ATHENS ACADEMY · Mathematics · Grade 12 · Unit 1

ATHENS ACADEMY · Mathematics · Grade 12 · Unit 1 · Functions and Graphs

Unit 1 — Functions and Graphs

A function is nothing more than a reliable machine: put one number in, get exactly one number out, every time. This unit teaches you to read that machine four ways — as a rule, as a domain-and-range, as a type (one-one, onto, both), and as a picture on the plane. Almost every mark in this chapter is won by writing the definition first and then working in clean, numbered steps. Move slowly through the eight worked examples; they are the exact shapes the board asks.

1. What a Function Is

Think of a function as a juice machine at a Lahore bazaar. You feed it one fruit and it hands back one glass of juice. It never hands back two different glasses for the same fruit, and it never refuses a fruit it is meant to accept. That single rule — one input, exactly one output — is the whole idea.

Definition — function A function

f:A\to B

is a rule that assigns to every element of set

A

exactly one element of set

B

. No input may have two outputs.

A

is the domain,

B

is the codomain.

Three sets travel with every function, and the exam separates marks across all three. The domain is the set of allowed inputs — "what can go in." The range is the set of outputs that actually come out. The codomain is the set of outputs that are allowed to come out. The range always sits inside the codomain: $\text{Range}\subseteq B$ .

Memory hook — three sets "In, out-actual, out-allowed." Domain is what goes in; range is what actually comes out; codomain is what is allowed out. The range can be smaller than the codomain, never bigger.

2. Finding the Domain

For most functions the domain is every real number. Only two things ever spoil that, and you must hunt for both:

The only two domain restrictions

No zero in a denominator. Set the denominator $\neq 0$ and remove those $x$ -values.
No negative inside a square root. Set the radicand $\geq 0$ and solve.

If neither danger is present, the domain is all reals:

\mathbb{R}=(-\infty,\infty)

Memory hook — domain "Down can't be zero, root can't be negative." Two checks, nothing else. If you see no fraction and no square root, write

\mathbb{R}

and move on.

The trap that costs the most marks When you solve

x^{2}=9

to find what to exclude, you must take both roots:

x=\pm 3

, not just

x=3

. Forgetting the negative root leaves half the answer — and half the marks — on the table.

3. Finding the Range

The range is harder because it asks which outputs are possible. Two dependable methods cover almost every Grade 12 question. For a square-root or a shifted graph, reason from the smallest value upward: the square root $\sqrt{\;\;}$ is never negative, so its smallest output is $0$ , and any constant added or subtracted just slides that floor. For a rational function, solve $y=f(x)$ back for $x$ and ask which $y$ -values keep that expression legal — the same two restriction rules, now applied to $y$ .

Figure 1. A function maps each input to exactly one output; the range is the part of the codomain actually reached.

4. Types of Functions

Grade 12 asks you to classify a function by how it maps. There are three properties to test, and the prize property — bijective — is the one that lets a function be reversed.

Type	Condition	One-line test
One-one (injective)	Distinct inputs give distinct outputs	$f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}$
Onto (surjective)	Every element of $B$ is reached	$\text{Range}=B$
Into	At least one element of $B$ is unused	$\text{Range}\neq B$
Bijective	One-one and onto together	Both tests pass $\Rightarrow f^{-1}$ exists

Memory hook — the two big words "ONTO — Output covers, Nothing Omitted." "ONE-ONE — Only Non-repeating Exits." If every codomain element is hit, the function is onto; if no two inputs share an exit, it is one-one.

Common error Calling

x^{2}

one-one. Any even power —

x^{2}

x^{4}

|x|

— sends

+x

and

-x

to the same output, so it is many-one over

\mathbb{R}

and has no inverse unless you first restrict the domain. Odd powers like

x^{3}

are one-one.

5. Inverse Functions

An inverse function runs the machine backwards. If $f$ turns $a$ into $b$ , then $f^{-1}$ turns $b$ back into $a$ . But a machine can only be run backwards if it never sent two inputs to the same output and never skipped an output — that is, only if it is bijective.

Definition — inverse function If

f:A\to B

is bijective, its inverse

f^{-1}:B\to A

is the rule with

f(a)=b \iff f^{-1}(b)=a

. Domain and range swap:

\text{Dom}\,f=\text{Range}\,f^{-1}

and

\text{Range}\,f=\text{Dom}\,f^{-1}

Two properties are marked again and again. First, composing a function with its inverse returns the input untouched: $f\big(f^{-1}(x)\big)=f^{-1}\big(f(x)\big)=x$ . Second, the graphs of $f$ and $f^{-1}$ are mirror images in the line $y=x$ — reflect any point $(a,b)$ on $f$ across that line and you land on $(b,a)$ , a point of $f^{-1}$ .

