Quadratic Functions — SL 2.2 Quadratic functions (vertex, intercepts, discriminant) and solving quadratics - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Humans have been solving problems involving squares and areas for thousands of years. Ancient civilizations needed to calculate land areas, design arched structures, and predict the paths of thrown objects — all tasks that require understanding relationships where a quantity is multiplied by itself. The mathematics behind these problems eventually became what we call quadratic functions, expressions where the highest power of the variable is 2. From Babylonian clay tablets to modern engineering, quadratics have remained one of the most practical and widely used tools in mathematics.

~2000 BCE • Babylonian Quadratic Problems

Babylonian scribes solved problems equivalent to quadratic equations using geometric methods on clay tablets, finding lengths and widths of rectangles given their area and perimeter.

~300 BCE • Euclid's Geometric Algebra

In his work Elements, Euclid formalized geometric techniques for solving what we now recognize as quadratic equations, relying on areas of squares and rectangles.

~820 CE • Al-Khwārizmī's Algebraic Solutions

The Persian mathematician al-Khwārizmī published systematic methods for solving quadratics, introducing the technique of 'completing the square' — from which we derive the word 'algebra.'

1637 • Descartes and Coordinate Geometry

René Descartes connected algebra with geometry by plotting equations on a coordinate plane, allowing quadratics to be visualized as parabolas for the first time.

Today, quadratic functions model everything from projectile motion in physics to profit optimization in business. The central question of this topic is straightforward: given a quadratic function, how do we identify its key features — its vertex, intercepts, and discriminant — and how do we solve quadratic equations using multiple methods?

SECTION 2

Core Principles & Definitions

A quadratic function is any function that can be written in the form f(x) = ax² + bx + c, where a, b, and c are real numbers and a ≠ 0. The graph of every quadratic function is a U-shaped curve called a parabola. Understanding a few foundational ideas will unlock your ability to analyze and solve any quadratic.

Standard Form

Written as f(x) = ax² + bx + c, this form immediately reveals the y-intercept at (0, c) and whether the parabola opens upward (a > 0) or downward (a < 0).

Vertex Form

Written as f(x) = a(x − h)² + k, this form directly shows the vertex at (h, k) — the turning point of the parabola. The vertex is either the minimum or maximum value of the function.

Factored Form

Written as f(x) = a(x − p)(x − q), this form reveals the x-intercepts (also called roots or zeros) at x = p and x = q. Not every quadratic can be factored over the reals.

Axis of Symmetry

Every parabola has a vertical line of symmetry passing through the vertex. In standard form, this line is x = −b/(2a). Points equidistant from this axis have the same y-value.

Discriminant

The value Δ = b² − 4ac determines the nature of the roots. If Δ > 0, there are two distinct real roots. If Δ = 0, there is one repeated root. If Δ < 0, there are no real roots.

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KEY TAKEAWAY

Think of a quadratic function like a ball thrown into the air. The vertex is the highest point the ball reaches. The x-intercepts are where the ball is at ground level (launch and landing). The discriminant tells you whether the ball ever touches the ground at all — two touchpoints, one bounce, or it never comes down (imagine throwing it from underground).

SECTION 3

Anatomy of a Parabola

The diagram below shows a parabola with all of its key features labeled. Study how the vertex, axis of symmetry, x-intercepts, and y-intercept relate to one another on the coordinate plane. Notice how the parabola is perfectly symmetric about the vertical line through the vertex.

The parabola f(x) = x² − 4x − 5 opens upward (since a = 1 > 0). The vertex at (2, −9) is the minimum point. The x-intercepts at (−1, 0) and (5, 0) are symmetric about the axis of symmetry x = 2. The y-intercept at (0, −5) is the constant term c in standard form.

In the diagram above, notice that the two x-intercepts are equidistant from the axis of symmetry. The x-intercept at x = −1 is 3 units to the left of x = 2, and the x-intercept at x = 5 is 3 units to the right. This symmetry is a defining property of every parabola and is extremely useful: if you know one x-intercept and the axis of symmetry, you can immediately find the other x-intercept.

SECTION 4

Mathematical Framework

This section presents the essential formulas you need for analyzing and solving quadratic functions. Each equation is a tool — know when and how to use each one.

STANDARD FORM

f(x) = ax² + bx + c

a = coefficient of x² (determines direction and width), b = coefficient of x, c = y-intercept. The parabola opens upward if a > 0 and downward if a < 0.

VERTEX FORMULA

Vertex = (h, k) where h = −b / (2a) and k = f(h)

The x-coordinate of the vertex is found using h = −b/(2a). Substitute h back into f(x) to find the y-coordinate k. In vertex form f(x) = a(x − h)² + k, you can read (h, k) directly.

