Quadratic functions, roots, and vertices

IB Mathematics: Applications & Interpretation SL· 45 min read

1. Forms of Quadratic Functions★★☆☆☆⏱ 10 min

📘 Definition

Quadratic function

$f (x) = a x^{2} + b x + c, a \neq = 0$

A second-degree polynomial function whose graph is a parabola. It opens upwards if $a > 0$ (minimum vertex) and downwards if $a < 0$ (maximum vertex).

Example:

$f (x) = 2 x^{2} - 5 x + 2$ is a quadratic opening upwards.

Standard form: $f (x) = a x^{2} + b x + c$ , directly gives the y-intercept $(0, c)$
Factored form: $f (x) = a (x - r_{1}) (x - r_{2})$ , directly gives the roots $r_{1}, r_{2}$
Vertex form: $f (x) = a (x - h)^{2} + k$ , directly gives the vertex $(h, k)$

📐 Worked Example

Write $f (x) = x^{2} - 4 x + 3$ in factored and vertex form, and state what feature each form gives directly.

1
Factor the quadratic: find two numbers that multiply to 3 and add to -4, which are -1 and -3. This gives factored form:
2
$f(x) = (x - 1)(x - 3)$
3
Convert to vertex form by completing the square: group x terms and complete the square:
4
$f(x) = (x^2 - 4x) + 3 = (x^2 - 4x + 4) - 4 + 3 = (x - 2)^2 - 1$
5
Factored form gives roots at $x = 1$ and $x = 3$ directly, and vertex form gives the vertex at $(2, - 1)$ directly.

Exam tip:

Always check what form of quadratic you are given to avoid unnecessary calculations for features you can read directly.

2. Finding Roots of Quadratic Functions★★★☆☆⏱ 15 min

📘 Definition

Root of a quadratic

A value of $x$ for which $f (x) = 0$ , corresponding to where the parabola crosses the x-axis. A quadratic can have 0, 1, or 2 real roots.

Example:

The roots of $f (x) = (x - 1) (x - 3)$ are $x = 1$ and $x = 3$ .

📐 Worked Example

Find all real roots of $f (x) = 2 x^{2} - 7 x + 3$ .

1
Identify coefficients from standard form: $a = 2$ , $b = - 7$ , $c = 3$ . Calculate the discriminant:
2
$\Delta = (-7)^2 - 4(2)(3) = 49 - 24 = 25$
3
$Δ = 25 > 0$ , so we have two distinct real roots. Apply the quadratic formula:
4
$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{7 \pm 5}{4}$
5
Calculate the two roots: $x = \frac{12}{4} = 3$ and $x = \frac{2}{4} = 0.5$ . The roots are $x = 0.5$ and $x = 3$ .

Exam tip:

Always write down your discriminant calculation first — you get method marks for it even if you make a mistake in calculating the final roots.

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

📘 Definition

Vertex of a parabola

The turning point of the parabola. It is the minimum point if the parabola opens upwards ( $a > 0$ ) and the maximum point if it opens downwards ( $a < 0$ ). The x-coordinate of the vertex is always halfway between the two roots.

Example:

For roots $x = 1$ and $x = 3$ , the x-coordinate of the vertex is $\frac{1 + 3}{2} = 2$ .

For a quadratic in standard form $a x^{2} + b x + c$ , the x-coordinate of the vertex can be calculated directly with the formula $x = - \frac{b}{2 a}$ . Substitute this value back into the function to get the y-coordinate.

📐 Worked Example

Find the vertex of $f (x) = 3 x^{2} - 12 x + 7$ , and state if it is a maximum or minimum.

1
Identify $a = 3$ , $b = - 12$ . Use the formula for x-coordinate of the vertex:
2
$x = -\frac{b}{2a} = -\frac{(-12)}{2(3)} = 2$
3
Substitute $x = 2$ back into the function to find the y-coordinate:
4
$f(2) = 3(2)^2 - 12(2) + 7 = 12 - 24 + 7 = -5$
5
The vertex is at $(2, - 5)$ . Since $a = 3 > 0$ , the parabola opens upwards, so this is a minimum turning point.

Exam tip:

For optimization problems (maximum profit, minimum cost, maximum height), the answer is almost always the y-coordinate of the vertex.

