Functions (Grade 10) — Mathematics Grades 8–11

Read, sketch and describe graphs of the straight line, parabola, hyperbola and exponential.

Gr 10 Term 2 - FunctionsTerm 2 - 4 weeks (CAPS)

Practise Functions in the app →

What gets asked

Decide whether a table, graph or equation represents a function.
Evaluate f(a) from a formula, or solve f(x) equal to a given value.
State the domain, range, asymptotes or intercepts of a given graph.
Describe how changing a or q moves or reshapes a graph.

You must be able to

Check that every input gives exactly one output before calling something a function.
Read f(3) as substitute 3 into f, not as f multiplied by 3.
Find where a graph crosses the axes by setting x to 0 or y to 0.
Tell a vertical shift (q) apart from a vertical stretch or reflection (a).

Traps that cost marks

Rejecting a relationship as a function just because two different inputs share one output. That is still allowed. A function only breaks if one input gives two different outputs.
Setting x to 0 when looking for the x-intercept, instead of setting y to 0. For the x-intercept, let y = 0. For the y-intercept, let x = 0. Easy to swap by mistake.
Reading a negative a as if it shifts the graph down, instead of flipping it over. A negative a reflects the graph in the x-axis. Only q moves a graph up or down.

Worked example

Given

f (x) = x^{2} - 4

, determine the x-intercepts.

Let $y = 0$ : $x^{2} - 4 = 0$ .
Factorise: $(x - 2) (x + 2) = 0$ .
$x = 2$ or $x = - 2$ .

The function concept

A rule you feed a number to, and the short way of writing what comes back out.

function: A rule that turns each input into exactly one output.
input: The number you feed into the rule.
output: The number the rule hands back to you.
function notation: Writing $f (x)$ for the output at input $x$ . The brackets never mean multiply.
table of values: Two rows: the inputs you picked, and the outputs the rule gave.

A function is a machine with a rule inside it. Put a number in, and exactly one number comes out. Prepaid electricity works like this. The rands you pay decide the units you get, and one amount can never give two different answers.

That is the whole test. One input may never give two different outputs. Two different inputs are allowed to share an output, and that breaks nothing. Check the inputs, not the outputs.

Function notation keeps things short. $f (3)$ means put $3$ in and see what comes out. It does not mean $f$ times $3$ . The letter $f$ is only the name of the rule.

Two questions look alike but run opposite ways. Working out $f (3)$ gives you an output. Solving $f (x) = 7$ hands you the output and asks for the input. Then the $7$ goes on the right, not into the rule.

One function can be shown four ways: in words, as a table of values, as a formula, and as a graph. The table is the bridge between them. Read words carefully. "Goes up by R5 each time" means add $5$ , not multiply by $5$ .

Rules to remember

$f (a)$ is the output when the input is $a$
One input, one output

Examples

A table has inputs

1

2

and

3

, giving outputs

4

4

and

9

. Is it a function?

Worked answer

Look at the inputs first. Each of $1$ , $2$ and $3$ appears once.
So no input is being asked for two different outputs.
Two inputs do share the output $4$ , and that is allowed.
So this table describes a function.

Answer: yes, it is a function

The test reads the inputs, and a repeated output breaks nothing.

Given

f (x) = 3 x - 5

, find

f (4)

Worked answer

$f (4)$ means put $4$ wherever $x$ stands.
So $f (4) = 3 (4) - 5$ .
$3 (4) = 12$ , and $12 - 5 = 7$ .
So the output is $7$ .

Answer: $f (4) = 7$

The brackets held the input, and nothing was multiplied by $f$ .

A tank holds

600

litres and loses

40

litres a day. Write a formula, then say when it runs dry.

Worked answer

It starts at $600$ and takes off $40$ for each day $d$ .
So the formula is $V = 600 - 40 d$ .
Dry means the volume is $0$ , so $600 - 40 d = 0$ .
Then $40 d = 600$ , so $d = 15$ .
A table agrees: at $d = 5$ the tank still holds $400$ litres.

Answer: $V = 600 - 40 d$ , and it is dry after $15$ days

"Loses $40$ a day" is a subtraction each day, so $40$ multiplies the days.