QUADRATIC FORMULA

x = (−b ± √(b² − 4ac)) / (2a)

This formula gives the x-intercepts (roots) of ax² + bx + c = 0. It works for every quadratic — factorable or not. The expression under the square root, b² − 4ac, is the discriminant (Δ).

DISCRIMINANT

Δ = b² − 4ac

If Δ > 0 → two distinct real roots. If Δ = 0 → one repeated real root (vertex touches x-axis). If Δ < 0 → no real roots (parabola does not cross the x-axis).

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IB Formula Booklet Reminder

The quadratic formula and the discriminant are provided in the IB formula booklet. However, you should know the vertex formula h = −b/(2a) by heart and be fluent in converting between standard, vertex, and factored forms.

SECTION 5

The Discriminant & Nature of Roots

The discriminant Δ = b² − 4ac is a single number that tells you everything about how many times a parabola crosses the x-axis. The diagram below shows three parabolas with different discriminant values, illustrating the three possible cases.

Left: When Δ > 0, the parabola crosses the x-axis at two distinct points. Center: When Δ = 0, the vertex sits exactly on the x-axis, giving one repeated root. Right: When Δ < 0, the entire parabola lies above the x-axis, so there are no real solutions.

Discriminant Value	Number of Real Roots	Graphical Meaning	Example (a = 1)
Δ > 0	Two distinct real roots	Parabola crosses x-axis twice	x² − 5x + 6 = 0 → Δ = 1 → x = 2, 3
Δ = 0	One repeated real root	Vertex touches x-axis	x² − 6x + 9 = 0 → Δ = 0 → x = 3
Δ < 0	No real roots	Parabola does not cross x-axis	x² + 2x + 5 = 0 → Δ = −16 → no real solutions

Summary of discriminant cases

SECTION 6

Worked Example — Full Analysis of a Quadratic

Let's completely analyze the quadratic function f(x) = 2x² − 8x + 3. We will find its vertex, axis of symmetry, y-intercept, discriminant, and x-intercepts.

Analyze f(x) = 2x² − 8x + 3

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Step 1 — Identify a, b, and c

Comparing f(x) = 2x² − 8x + 3 with the standard form ax² + bx + c, we identify a = 2, b = −8, and c = 3. Since a = 2 > 0, the parabola opens upward.

a = 2, b = −8, c = 3 → Opens upward
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Step 2 — Find the vertex

Use h = −b/(2a) = −(−8)/(2 × 2) = 8/4 = 2. Now find k by substituting h = 2 into f(x): k = f(2) = 2(2)² − 8(2) + 3 = 8 − 16 + 3 = −5. The vertex is at (2, −5).

Vertex = (2, −5)
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Step 3 — Axis of symmetry and y-intercept

The axis of symmetry is the vertical line x = h = 2. The y-intercept is found by evaluating f(0) = 2(0)² − 8(0) + 3 = 3, so the y-intercept is at (0, 3).

Axis of symmetry: x = 2 | y-intercept: (0, 3)
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Step 4 — Calculate the discriminant

Δ = b² − 4ac = (−8)² − 4(2)(3) = 64 − 24 = 40. Since Δ = 40 > 0, the function has two distinct real roots.

Δ = 40 > 0 → Two distinct real roots
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Step 5 — Find the x-intercepts using the quadratic formula

x = (−b ± √Δ) / (2a) = (8 ± √40) / 4 = (8 ± 2√10) / 4 = (4 ± √10) / 2. So x = (4 + √10)/2 ≈ 3.58 and x = (4 − √10)/2 ≈ 0.42.

x-intercepts: x = (4 ± √10)/2 ≈ 0.42 and ≈ 3.58
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Step 6 — Write in vertex form

Using the vertex (h, k) = (2, −5) and a = 2, the vertex form is f(x) = 2(x − 2)² − 5. You can verify by expanding: 2(x² − 4x + 4) − 5 = 2x² − 8x + 8 − 5 = 2x² − 8x + 3 ✓

Vertex form: f(x) = 2(x − 2)² − 5

SECTION 7

Solving Quadratics — Three Methods Compared

There are three primary methods for solving a quadratic equation ax² + bx + c = 0. Each method has its own strengths and ideal use cases. The IB may ask you to use a specific method, so you need to be confident with all three.

