Functions and Transformations | IEB Grade 12 Maths

Functions are the backbone of Paper 1, worth about 35 marks. This topic covers notation, how a graph moves when its equation changes, and the key features of the parabola, hyperbola and exponential graph that every other function question builds on.

Functional notation

$f (x)$ is the output of the function $f$ when the input is $x$ . To evaluate, replace every $x$ with the input, in brackets.

Example. $f (x) = 2 x^{2} - 3$

$f (- 2) = 2 (- 2)^{2} - 3 = 5$
$f (a + 1) = 2 (a + 1)^{2} - 3 = 2 a^{2} + 4 a - 1$

A function gives exactly one output for each input. On a graph, a vertical line cuts a function at most once.

Domain: the allowed $x$ -values.
Range: the $y$ -values the function produces.

Transformations

Starting from $y = f (x)$ :

$y = f (x - a)$ moves the graph $a$ units right (left if $a < 0$ ).
$y = f (x) + a$ moves the graph $a$ units up.
$y = a f (x)$ stretches the graph vertically by a factor $a$ . If $a < 0$ it also reflects in the $x$ -axis.
$y = f (- x)$ reflects the graph in the $y$ -axis.

The horizontal shift feels backwards: $f (x - 3)$ moves right by 3, because $x$ must be 3 bigger to give the same output.

Example. $f (x) = x^{2}$ has its turning point at $(0; 0)$ . Then $g (x) = f (x + 2) - 5 = (x + 2)^{2} - 5$ has its turning point at $(- 2; - 5)$ .

Parabola, hyperbola and exponential graphs

Parabola $y = a (x + p)^{2} + q$

Turning point $(- p; q)$ , axis of symmetry $x = - p$
$a > 0$ opens up (minimum); $a < 0$ opens down (maximum)
Range: $y \geq q$ if $a > 0$ , or $y \leq q$ if $a < 0$

Hyperbola $y = \frac{a}{x + p} + q$

Asymptotes $x = - p$ and $y = q$
Axes of symmetry: $y = x + p + q$ and $y = - x - p + q$ , through the point where the asymptotes meet

Exponential $y = a b^{x + p} + q$ , with $b > 0$ , $b \neq = 1$

Horizontal asymptote $y = q$
Increasing if $b > 1$ and $a > 0$

For every sketch, show intercepts with both axes, asymptotes as dotted lines, and turning points, all labelled.

Example. $g (x) = \frac{3}{x - 1} + 2$ : asymptotes $x = 1$ and $y = 2$ . $y$ -intercept: $g (0) = - 1$ . $x$ -intercept: $\frac{3}{x - 1} = - 2 \Rightarrow x = - \frac{1}{2}$ .

Key terms

Domain: The set of input values for which a function is defined.
Range: The set of output values a function produces.
Asymptote: A line a graph approaches but never touches.
Axis of symmetry: A line that divides a graph into two mirror images.
Translation: A shift of a graph left, right, up or down without changing its shape.

Common mistakes

Shifting $f (x - 3)$ to the left instead of the right.
Writing the turning point of $y = a (x + p)^{2} + q$ as $(p; q)$ instead of $(- p; q)$ .
Drawing a hyperbola or exponential graph that crosses its asymptote.
Forgetting brackets when substituting a negative number, so $- 2^{2}$ becomes $- 4$ .
Stating the range of a parabola with the wrong inequality direction.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Given $f (x) = 2 x^{2} - 3$ . Determine:

(a) $f (- 2)$ (1)

(b) $f (a + 1)$ , simplified (2)

Worked solution

Hidden - try the question first

Solution

(a) $f (- 2) = 2 (4) - 3 = 5$
(b) $f (a + 1) = 2 (a + 1)^{2} - 3 = 2 (a^{2} + 2 a + 1) - 3$
$= 2 a^{2} + 4 a - 1$

Question 2

5 marks total

routine

Given $g (x) = \frac{3}{x - 1} + 2$ .

(a) Write down the equations of the asymptotes of $g$ . (2)

(b) Determine the intercepts of $g$ with the axes. (3)

Worked solution

Hidden - try the question first

Solution

(a) $f (- 2) = 2 (4) - 3 = 5$
$y = 2$
(b) $y$ -intercept: $g (0) = - 3 + 2 = - 1$
$x$ -intercept: $\frac{3}{x - 1} = - 2 \Rightarrow 3 = - 2 x + 2$
$x = - \frac{1}{2}$

Question 3

7 marks total

complex

Given $h (x) = 2^{x + 1} - 4$ .

(a) Write down the equation of the asymptote and the range of $h$ . (2)

(b) Determine the intercepts of $h$ with the axes. (3)

(c) The graph of $h$ is translated so that its asymptote becomes $y = 0$ and its $y$ -intercept becomes $(0; 1)$ . Write down the equation of the new graph and describe the translation. (2)

Worked solution

Hidden - try the question first

Solution

(a) Asymptote $f (- 2) = 2 (4) - 3 = 5$ ; range $y = 2$
(b) $y$ -intercept: $h (0) = 2 - 4 = - 2$
$x$ -intercept: $\frac{3}{x - 1} = - 2 \Rightarrow 3 = - 2 x + 2$
$x + 1 = 2 \Rightarrow x = 1$
(c) $y = 2^{x}$ (asymptote $y = 0$ , $y$ -intercept $(0; 1)$ )
4 units up (removes the $- 4$ ) and 1 unit to the right (replaces $x + 1$ with $x$ )

Question 4

9 marks total

problem-solving

The parabola $f$ has a turning point at $(2; 9)$ and passes through $(5; 0)$ .

(a) Show that $f (x) = - (x - 2)^{2} + 9$ . (3)

(b) Determine the $x$ -intercepts of $f$ . (2)

(d) Write down the range of $h (x) = 2 f (x)$ and the turning point of $p (x) = f (- x)$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $f (- 2) = 2 (4) - 3 = 5$
$0 = a (3)^{2} + 9$
$a = - 1$ , so $f (x) = - (x - 2)^{2} + 9$
(b) $(x - 2)^{2} = 9 \Rightarrow x - 2 = \pm 3$
$x = 5$ or $x = - 1$
(c) $g (x) = - (x - 3)^{2}$
Its turning point $(3; 0)$ lies on the $x$ -axis, so it touches the axis once
(d) Range of $h$ : $y \leq 18$
Turning point of $p$ : $(- 2; 9)$

FAQs

Which graphs are examined in IEB Grade 12 Paper 1?

Straight lines, parabolas, hyperbolas, exponential and logarithmic graphs, and cubics in the calculus section. Trigonometric graphs are examined in Paper 2.

How do I remember which way a graph shifts?

Anything added inside the bracket with $x$ moves the graph horizontally and works opposite to the sign. Anything added outside moves it vertically in the direction of the sign.

Do I need to show asymptotes on my sketch?

Yes. Draw them as dotted lines and label their equations. Missing asymptotes is one of the most common ways to lose sketch marks.

Functions, notation and transformations

Functional notation

Transformations

Parabola, hyperbola and exponential graphs

Key terms

Common mistakes

Practice questions

FAQs

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