Traps

Rejecting a relationship as a function just because two different inputs share one output. That is still allowed. A function only breaks if one input gives two different outputs.
Reading $f (3)$ as $f$ multiplied by $3$ . The brackets hold the input. $f (3)$ means put $3$ into the rule.
Being asked to solve $f (x) = 10$ and working out $f (10)$ instead. Here $10$ is the output. Set the rule equal to $10$ and find the input.
Reading "goes up by R5 each time" as a multiplication by $5$ . "Goes up by" means add. It joins the rule as $+ 5 n$ , not as $\times 5$ .

Learn and practise “The function concept” in the app →

Key features of graphs

The handful of facts that pin a curve down, even when you cannot see the picture.

domain: Every input value the function is allowed to take.
range: Every output value the function actually reaches.
asymptote: A line these graphs creep towards but never touch.
turning point: Where a parabola stops falling and starts rising, or the other way about.
axis of symmetry: The line that folds a graph exactly onto itself.
intercept: A point where the graph cuts one of the axes.

Grade 10 uses four shapes: the line $y = m x + c$ , the parabola $y = a x^{2} + q$ , the hyperbola $y = \frac{a}{x} + q$ and the exponential $y = a b^{x} + q$ . Build a table of values and the shape shows itself.

Fill that table with care. Squaring a negative gives a positive: $(- 3)^{2} = 9$ . Nothing may be divided by $0$ , so for $y = \frac{6}{x}$ the cell at $x = 0$ stays empty. We call it undefined.

The domain is the inputs allowed. The range is the outputs reached. The parabola $y = x^{2} - 4$ takes any input at all. Its turning point sits at $- 4$ , and that is a floor, so the range is $y \geq - 4$ .

An asymptote is a line, so give it as an equation. For $y = \frac{6}{x} + 2$ the flat one is $y = 2$ , not $y = 0$ . The $+ 2$ lifted it. The upright one is $x = 0$ .

For a $y$ -intercept let $x = 0$ ; for an $x$ -intercept let $y = 0$ . A parabola $y = a x^{2} + q$ turns at $(0; q)$ , with $x = 0$ as its axis of symmetry. $y = 2^{x}$ stays above the $x$ -axis, so it has no $x$ -intercept.

Rules to remember

For $y = a x^{2} + q$ the turning point is $(0; q)$
For $y = \frac{a}{x} + q$ the asymptotes are $x = 0$ and $y = q$

Examples

Complete a table for

y = x^{2} - 4

x = - 2

x = 0

and

x = 2

, then give the range.

Worked answer

At $x = - 2$ : $(- 2)^{2} = 4$ , then $4 - 4 = 0$ .
At $x = 0$ : $0 - 4 = - 4$ .
At $x = 2$ : $2^{2} - 4 = 0$ .
The outputs are $0$ , $- 4$ and $0$ . The lowest is $- 4$ .
That low point is the turning point, so the range is $y \geq - 4$ .

Answer: outputs $0$ , $- 4$ , $0$ ; range $y \geq - 4$

A parabola bottoms out at its turning point, which gives the range a floor.

For

y = \frac{6}{x} + 2

give the domain and both asymptotes.

Worked answer

Dividing by $0$ is not allowed, so $x$ may not be $0$ .
The domain is every real number except $0$ .
As $x$ grows very large, $\frac{6}{x}$ shrinks towards $0$ .
So the outputs close in on $2$ , and the flat asymptote is $y = 2$ .
The upright asymptote is the line $x = 0$ .

Answer: domain: all real $x$ except $0$ ; asymptotes $x = 0$ and $y = 2$

The $+ 2$ carried the flat asymptote up off the $x$ -axis.

For

y = - 2 x^{2} + 8

give the turning point, the axis of symmetry and the

x

-intercepts.

Worked answer

Here $a = - 2$ and $q = 8$ , so the turning point is $(0; 8)$ .
The axis of symmetry is the line $x = 0$ .
For the intercepts on the $x$ -axis let $y = 0$ : $- 2 x^{2} + 8 = 0$ .
Divide by $- 2$ : $x^{2} - 4 = 0$ , so $(x - 2) (x + 2) = 0$ .
That gives $x = 2$ or $x = - 2$ .

