Quadratic functions: roots, vertices, inequalities

IB Mathematics AI HL· 25 min read

1. Roots of Quadratics and the Discriminant★★☆☆☆⏱ 15 min

📘 Definition

Root of a quadratic function

An input value $x$ for which $f (x) = 0$ , corresponding to the x-intercept(s) of the parabola on the coordinate plane.

Example:

For $f (x) = x^{2} - 4$ , the roots are $x = 2$ and $x = - 2$ .

Roots can be found via three main methods: factorisation, completing the square, and the quadratic formula. The discriminant tells us how many real roots a quadratic has before we solve, saving time in exam questions.

\Delta = b^2 - 4ac \quad \text{for } ax^2 + bx + c = 0

If $Δ > 0$ : Two distinct real roots
If $Δ = 0$ : One repeated real root (parabola touches the x-axis)
If $Δ < 0$ : No real roots

📐 Worked Example

Find the number of real roots of $f (x) = 2 x^{2} - 3 x + 1$ , and state their values.

1
Identify $a, b, c$ from standard form:
$a = 2, b = -3, c = 1$
2
Calculate the discriminant:
$\Delta = (-3)^2 - 4(2)(1) = 9 - 8 = 1 > 0$
3
We have two distinct real roots. Use the quadratic formula:
$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{3 \pm 1}{4}$
4
Calculate the final roots:
$x = 1, \quad x = 0.5$

Exam tip:

Always calculate the discriminant first if asked for the number of roots, it lets you stop early if there are no real roots.

2. Finding the Vertex of a Quadratic Parabola★★☆☆☆⏱ 15 min

📘 Definition

Vertex of a quadratic

The turning point of the parabola: a minimum if the coefficient of $x^{2}$ is positive, a maximum if the coefficient is negative. It lies on the axis of symmetry of the parabola.

There are two common methods to find the vertex coordinates: completing the square, and using the axis of symmetry formula for standard form. For completed square form $f (x) = a (x - h)^{2} + k$ , the vertex is directly at $(h, k)$ .

For quadratics in standard form $a x^{2} + b x + c$ , the x-coordinate of the vertex is given by:

x = -\frac{b}{2a}

Substitute this x-value back into $f (x)$ to find the corresponding y-coordinate of the vertex.

📐 Worked Example

Find the coordinates of the vertex of $f (x) = 3 x^{2} - 12 x + 5$ , and state if it is a minimum or maximum.

1
Identify $a = 3$ , $b = - 12$ , $c = 5$ . Calculate the x-coordinate:
$x = -\frac{b}{2a} = -\frac{(-12)}{2(3)} = 2$
2
Substitute $x = 2$ back into $f (x)$ to find the y-coordinate:
$f(2) = 3(2)^2 - 12(2) + 5 = 12 - 24 + 5 = -7$
3
Check concavity: $a = 3 > 0$ , so the parabola opens upwards.

Exam tip:

Almost all optimisation problems with quadratic models use the vertex to find the maximum or minimum value, so remember this formula!

3. Quadratic Inequalities: Algebraic Method★★★☆☆⏱ 20 min

To solve a quadratic inequality algebraically, follow a structured step-by-step process to avoid common mistakes:

Rearrange the inequality to get all terms on one side, and $0$ on the other.
Find the roots of the corresponding quadratic equation.
Test the sign of the quadratic in each interval divided by the roots.
Select the intervals that satisfy the original inequality.

📐 Worked Example

Solve the inequality $x^{2} - 2 x - 8 \leq 0$ algebraically.

1
The inequality is already rearranged. Factorise to find roots:
$(x - 4)(x + 2) = 0 \implies x = 4, x = -2$
2
Test the sign of the product in each interval divided by the roots:
$x < -2: (-)(-) = + > 0 \\ -2 < x < 4: (-)(+) = - < 0 \\ x > 4: (+)(+) = + > 0$
3
We need values where the quadratic is $\leq 0$ , so we take the negative interval and include the roots (for the $\leq$ inequality).

4. Quadratic Inequalities: Graphical Method★★★☆☆⏱ 15 min

The graphical method uses the shape of the parabola to quickly identify the solution interval, and is an excellent check for algebraic solutions. The core idea is simple: the quadratic is positive when it is above the x-axis, and negative when it is below the x-axis.

📐 Worked Example

Solve $- 2 x^{2} + 4 x + 16 > 0$ using the graphical method.

1
Simplify the inequality by dividing by $- 2$ , and reverse the inequality sign:
$x^2 - 2x - 8 < 0$
2
Find the roots: $x = - 2$ and $x = 4$ . The coefficient of $x^{2}$ is positive, so the parabola opens upwards.
3
We need the quadratic to be less than zero, which is where the parabola is below the x-axis. For an upward opening parabola, this is between the two roots. The inequality is strict, so we do not include the roots.

5. Common Pitfalls

Wrong move:

Forgetting to reverse the inequality sign when dividing/multiplying by a negative number

Why:

Multiplying or dividing by a negative reverses the order of values on the number line, so the inequality direction must change

Correct move:

Always reverse the inequality sign if you multiply or divide both sides by a negative number

Wrong move:

Including roots when the inequality is strict ( $<$ or $>$ )

Why:

Strict inequalities do not allow $f (x) = 0$ , so roots are not part of the solution set

Correct move:

Use open interval notation (e.g. $(- 2, 4)$ ) for strict inequalities, closed brackets for $\leq$ or $\geq$

Wrong move:

Getting the vertex x-coordinate wrong for completed square form

Why:

The form is $a (x - h)^{2} + k$ , so $x + 3 = x - (- 3)$ meaning $h$ is negative

Correct move:

For $f (x) = (x + 3)^{2} - 2$ , the vertex is at $(- 3, - 2)$ , not $(3, - 2)$

Wrong move:

Writing the solution to $x^{2} > 9$ as $x > \pm 3$

Why:

This incorrectly includes values between $- 3$ and $3$ which do not satisfy the inequality

Correct move:

The correct solution is $x < - 3$ or $x > 3$

Wrong move:

Leaving the inequality with terms on both sides when solving

Why:

This leads to incorrect sign tests and wrong interval selection

Correct move:

Always rearrange to get all terms on the left side and 0 on the right first

6. Quick Reference Cheatsheet

Concept	Key Rule/Formula
Roots of $a x^{2} + b x + c = 0$	$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
Discriminant rules	$Δ > 0$ : 2 roots; $Δ = 0$ : 1 root; $Δ < 0$ : 0 roots
Vertex (standard form)	$x = - \frac{b}{2 a}$ , substitute to find $y$
Vertex (completed square)	$a (x - h)^{2} + k \to$ vertex at $(h, k)$
$a x^{2} + b x + c > 0$ , $a > 0$	Solution: $x < r_{1}$ or $x > r_{2}$
$a x^{2} + b x + c < 0$ , $a > 0$	Solution: $r_{1} < x < r_{2}$
Multiply/divide by negative	Always reverse the inequality sign

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.

2025 · 1
Find roots and vertex of quadratic
2024 · 2
Optimisation with quadratic inequality
2023 · 1
Solve quadratic inequality

What's Next

Quadratic functions are foundational for almost all higher topics in IB AI HL. The skills you learned here for finding roots, turning points, and solving inequalities will be extended to higher-degree polynomials, non-linear functions, and optimisation problems using calculus. Quadratic models are also commonly used in statistics for quadratic regression, where the vertex gives the optimal value for real-world data. Mastery of this subtopic is essential for success in both Paper 1 and Paper 2 exams, as it frequently appears in both short and extended response questions.