4. Interpreting Roots and Vertices in Context★★★★☆⏱ 15 min

Almost all quadratic questions in IB AI SL are context-based, so you need to interpret calculated values in the problem's scenario, not just give mathematical results.

📐 Worked Example

The height of a ball $t$ seconds after being thrown is $h (t) = - 5 t^{2} + 20 t + 2$ (height in meters). When does the ball hit the ground, and what is its maximum height?

1
The ball hits the ground when height is 0, so we need the positive root of $h (t) = 0$ (negative time is impossible here). Maximum height is the y-coordinate of the vertex, since $a = - 5 < 0$ (maximum vertex).
2
Calculate the vertex: time of maximum height is $t = - \frac{20}{2 ( - 5 )} = 2$ seconds. Maximum height is:
3
$h(2) = -5(2)^2 + 20(2) + 2 = 22 \text{ meters}$
4
Solve for the positive root using the quadratic formula:
5
$t = \frac{20 \pm \sqrt{440}}{10} \approx 4.1 \text{ seconds (positive root)}$
6
Interpretation: The ball reaches a maximum height of 22 m after 2 seconds, and hits the ground after approximately 4.1 seconds.

5. Common Pitfalls

Wrong move:

Treating $0 x^{2} + 2 x + 1$ as a quadratic and trying to apply the quadratic formula

Why:

If the coefficient of $x^{2}$ is zero, the function is linear, not quadratic, so it only has one root

Correct move:

Always confirm that the coefficient of $x^{2}$ is non-zero before using any quadratic methods

Wrong move:

Mismanaging signs when calculating $x = - \frac{b}{2 a}$ , getting the wrong x-coordinate for the vertex

Why:

It is easy to forget the negative sign out front when $b$ is already negative

Correct move:

Write down $b$ and its sign explicitly before substitution, e.g. for $x^{2} - 4 x + 3$ , $b = - 4$ so $x = - \frac{( - 4 )}{2 ( 1 )} = 2$

Wrong move:

Leaving both positive and negative roots in a context problem like projectile motion

Why:

Examiners expect you to interpret results in context, not just output all mathematical solutions

Correct move:

Always discard any results that do not make sense in the problem's scenario before giving your final answer

Wrong move:

Giving the x-coordinate of the vertex as the answer when asked for maximum height

Why:

It is easy to stop after calculating the x-coordinate, but questions usually ask for the maximum value, not the time it occurs at

Correct move:

Always check what the question asks for: maximum value = y-coordinate, time of maximum = x-coordinate

6. Quick Reference Cheatsheet

Feature	Standard Form: $a x^{2} + b x + c$	Factored Form: $a (x - r_{1}) (x - r_{2})$	Vertex Form: $a (x - h)^{2} + k$
Roots	$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$	Read as $r_{1}, r_{2}$ directly	Set equal to 0 and solve for $x$
Number of real roots	$Δ > 0$ : 2; $Δ = 0$ : 1; $Δ < 0$ : 0	2 distinct or 1 repeated	Same as standard form
Vertex x-coordinate	$- \frac{b}{2 a}$	$\frac{r _{1} + r _{2}}{2}$	Read as $h$ directly
Vertex y-coordinate	Substitute $x = - \frac{b}{2 a}$	Substitute $x = \frac{r _{1} + r _{2}}{2}$	Read as $k$ directly
Y-intercept	Read as $c$ directly	Evaluate at $x = 0$	Evaluate at $x = 0$
Max/Min	Max if $a < 0$ , Min if $a > 0$	Same rule	Same rule

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 vertex and roots of quadratic
2024 · 2
Projectile height context problem
2023 · 1
Number of roots via discriminant

What's Next

Quadratic functions are the foundation for more advanced non-linear modeling in IB AI SL, and they appear in almost every exam paper, often in 5-8 mark context-based questions. Understanding how to quickly identify and calculate roots and vertices allows you to solve common exam problems including optimization, projectile motion, and profit modeling efficiently. This sub-topic is also a prerequisite for working with other non-linear functions like exponential and cubic functions later in the course. Next, you will build on this knowledge to solve quadratic inequalities and use quadratic regression to model real-world bivariate data.