Answer: $(0; 8)$ ; $x = 0$ ; $x = 2$ or $x = - 2$

Letting $y = 0$ found where the curve cuts across, and factorising split it in two.

Traps

Setting x to 0 when looking for the x-intercept, instead of setting y to 0. For the x-intercept, let y = 0. For the y-intercept, let x = 0. Easy to swap by mistake.
Giving the asymptote of $y = \frac{6}{x} + 2$ as just $2$ . An asymptote is a line, so it needs an equation: $y = 2$ .
Saying the range of $y = x^{2} - 4$ is every real number. The turning point stops it. The lowest output is $- 4$ , so $y \geq - 4$ .
Writing the turning point of $y = 2 x^{2} - 5$ as $(- 5; 0)$ . It sits on the $y$ -axis, so the turning point is $(0; - 5)$ .

Learn and practise “Key features of graphs” in the app →

Effect of a and q

Two letters in the equation decide whether a curve flips, steepens or simply rises.

parameter: A letter in the equation whose value you may change, like $a$ or $q$ .
vertical shift: A move of the whole graph straight up or straight down.
reflection: A flip of the graph across a line, into its mirror image.
stretch: A pull that makes a graph steeper, further from the $x$ -axis.
compression: A squash that makes a graph flatter, closer to the $x$ -axis.

Every graph in this section is written $y = a f (x) + q$ . The shape $f$ stays as it is. Two parameters do all the changing: $a$ in front and $q$ at the end. Learn what each does and you can read any of these graphs.

The parameter $q$ lifts or drops the graph — a vertical shift, nothing else. The turning point, the asymptote and the $y$ -intercept travel with it. The $x$ -intercepts do NOT — each old one now sits $q$ above the axis. Set $y = 0$ again to find them.

The parameter $a$ sets steepness and direction. A negative $a$ is a reflection in the $x$ -axis: the graph flips over, it does not drop. An $a$ bigger than $1$ gives a stretch, and an $a$ between $0$ and $1$ gives a compression.

Reading a sketch backwards uses the same facts. Find $q$ first, from the asymptote or the turning point. Then choose a point the curve really passes through, and put it in to find $a$ . A point on the asymptote is not on the curve.

Rules to remember

$q$ moves a graph up or down, never sideways
A negative $a$ reflects the graph in the $x$ -axis

Examples

Describe how

y = - 3 x^{2} + 4

differs from

y = x^{2}

Worked answer

Here $a = - 3$ and $q = 4$ .
The minus sign flips the parabola, so it opens downwards.
The $3$ pulls it steeper than $y = x^{2}$ , which is a stretch.
The $+ 4$ lifts every point by $4$ , a vertical shift.
So the turning point moves from $(0; 0)$ up to $(0; 4)$ .

Answer: flipped over, stretched, and shifted $4$ up

The sign of $a$ and the size of $a$ do two separate jobs.

Does

y = - x^{2} + 2

move down, or flip over?

Worked answer

The minus sign in front of $x^{2}$ is $a$ , not $q$ .
A negative $a$ always means a reflection in the $x$ -axis.
Here $q = 2$ , and $q$ is the only part that shifts.
So the graph flips over first, then rises by $2$ .
It opens downwards, with its turning point at $(0; 2)$ .

Answer: it flips over, then shifts $2$ up

A reflection can hide inside a shift, so check the sign of $a$ on its own.

A hyperbola

y = \frac{a}{x} + q

has asymptote

y = 3

and passes through

(1; 5)

. Find

a

and

q

Worked answer

The asymptote hands you $q$ at once: $q = 3$ .
So the equation so far is $y = \frac{a}{x} + 3$ .
The point $(1; 5)$ lies on the curve, so put $x = 1$ and $y = 5$ .
That gives $5 = \frac{a}{1} + 3$ , so $a = 2$ .
The equation is $y = \frac{2}{x} + 3$ .

Answer: $a = 2$ and $q = 3$

The asymptote fixed $q$ , and one genuine point on the curve fixed $a$ .