Three steps to find

f^{-1}(x)

Write $y=f(x)$ .
Make $x$ the subject — solve for $x$ in terms of $y$ .
Swap the letters: replace $y$ with $x$ . That expression is $f^{-1}(x)$ .

The single most common inverse error The inverse is not the reciprocal:

f^{-1}(x)\neq\dfrac{1}{f(x)}

. For

f(x)=3x+2

, the inverse is

\dfrac{x-2}{3}

, never

\dfrac{1}{3x+2}

. The little

-1

means "undo," not "one over."

6. Standard Graphs You Must Know

The board expects you to recognise these eight graphs instantly, together with their domain and range. Learn the shape and the two sets as one package.

Function	Shape	Domain	Range
$y=ax+b$	Straight line	$\mathbb{R}$	$\mathbb{R}$
$y=ax^{2}+bx+c$	Parabola	$\mathbb{R}$	$[k,\infty)$ or $(-\infty,k]$
$y=\sqrt{x}$	Half-parabola	$[0,\infty)$	$[0,\infty)$
$y=\|x\|$	V-shape	$\mathbb{R}$	$[0,\infty)$
$y=a^{x}\ (a>1)$	Growth curve	$\mathbb{R}$	$(0,\infty)$
$y=\log_{a}x$	Log curve	$(0,\infty)$	$\mathbb{R}$
$y=\sin x,\ \cos x$	Wave	$\mathbb{R}$	$[-1,1]$
$y=\tan x$	Branches	$x\neq(2n+1)\dfrac{\pi}{2}$	$\mathbb{R}$

Notice a beautiful pattern: $y=a^{x}$ and $y=\log_{a}x$ are inverses of each other, so their domain and range are swapped — the exponential has range $(0,\infty)$ , the log has domain $(0,\infty)$ . That is the mirror-in- $y=x$ rule showing up in the tables.

Figure 2. The exponential and logarithm are inverse functions — mirror images in the line

y=x

7. Laws of Logarithm

These are guaranteed one-mark recall lines. Write them exactly; a dropped base or a wrong sign loses the mark.

\log_{a}(xy)=\log_{a}x+\log_{a}y\qquad\qquad \log_{a}\!\left(\frac{x}{y}\right)=\log_{a}x-\log_{a}y

\log_{a}(x^{n})=n\log_{a}x\qquad\quad \log_{a}x=\frac{\log_{b}x}{\log_{b}a}\qquad\quad \log_{a}1=0

Common error The log of a product is a sum, not a product:

\log_{a}(xy)=\log_{a}x+\log_{a}y

, never

\log_{a}x\cdot\log_{a}y

. And the log of

0

or of a negative number does not exist — the domain of

\log_{a}x

is strictly

x>0

8. Transformations of Graphs

Once you know a parent graph, you can move it around the plane without re-plotting a single point. Everything reduces to one idea about where the change sits.

Change	Effect on the graph
$f(x)+c$	Up $c$ units
$f(x)-c$	Down $c$ units
$f(x+c)$	Left $c$ units (opposite of the sign!)
$f(x-c)$	Right $c$ units
$c\,f(x)$	Vertical stretch when $c>1$
$f(cx)$	Horizontal compression when $c>1$

Memory hook — shifts "Inside lies, outside tells the truth." A change inside the bracket moves the graph the opposite way and horizontally:

f(x+3)

goes left 3. A change outside the bracket moves it the obvious way and vertically:

f(x)-3

goes down 3.

9. Applications — Growth and Decay

Exponential and logarithmic functions are working mathematics. A bank deposit growing at 8% a year, the population of Karachi, the cooling of a cup of chai, the decay of a medicine in the blood — all follow the same three models.

\text{Growth: } P(t)=P_{0}e^{rt}\qquad \text{Decay: } P(t)=P_{0}e^{-rt}\qquad \text{Compound: } A(t)=P(1+r)^{t}

To find a time $t$ hidden in the exponent, isolate the exponential term and then take a logarithm of both sides — that is the only move that brings the power down to ground level, using $\log_{a}(x^{n})=n\log_{a}x$ . For an $e^{rt}$ model use the natural log $\ln$ ; for a base like $1.08$ either $\ln$ or $\log$ works, as long as you use the same base on both sides.

Examiner tip Keep every rupee, every year, and every rate in the same units, and carry them into the final line. A time answer is in years; an amount answer is in rupees. Marks are lost for a bare number with no unit.

10. Worked Examples

Worked Example 1 — Domain of a rational function

Statement

Find the domain of

h(x)=\dfrac{x+4}{x^{2}-9}

Working

The only danger is a zero denominator, so set it not equal to zero:

x^{2}-9\neq 0 \;\Longrightarrow\; x^{2}\neq 9 \;\Longrightarrow\; x\neq\pm 3.

