# Quadratic functions: roots, vertices, inequalities

> IB Mathematics AI HL · Unit 2: Functions
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u2-quadratic-functions-roots-vertices-inequalities/

This subtopic explores core properties of quadratic functions, including finding roots, calculating vertex coordinates for optimisation problems, and solving quadratic inequalities using both algebraic and graphical methods common to IB AI HL assessments.

**Prerequisites:** [Algebraic manipulation of quadratics (expanding, factorising)](https://www.owlsprep.com/study/ib-math-ai-hl-algebra-quadratic-manipulation/); [Solving linear inequalities](https://www.owlsprep.com/study/ib-math-ai-hl-linear-inequalities/)

## Learning objectives

- Identify the number of real roots of a quadratic using the discriminant
- Calculate vertex coordinates and interpret them as minima/maxima
- Solve quadratic inequalities using algebraic and graphical methods
- Apply these properties to problem-solving and optimisation questions

## Roots of Quadratics and the Discriminant

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

**Worked example:** Find the number of real roots of $f(x) = 2x^2 - 3x + 1$, 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.

## Finding the Vertex of a Quadratic Parabola

**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 $ax^2 + bx + c$, 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 $f(x) = 3x^2 - 12x + 5$, 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.

*Conclusion:* The vertex is at $(2, -7)$, and it is a minimum turning point.

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

## Quadratic Inequalities: Algebraic Method

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

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

**Worked example:** Solve the inequality $x^2 - 2x - 8 \leq 0$ 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).

*Conclusion:* The solution is $-2 \leq x \leq 4$.

## Quadratic Inequalities: Graphical Method

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.

> **tip**
>
> Always use the graphical method to check your answer after solving algebraically in exams. It catches almost all common mistakes in less than a minute.

**Worked example:** Solve $-2x^2 + 4x + 16 > 0$ 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.

*Conclusion:* The solution is $-2 < x < 4$, which matches the algebraic result.

## Common pitfalls

- **Wrong:** Forgetting to reverse the inequality sign when dividing/multiplying by a negative number
  - Why it fails: Multiplying or dividing by a negative reverses the order of values on the number line, so the inequality direction must change
  - Correct: Always reverse the inequality sign if you multiply or divide both sides by a negative number
- **Wrong:** Including roots when the inequality is strict ($<$ or $>$)
  - Why it fails: Strict inequalities do not allow $f(x) = 0$, so roots are not part of the solution set
  - Correct: Use open interval notation (e.g. $(-2, 4)$) for strict inequalities, closed brackets for $\leq$ or $\geq$
- **Wrong:** Getting the vertex x-coordinate wrong for completed square form
  - Why it fails: The form is $a(x - h)^2 + k$, so $x + 3 = x - (-3)$ meaning $h$ is negative
  - Correct: For $f(x) = (x + 3)^2 - 2$, the vertex is at $(-3, -2)$, not $(3, -2)$
- **Wrong:** Writing the solution to $x^2 > 9$ as $x > \pm 3$
  - Why it fails: This incorrectly includes values between $-3$ and $3$ which do not satisfy the inequality
  - Correct: The correct solution is $x < -3$ or $x > 3$
- **Wrong:** Leaving the inequality with terms on both sides when solving
  - Why it fails: This leads to incorrect sign tests and wrong interval selection
  - Correct: Always rearrange to get all terms on the left side and 0 on the right first

## Cheatsheet

| Concept | Key Rule/Formula |
| --- | --- |
| Roots of $ax^2 + bx + c = 0$ | $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ |
| Discriminant rules | $\Delta > 0$: 2 roots; $\Delta = 0$: 1 root; $\Delta < 0$: 0 roots |
| Vertex (standard form) | $x = -\frac{b}{2a}$, substitute to find $y$ |
| Vertex (completed square) | $a(x - h)^2 + k \to$ vertex at $(h, k)$ |
| $ax^2+bx+c > 0$, $a>0$ | Solution: $x < r_1$ or $x > r_2$ |
| $ax^2+bx+c < 0$, $a>0$ | Solution: $r_1 < x < r_2$ |
| Multiply/divide by negative | Always reverse the inequality sign |

## 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.

- [Exponential and logarithmic functions](https://www.owlsprep.com/study/ib-math-ai-hl-u2-exponential-and-logarithmic-functions/)

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