Linear and quadratic functions

IB Mathematics: Analysis and Approaches HL· 6 min read

1. Linear Functions★★☆☆☆HL only⏱ 15 min

📘 Definition

Linear function

f(x) = mx + c

A first-degree polynomial function that graphs to a straight line. $m$ is the constant gradient (slope), and $c$ is the y-intercept (the point where the line crosses the y-axis).

Example:

f(x) = 3x - 2 has gradient 3 and crosses the y-axis at (0, -2).

To find the equation of a line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , first calculate the gradient as $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ , then substitute one point and $m$ into $y = m x + c$ to solve for $c$ .

📐 Worked Example

Find the equation of the line passing through $(3, 7)$ and $(1, 1)$ .

1
Calculate the gradient from the two points:
2
$m = \frac{1 - 7}{1 - 3} = \frac{-6}{-2} = 3$
3
Substitute $m = 3$ and $(x, y) = (1, 1)$ to find $c$ :
4
$1 = 3(1) + c \implies c = 1 - 3 = -2$
5
Write the final equation of the line:
6
$y = 3x - 2$

Exam tip:

Always check your answer by substituting the second point into your final equation to confirm it is correct.

2. Quadratic Functions: Forms and Key Features★★★☆☆HL only⏱ 25 min

📘 Definition

Quadratic function

Standard: $f (x) = a x^{2} + b x + c$ ; Vertex: $f (x) = a (x - h)^{2} + k$ ; Factored: $f (x) = a (x - p) (x - q)$

A second-degree polynomial function that graphs to a parabola. The leading coefficient $a$ determines the direction it opens: $a > 0$ opens upwards, $a < 0$ opens downwards.

Each form of a quadratic is designed to show different key features: standard form gives the y-intercept $(0, c)$ directly, vertex form gives the turning point (vertex) at $(h, k)$ directly, and factored form gives the roots (x-intercepts) at $x = p$ and $x = q$ directly. To convert from standard form to vertex form, you use the method of completing the square.

📐 Worked Example

Convert $f (x) = 3 x^{2} - 12 x + 5$ to vertex form and state the vertex coordinates.

1
Factor the leading coefficient out of the x terms:
2
$f(x) = 3(x^2 - 4x) + 5$
3
Complete the square inside the bracket by adding and subtracting the square of half the x coefficient:
4
$x^2 - 4x = (x^2 - 4x + 4) - 4 = (x-2)^2 - 4$
5
Substitute back and simplify, remembering to multiply the constant term by 3:
6
$f(x) = 3\left((x-2)^2 - 4\right) + 5 = 3(x-2)^2 - 12 + 5 = 3(x-2)^2 - 7$
7
State the vertex from vertex form:
8
$\text{Vertex} = (2, -7)$

3. Intersections and the Discriminant★★★★☆HL only⏱ 20 min

To find intersections between a line and a quadratic, equate the two expressions for $y$ to get a single quadratic equation in $x$ . The number of real intersections is determined by the discriminant of this quadratic.

📘 Definition

Discriminant

\Delta = b^2 - 4ac $f or$ ax^2 + bx + c = 0$

A value that tells you how many real roots a quadratic equation has. The rules are: $Δ > 0$ = 2 distinct real roots, $Δ = 0$ = 1 repeated real root, $Δ < 0$ = 0 real roots.

📐 Worked Example

Find how many intersections there are between $y = 2 x + 3$ and $y = 2 x^{2} - 3 x + 1$ .

1
Equate the two expressions for y:
2
$2x + 3 = 2x^2 - 3x + 1$
3
Rearrange into standard quadratic form:
4
$2x^2 - 5x - 2 = 0$
5
Calculate the discriminant with $a = 2$ , $b = - 5$ , $c = - 2$ :
6
$\Delta = (-5)^2 - 4(2)(-2) = 25 + 16 = 41$
7
Interpret the result:
8
$\Delta = 41 > 0 \implies \text{2 distinct real intersections}$

✓ Quick check

Test your understanding of discriminant rules:

A line is tangent to a parabola. What is the discriminant of the resulting quadratic?
- Negative
- Zero
- Positive
Reveal answer
Zero —
A tangent touches the parabola at exactly one point, so there is one repeated root, which gives a discriminant of zero.
The discriminant of a quadratic is -9. How many real roots does it have?
- 0
- 1
- 2
Reveal answer
0 —
Negative discriminant means no real roots, only two complex roots.

4. Common Pitfalls

Wrong move:

Writing the gradient formula as $\frac{x _{2} - x _{1}}{y _{2} - y _{1}}$ instead of $\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Why:

This gives the reciprocal of the correct gradient, leading to a wrong line equation and lost marks

Correct move:

Remember gradient = rise over run, which is change in y divided by change in x

Wrong move:

Forgetting to multiply the constant term by the leading coefficient when completing the square

Why:

When you factor $a$ out of the x terms, the constant you subtract inside the bracket must be multiplied by $a$ when you expand

Correct move:

Always expand your final vertex form back to standard form to check it matches the original quadratic

Wrong move:

Reading the x-coordinate of the vertex as $- h$ from vertex form $a (x - h)^{2} + k$

Why:

The vertex occurs when the bracket equals zero: $x - h = 0 ⟹ x = h$ , not $x = - h$

Correct move:

Always set the bracket term equal to zero to find the x-coordinate of the vertex to avoid sign errors

Wrong move:

Stating there are two intersections even when the discriminant is negative

Why:

IB exam questions regularly test whether you remember that negative discriminant means no real intersections

Correct move:

Always calculate and explicitly reference the discriminant when asked for the number of intersections or roots

Wrong move:

Taking the y-intercept from standard form $a x^{2} + b x + c$ as $b$ instead of $c$

Why:

When $x = 0$ , all terms with $x$ disappear, so $f (0) = c$ , not $b$

Correct move:

Substitute $x = 0$ into the quadratic to confirm the y-intercept if you are unsure

5. Quick Reference Cheatsheet

Concept	Form/Rule	Key Feature
Linear function	$y = m x + c$	$m$ = gradient, $c$ = y-intercept
Quadratic (standard)	$y = a x^{2} + b x + c$	Y-intercept at $(0, c)$
Quadratic (vertex)	$y = a (x - h)^{2} + k$	Vertex at $(h, k)$
Quadratic (factored)	$y = a (x - p) (x - q)$	Roots at $x = p, x = q$
Discriminant	$Δ = b^{2} - 4 a c$	$Δ > 0$ : 2 roots; $Δ = 0$ : 1 root; $Δ < 0$ : 0 real roots

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 · P1
Find quadratic vertex by completing square
2023 · P2
Find number of line-parabola intersections

What's Next

Linear and quadratic functions are the foundation for all higher-degree polynomials and more complex functions you will encounter in IB AA HL. The skills you learned here, from completing the square to using the discriminant, will be used repeatedly in topics like calculus, complex numbers, and general polynomial functions. Understanding how to find intersections and key graph features is also essential for solving systems of equations and optimization problems, which are common on both Paper 1 and Paper 2 exams. Next, you will build on this knowledge to explore more complex polynomial functions and their properties.