There is no square root, so every other real number is allowed.

Result

Domain

=\mathbb{R}-\{-3,\,3\}

Watch out

Both roots must be removed. Writing only

x\neq 3

and forgetting

x=-3

is the classic half-answer.

Worked Example 2 — Domain and range of a shifted root

Statement

Find the domain and range of

f(x)=3+\sqrt{x-2}

Working

Domain — the radicand must be non-negative:

x-2\geq 0 \;\Longrightarrow\; x\geq 2,\quad\text{so Domain}=[2,\infty).

Range — the square root

\sqrt{x-2}

is smallest, namely

0

, when

x=2

, giving

f=3

. As

x

increases,

\sqrt{x-2}

grows without bound, so

f\geq 3

Result

Domain

=[2,\infty)

; Range

=[3,\infty)

Watch out

The question asks for both sets. The range is not

\mathbb{R}

— the

+3

lifts the floor to

3

Worked Example 3 — Classify the type of a function

Statement

For

f(x)=x^{2}-1

with domain

\{1,-1,2,-2\}

and codomain

\{0,3,-3\}

, state the type.

Working

Compute every image:

f(1)=0,\quad f(-1)=0,\quad f(2)=3,\quad f(-2)=3.

So the range is

\{0,3\}

. Compare with the codomain

\{0,3,-3\}

: the element

-3

is never reached, so

\text{Range}\neq

codomain. Also,

f(1)=f(-1)

and

f(2)=f(-2)

, so distinct inputs share outputs.

Result

f

is into (since

-3

is unused) and many-one (since inputs share outputs). It is therefore not bijective and has no inverse.

Watch out

"Into" and "onto" are opposites here: onto would need every codomain element used. One missing element is enough to make it into.

Worked Example 4 — Inverse of a rational function

Statement

Find

f^{-1}(x)

for

f(x)=\dfrac{x}{x-2}

Working

Step 1 — write

y=\dfrac{x}{x-2}

, so

y(x-2)=x

.Step 2 — make

x

the subject:

xy-2y=x \;\Longrightarrow\; xy-x=2y \;\Longrightarrow\; x(y-1)=2y \;\Longrightarrow\; x=\frac{2y}{y-1}.

Step 3 — swap

y

and

x

Result

f^{-1}(x)=\dfrac{2x}{x-1}

Watch out

Collect every

x

on one side before factoring. And this is not the reciprocal

\dfrac{x-2}{x}

Worked Example 5 — One-one test

Statement

Determine whether

f(x)=2x^{2}+1

is one-one on

\mathbb{R}

Working

Assume two inputs give the same output:

2x_{1}^{2}+1=2x_{2}^{2}+1 \;\Longrightarrow\; x_{1}^{2}=x_{2}^{2}.

This does not force

x_{1}=x_{2}

; for instance

x_{1}=2

and

x_{2}=-2

both give

f=9

Result

f

is not one-one; it is many-one, so it has no inverse over

\mathbb{R}

Watch out

x_{1}^{2}=x_{2}^{2}

gives

x_{1}=\pm x_{2}

, not

x_{1}=x_{2}

. The minus sign is the whole reason even powers fail.

Worked Example 6 — Sketching a parabola

Statement

For

y=2x^{2}+3x-2

, find the opening direction, the axis of symmetry, and the vertex.

Working

Direction:

a=2>0

, so the parabola opens upward.Axis of symmetry:

x=-\frac{b}{2a}=-\frac{3}{2(2)}=-\frac{3}{4}.

Vertex

y

-value: substitute

x=-\tfrac{3}{4}

y=2\left(-\tfrac{3}{4}\right)^{2}+3\left(-\tfrac{3}{4}\right)-2=2\cdot\tfrac{9}{16}-\tfrac{9}{4}-2=\tfrac{9}{8}-\tfrac{9}{4}-2=-\frac{25}{8}.

The

y

-intercept is

(0,\,c)=(0,-2)

Result

Opens upward; axis

x=-\tfrac{3}{4}

; vertex

\left(-\tfrac{3}{4},\,-\tfrac{25}{8}\right)

;

y

-intercept

(0,-2)

Watch out

Watch the sign in

-\dfrac{b}{2a}

: with

b=+3

the axis is negative. And square the fraction before multiplying.

Worked Example 7 — Applied: compound growth

Understand

Rs. 30,000 is deposited at 8% compounded annually, so the amount after

t

years is

A(t)=30000(1.08)^{t}

. How long until it reaches Rs. 150,000?

