1 (2)
4. a > 0

$25$

intercepts
asymptotes
shape (3)
[11]

Solution
2.

3 x = –1 y = –2 (2)
y – intercept
y = 3 – 2 = –5
0 – 1
(0; –5)
x – intercept 0 = 3 –2
x – 1
2 = 3
x – 1
2(x – 1) = 3
2x – 2 = 3
2x = 5
x = 5
2
( ⁵/₂ ; 0 ) (3)
intercepts
asymptotes
shape (3)
a > 0
$26$
f(x) = 3 – 2
x – 1
– f(x) = – ( 3 – 2 )
x – 1
– f(x) = – 3 + 2
x – 1
Range: y ∈ R; y ≠ 2 (1)
g(x) = – 3 – 2
x + 1
g(x) = 3 – 2
-x -1
Since x is negative this is the reflectionof f about the y-axis (2)
[11]

In the graph 1 (d) the points (4; 3), x = 4 was chosen because it has x-coordinate greater than
x = 3 the vertical asymptote. The point (2; –1), was chosen because has x-coordinate x = 2 is less than x = 3 the vertical asymptote. These points can also be used to help determining in which quadrants the graph must be drawn. The points (2; 1) and (–2; –3) on graph 2 (iii) were chosen similarly.

Activity 7
The diagram below represents the graph of f(x) = a + q. T(5; 3) is a point on f.
x + p
$27$
4.1 Determine the values of a, p and q (4)
4.2 If the graph of f is reflected across the line having the equation y = –x + c, the new graph coincides with the graph of y = f(x).
Determine the value of c. (3)
[7]

Solutions
4.1 p = 4 and q = 2 using the asymptotes
Substitute T(5; 3) into y = a + 2
x – 4
3 = a + 2
5 – 4
3 = a + 2
a = 1 (4)
4.2 Substitute (4; 2) 3into y = –x + c
2 = –(4) + c ∴ c = 6 (3)
[7]

Activity 8
Sketched below are the graphs of f(x) = (x + p)² + q and g(x) = a + c A(2½;0) is a point on the graph of f. P is the turning point of f. The asymptotes of g are
x + b
represented by the dotted lines. The graph of g passes through the origin
$28$
5.1 Determine the equation of g. (4)
5.2 Determine the coordinates of P, the turning point of f. (4)
5.3 Write down the equation of the asymptotes of g(x – 1). (2)

5.4 Write down the equation of h, if h is the image of f reflected about the x–axis. (1)
[11]

Solutions
5.1 Using the asymptotes 3b = 1 and c = 2
Substitute (0; 0) into y =   a + 2
x – 1
0 = a   + 2 ⇒ 0 = –a + 2 ∴ a = 2
0 – 1
y = 2   + 2 (4)
x – 1
5.2 Axis of symmetry p = 1
f(x) = (x – 1)² + q
( ⁵/₂;0 )
0 = ( ⁵/₂– 1)² + q
0 = ⁹/₄+ q
q = – ⁹/₄ ∴ P ( 1; –⁹/₄) (4)
5.3 g(x) =   2   + 2
x – 1
g(x – 1) = 2     + 2 substitute x with (x – 1)
(x – 1) – 1
g(x – 1) = 2 + 2
x – 2
x = 2 and y = 2 (2)
5.4 f(x) = (x – 1)2 – ⁹/₄
Reflection about the x – axis y changes the sign
– y = (x – 1)² – ⁹/₄
y =– [(x – 1)² – ⁹/₄]
y = –(x – 1)² + ⁹/₄ (1)
[11]

Activity 9
The curve of an exponential function is given by f(x) = k^x and cuts the y–axis at A (0; 1) while B ( 2:⁹/₄) lies on the curve.
$34$
Determine
1.1 the equation of the function f. (3)
1.2 the equation of the asymptote of h if h(x) = –f(x). (2)
1.3 the range of h. (1)
1.4 The equation of the function g of which the curve is the reflection of the curve of f in the line y = x. (2)

Solutions
1.1 f(x) = k^x
⁹/₄ = a²
(³/₂)² = a²
a = ³/₂ ∴f(x) = ( ³/₂)^x(3)
1.2 y = 0 (2)
1.3 y ≤ 0 (1)
1.4 g(x) = log _³/2 x (2)
[8]

Activity 10

1. If f(x) = –3x² ,write down the equation for the inverse function in the form y = ……………….. (2)
2. Determine the domain and range of f(x) and f –1 (x) (4)
3. Determine the points of intersection of f(x) and f –1 (x) (4)
1. If g(x) = 3x + 2, find g –1(x) (2)
2. Sketch g, g –1 and the line y = x on the same set of axes. (3)
  [15]

Solutions

1. For f(x) = – 3x^2.f ^–1 (x): x = –3y²^-x/₃ = y²y = ± √^-x/₃(2)
2. f(x) f^–1(x)
  Domain x ∈ R x ≥ 0
  Range y ≥ 0 y ∈ R
  (4)
3. To determine the points of intersection, we equate the two equations.
  The line y = x, the axis of symmetry of f(x) and f –1(x), can also be used to determine the points of intersection of f(x) and f –1(x).
  y = x and f(x) = − 3x²
  ∴ x = −3x²
  ∴ 3x² + x = 0
  ∴ x(3x + 1) = 0
  ∴ x = 0 or x =^-1/₃
  Substitute x = 0 in y = x ∴ y = 0 ∴ (0; 0)
  Substitute x =^-1/₃ in y = x ∴ y = ^-1/₃∴ (^-1/₃; ¹/₃ ) (4)
1. g(x) = 3x + 2
  For g −1 (x), x = 3y + 2
  x – 2 = 3y
  y = x − 2
  3
  y = x − 2
  3 3
  $37$
  (4)
  [15]

Given: g (x) = – x² where x < 0 and y < 0

Write down the inverse of g, g^–1 in the form h(x) = ……….. (3)
Sketch the graphs of g, h and y = x on the same set of axis. (4)

Solutions

y = –x²
x = –y²
– x = y²
± √–x = y²
– √–x = y where x < 0 and y < 0
∴ h(x) = – √–x (3)
$38$
For g correct shape 3and the intercept 3
For h correct shape 3and the intercept 3 (4)
[7]

Activity 11
The graph of h(x) = ax is sketched below. A ( –1; ½ ) is a point on the graph of h.

$42$

Solutions

h(0) = a⁰ = 1. Any base raised to the power of 0 is 1. (2)
h(x) = a^x and A(–1; ½) so a –1 = ½
a ^–1 = 2 ^–1 so a = 2 and y = 2^x (2)
Interchange x and y, so x = 2^y and y = log₂ x (1)
(2)
x > 0.5 (1)
[8]

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