Traps

Reading a negative a as if it shifts the graph down, instead of flipping it over. A negative a reflects the graph in the x-axis. Only q moves a graph up or down.
Calling $y = \frac{1}{2} x^{2}$ a stretch. Between $0$ and $1$ , the parameter $a$ flattens a graph: a compression.
Leaving the asymptote at $y = 0$ after a $q$ has been added. The asymptote travels with the graph. For $y = \frac{2}{x} + 3$ it is $y = 3$ .
Using a point read off the asymptote as though it were on the curve. These graphs never touch their asymptotes. Pick a point the curve really goes through.

Learn and practise “Effect of a and q” in the app →

Interpreting graphs

Reading the story a graph tells, and deciding whether its points should be joined up at all.

discrete: Made of separate, countable values only, like whole loaves of bread.
continuous: Able to take any value in between, like time, mass or distance.
point of intersection: Where two graphs cross. There the two outputs are equal.
scale: The size of one step along an axis. Two graphs may use different ones.

A graph in a story is more than a picture. Read the label and units on both axes first. Then decide which axis carries the quantity you want.

Ask what the input can be. Loaves, learners, taxis: nobody has half of one, so those inputs are discrete. Their graph is a row of separate dots, never joined up. Time, mass and distance are continuous, so those graphs are unbroken lines.

How many dots were drawn tells you nothing. A graph of one day's temperature may show only five points, and temperature is still continuous. Judge from the quantity, not the picture.

When two graphs are compared, the point of intersection is where they give the same value. That is a boundary, not the answer to which is bigger. Test a value on each side. Check the scale on each axis first.

Rules to remember

At the point of intersection, both graphs give the same output
Whole things only means dots; anything in between means a line

Examples

A spaza shop sells cold drinks at R12 each. A graph shows cost against number of cans. Dots, or a line?

Worked answer

The input here is the number of cans.
Nobody buys half a can, so only whole numbers work.
The graph is separate dots at $1$ , $2$ , $3$ and onwards.
Do not join them, even though they line up neatly.

Answer: separate dots, because the number of cans is discrete

The input decided it. Nothing exists between one can and two.

Shop A charges R20 to set up a print job, then R3 a page. Shop B charges R5 a page and no set-up. When is A cheaper?

Worked answer

For $p$ pages, A costs $20 + 3 p$ and B costs $5 p$ .
They cross where $20 + 3 p = 5 p$ .
That gives $20 = 2 p$ , so $p = 10$ .
Test further along: at $p = 20$ , A costs $80$ and B costs $100$ .
So they tie at $10$ pages, and A wins past that.

Answer: A is cheaper for more than $10$ pages

The crossing point is only the boundary; a test past it settled the rest.

A graph shows litres up the side and hours along the bottom. A tank holds

600

litres and empties in

15

hours. How much is left after

9

hours?

Worked answer

Over $15$ hours the tank loses all $600$ litres.
So each hour it loses $600 \div 15 = 40$ litres.
After $9$ hours it has lost $9 \times 40 = 360$ .
Left in the tank: $600 - 360 = 240$ .
Read the $9$ along the bottom, and the $240$ up the side.

Answer: $240$ litres

Hours and litres sit on different axes, so each came off its own.

Traps

Drawing a line through the dots for the cost of $1$ , $2$ and $3$ cans. Half a can does not exist, so nothing sits between the dots. Leave them apart.
Calling a graph discrete because only five dots were plotted. Ask what the input is. Time stays continuous even when five points are shown.
Answering " $10$ pages" when asked where shop A is cheaper. At $10$ pages the two shops cost the same. A only wins past that point.
Reading the $9$ off the litres axis instead of the hours axis. Check the label and the scale before you read a number off an axis.

Learn and practise “Interpreting graphs” in the app →

Grade 10 Functions

What gets asked

You must be able to

Traps that cost marks

Worked example

The function concept

Rules to remember

Examples

Traps

Key features of graphs

Rules to remember

Examples

Traps

Effect of a and q

Rules to remember

Examples

Traps

Interpreting graphs

Rules to remember

Examples

Traps

About this material