Set up

Set

A(t)=150000

and isolate the exponential:

150000=30000(1.08)^{t} \;\Longrightarrow\; (1.08)^{t}=5.

Solve

Take the natural log of both sides to bring the power down:

\ln\!\big((1.08)^{t}\big)=\ln 5 \;\Longrightarrow\; t\ln 1.08=\ln 5 \;\Longrightarrow\; t=\frac{\ln 5}{\ln 1.08}=\frac{1.6094}{0.07696}\approx 20.9.

∴ Answer

About

20.9

years for the deposit to reach Rs. 150,000.

Watch out

Use the same base on both sides. Keep the unit — the answer is in years, not a bare number.

Worked Example 8 — Describing a transformation

Statement

Describe the transformation from

y=|x|

y=|x-1|-1

, and give the new vertex.

Working

Inside the bracket:

x-1

shifts the graph right 1 (inside lies — a minus means right).Outside the bracket: the

-1

shifts the graph down 1 (outside tells the truth).

Result

Right 1, then down 1. The vertex moves from

(0,0)

(1,-1)

Watch out

Do not flip the outside shift — only the inside change reverses.

-1

outside is plainly down.

11. Key Formulae & Terminology

Formula / Rule	What it means in words	Common exam mistake
$f:A\to B$ , one input → one output	Every domain element maps to exactly one codomain element.	Allowing an input to have two outputs, or leaving an input unmapped.
Denominator $\neq 0$ ; radicand $\geq 0$	The only two things that restrict a domain.	Forgetting one check, or missing the negative root when solving $x^{2}=9$ .
$f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}$	The one-one (injective) test.	Calling an even power $x^{2},x^{4},\|x\|$ one-one.
$\text{Range}=B$	The onto (surjective) test — nothing omitted.	Confusing onto with one-one, or codomain with range.
$f^{-1}$ exists $\iff f$ is bijective	An inverse needs one-one and onto together.	Finding an inverse for a many-one function without restricting the domain.
$f^{-1}(x)\neq\dfrac{1}{f(x)}$	The inverse undoes $f$ ; it is not the reciprocal.	Writing $\dfrac{1}{f(x)}$ as the inverse.
Graphs of $f$ and $f^{-1}$ mirror in $y=x$	Reflecting $(a,b)$ across $y=x$ gives $(b,a)$ .	Reflecting in the $x$ -axis or $y$ -axis instead.
$x=-\dfrac{b}{2a}$	Axis of symmetry and $x$ -coordinate of a parabola's vertex.	Dropping the minus sign, or the sign of $b$ .
$a>0$ up, $a<0$ down	The leading coefficient sets a parabola's opening.	Using $c$ or $b$ to decide the opening direction.
$\log_{a}(xy)=\log_{a}x+\log_{a}y$	Log of a product is a sum of logs.	Writing it as a product $\log_{a}x\cdot\log_{a}y$ .
$\log_{a}(x^{n})=n\log_{a}x$	A power inside a log comes down as a multiplier.	Leaving the exponent in place when solving for $t$ .
Domain of $\log_{a}x$ is $x>0$	Logarithms accept strictly positive inputs only.	Taking the log of $0$ or a negative number.
$f(x+c)$ left; $f(x)+c$ up	Inside the bracket reverses and is horizontal; outside is obvious and vertical.	Shifting $f(x+3)$ right instead of left.
$P(t)=P_{0}e^{rt}$	Exponential growth model; use $\ln$ to solve for $t$ .	Mixing bases, or dropping the unit from the final answer.

12. Last-Minute Reminders

Write the definition first — the examiner awards a mark before any computing.
Domain: down can't be zero, root can't be negative; otherwise $\mathbb{R}$ .
Solving $x^{2}=9$ gives $x=\pm 3$ — never just $+3$ .
Even powers $(x^{2},x^{4},|x|)$ are many-one; odd powers $(x^{3})$ are one-one.
An inverse exists only for a bijective function; find it by the three-step swap.
$f^{-1}$ is not $\dfrac{1}{f}$ , and the mirror line is $y=x$ .
Parabola: direction from $a$ , axis at $x=-\dfrac{b}{2a}$ , then substitute for the vertex $y$ .
Log domain is $x>0$ ; the log of a product is a sum.
Shifts: inside lies, outside tells the truth.
Growth problems: isolate the exponential, take $\ln$ , keep the unit.

Final word from Athens Academy

A function is just a promise — one input, one output — and every topic in this chapter is a way of keeping or reading that promise. Recognise the eight standard graphs on sight, learn the two domain checks and the three-step inverse, and write your working in clean numbered lines. Do that, and the long questions stop being puzzles and start being routines you already own. Steady hands, clean steps. You can do this.

📐 Chapter 1 · Functions and Graphs