Linear functions and their graphs

IB Mathematics: Analysis and Approaches SL· 5 min read

1. Core definitions and standard forms★☆☆☆☆⏱ 10 min

📘 Definition

Linear function

$f (x) = m x + c$

A first-order polynomial function that graphs to a straight line, where the highest power of the independent variable is 1.

Example:

$f (x) = 3 x - 2$ is linear; $f (x) = x^{2} + 1$ is non-linear

Linear functions can be written in three common standard forms, each suited to different use cases:

Gradient-intercept form: $y = m x + c$ , used when you know gradient and y-intercept
Point-gradient form: $y - y_{1} = m (x - x_{1})$ , used when you know one point and gradient
General form: $a x + b y + d = 0$ , where $a, b, d$ are integer constants

📐 Worked Example

Write $3 y - 6 x = 9$ in gradient-intercept form

1
Isolate the $y$ term on the left side by adding $6 x$ to both sides:
2
$3y = 6x + 9$
3
Divide all terms by the coefficient of $y$ (which is 3):
4
$y = 2x + 3$
5
This is gradient-intercept form, with gradient $m = 2$ and y-intercept $c = 3$ .

2. Gradient, intercepts and graph sketching★★☆☆☆⏱ 15 min

The gradient of a line measures how much $y$ changes for every 1 unit change in $x$ . It is calculated from any two points on the line.

📘 Definition

Gradient formula

$m = \frac{Δ y}{Δ x} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

For two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on a line, gradient is the ratio of vertical change to horizontal change.

📐 Worked Example

Find the gradient of the line through $(2, 5)$ and $(5, 11)$ , then find its x-intercept.

1
Label the points and substitute into the gradient formula:
2
$m = \frac{11 - 5}{5 - 2} = \frac{6}{3} = 2$
3
Use point-gradient form to find the full line equation:
4
$y - 5 = 2(x - 2) \implies y = 2x + 1$
5
Find x-intercept by setting $y = 0$ and solving for $x$ :
6
$0 = 2x + 1 \implies x = -\frac{1}{2}$
7
The x-intercept is at $(- \frac{1}{2}, 0)$ .

3. Parallel and perpendicular lines★★☆☆☆⏱ 12 min

There are simple rules relating the gradients of parallel and perpendicular lines, which are commonly tested in IB exams.

📘 Definition

Gradient rules for parallel/perpendicular lines

If two lines have gradients $m_{1}$ and $m_{2}$ : (1) Parallel lines: $m_{1} = m_{2}$ , (2) Perpendicular lines: $m_{1} m_{2} = - 1$

Example:

A line with gradient 4 is perpendicular to a line with gradient $- \frac{1}{4}$

📐 Worked Example

Find the equation of the line through $(3, - 2)$ that is perpendicular to $y = 3 x + 1$ .

1
The original line has gradient $m_{1} = 3$ . For a perpendicular line:
2
$m_2 = -\frac{1}{m_1} = -\frac{1}{3}$
3
Substitute the point $(3, - 2)$ into point-gradient form:
4
$y - (-2) = -\frac{1}{3}(x - 3)$
5
Rearrange to gradient-intercept form:
6
$y + 2 = -\frac{1}{3}x + 1 \implies y = -\frac{1}{3}x - 1$

4. Real-world applications of linear functions★★☆☆☆⏱ 15 min

Many real-world relationships with constant rates of change can be modeled with linear functions. In these problems, the gradient is the constant rate, and the intercept is the initial value when the input is zero.

📐 Worked Example

A coffee shop charges a fixed monthly subscription fee plus \ $2 p er co f f ee b o ug h t . I f 15 co f f eescos t$ 38 in a month, find the cost of 25 coffees.

1
Define variables: Let $C$ = total monthly cost, $n$ = number of coffees. The linear model is $C = mn + f$ , where $f$ = fixed fee, $m$ = cost per coffee.
2
We know $m = 2$ , substitute the given values to find $f$ :
3
$38 = 2(15) + f \implies 38 = 30 + f \implies f = 8$
4
The full model is $C = 2 n + 8$ . Calculate cost for 25 coffees:
5
$C = 2(25) + 8 = 50 + 8 = 58$
6
The total cost for 25 coffees is \$58.

5. Common Pitfalls

Wrong move:

Calculating gradient as $\frac{x _{2} - x _{1}}{y _{2} - y _{1}}$

Why:

Swapping numerator and denominator gives the reciprocal of the correct gradient

Correct move:

Always remember gradient is change in y over change in x: $m = \frac{Δ y}{Δ x}$

Wrong move:

Reading gradient and intercept directly from general form without rearranging

Why:

The coefficients of $x$ and $y$ in general form are not the same as $m$ and $c$

Correct move:

Always rearrange general form to $y = m x + c$ before reading off gradient and intercept

Wrong move:

Stating perpendicular lines have gradient $- m$ instead of $- \frac{1}{m}$

Why:

Confusing the negative sign with the requirement to take the reciprocal

Correct move:

Perpendicular gradients multiply to $- 1$ , so always take the negative reciprocal of the original gradient

Wrong move:

Writing gradient 0 for a vertical line $x = k$

Why:

Confusing the properties of horizontal and vertical lines

Correct move:

Horizontal lines $y = k$ have gradient 0; vertical lines $x = k$ have undefined gradient

Wrong move:

Mixing up input and output variables when calculating gradient for real-world problems

Why:

Swapping variables reverses the rate of change, leading to an incorrect gradient

Correct move:

Always explicitly define your variables at the start of the problem to avoid this mistake

6. Quick Reference Cheatsheet

Property	Rule/Formula	Note
Gradient between two points	$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$	Rise over run
Gradient-intercept form	$y = m x + c$	$m$ = gradient, $c$ = y-intercept
Point-gradient form	$y - y_{1} = m (x - x_{1})$	For known point + gradient
Parallel lines	$m_{1} = m_{2}$	Equal gradients
Perpendicular lines	$m_{1} m_{2} = - 1$	Negative reciprocal gradients
Horizontal line	$y = k$ , $m = 0$	Parallel to x-axis
Vertical line	$x = k$ , $m$ undefined	Parallel to y-axis
General form	$a x + b y + c = 0$	Rearrange to $y = m x + c$ to read values

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Find perpendicular line equation
2023 · 1
Real-world linear cost problem
2021 · 2
Gradient between two points

What's Next

Linear functions are the foundational building block for nearly all further topics in IB AA SL. The concept of constant rate of change for linear functions directly prepares you to study instantaneous rate of change for non-linear functions, which is the core idea of differential calculus. You will also apply linear functions to solve systems of linear equations, model bivariate data with linear regression in statistics, and solve kinematics problems involving constant speed or acceleration. Mastery of linear graphs and their properties is essential for interpreting and working with all other function types you will encounter, including quadratics, exponentials, and trigonometric functions.