Method	When to Use	Advantages	Limitations
Factoring	When the quadratic factors neatly over the integers (e.g., x² − 5x + 6)	Fast, elegant, no formula needed; directly gives exact roots	Not all quadratics factor neatly; requires trial and error or pattern recognition
Completing the Square	When converting to vertex form or deriving the quadratic formula; useful for exact answers	Always works; gives the vertex form directly; builds algebraic understanding	More steps than other methods; arithmetic can get messy with fractions
Quadratic Formula	When factoring fails or when you need to find roots quickly for any quadratic	Universal method — works for every quadratic; immediately identifies nature of roots via discriminant	Requires careful substitution; does not directly give vertex form

Comparison of the three methods for solving quadratic equations

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KEY TAKEAWAY

Think of these three methods like tools in a toolbox. Factoring is like a screwdriver — quick and efficient for the right screws (nice integer roots). Completing the square is like an adjustable wrench — versatile and revealing, but takes more effort. The quadratic formula is like a power drill — it handles every job, even the ugly ones. A skilled IB student uses all three fluently and chooses the best tool for the task at hand.

SECTION 8

Connection to Advanced Topics

Quadratic functions are a springboard to more advanced mathematics. The concepts you have learned here — vertex, roots, and discriminant — will appear in different guises as you progress through the IB course and beyond. Understanding quadratics deeply now will pay dividends when you encounter these more complex topics.

SL 2.2 Concept	Advanced Extension	Where You'll See It
Discriminant Δ < 0 (no real roots)	Complex / imaginary roots using i = √(−1)	HL Topic 1 (complex numbers)
Vertex as maximum/minimum	Optimization using calculus (finding turning points of any function)	SL Topic 5 (calculus)
Factored form f(x) = a(x − p)(x − q)	Polynomial factoring for cubics, quartics, and beyond	SL 2.5 (polynomial functions)
Completing the square	Equations of circles, ellipses, and integration by completing the square	HL Topic 2 & 5

How SL 2.2 concepts connect to higher-level topics

In calculus, you will learn a general method for finding the maximum or minimum of any smooth function, not just parabolas. But the technique of setting a derivative equal to zero is the generalized version of what you're already doing with h = −b/(2a). The quadratic formula itself can be derived by applying completing the square to the general equation ax² + bx + c = 0 — a beautiful connection between two methods.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A quadratic function has a = −3, b = 6, and c = 1. Without performing any calculations, explain whether the parabola opens upward or downward and whether the vertex represents a maximum or minimum value.

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Since a = −3 < 0, the parabola opens downward. When a parabola opens downward, its vertex is the highest point on the curve, meaning the vertex represents a maximum value of the function. This tells you that f(x) has no upper bound as x → ±∞ (it goes to −∞), but it has a greatest value at the vertex. The y-intercept is c = 1, which tells you the graph passes through (0, 1).

Question 2: BASIC CALCULATION

Find the vertex and axis of symmetry of f(x) = x² + 6x + 5.

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Using h = −b/(2a) = −6/(2 × 1) = −3. Then k = f(−3) = (−3)² + 6(−3) + 5 = 9 − 18 + 5 = −4. The vertex is (−3, −4). The axis of symmetry is the vertical line x = −3. In vertex form, f(x) = (x + 3)² − 4. You can verify by noting the x-intercepts are at x = −1 and x = −5 (since x² + 6x + 5 = (x + 1)(x + 5)), and their midpoint is (−1 + (−5))/2 = −3, confirming the axis of symmetry.

Question 3: INTERMEDIATE

Solve 3x² − 2x − 5 = 0 using the quadratic formula. State the discriminant and the nature of the roots.

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Here a = 3, b = −2, c = −5. The discriminant is Δ = (−2)² − 4(3)(−5) = 4 + 60 = 64. Since Δ = 64 > 0, there are two distinct real roots. Applying the quadratic formula: x = (−(−2) ± √64)/(2 × 3) = (2 ± 8)/6. So x = (2 + 8)/6 = 10/6 = 5/3, and x = (2 − 8)/6 = −6/6 = −1. The two roots are x = 5/3 and x = −1. You could verify by factoring: 3x² − 2x − 5 = (3x − 5)(x + 1).

Question 4: APPLIED

A ball is launched upward from a platform 2 meters high with an initial velocity of 14 m/s. Its height (in meters) after t seconds is modeled by h(t) = −5t² + 14t + 2. Find the maximum height reached by the ball and the time at which it hits the ground.

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For the maximum height, find the vertex: t = −b/(2a) = −14/(2 × (−5)) = 14/10 = 1.4 seconds. The maximum height is h(1.4) = −5(1.4)² + 14(1.4) + 2 = −5(1.96) + 19.6 + 2 = −9.8 + 19.6 + 2 = 11.8 meters. For when the ball hits the ground, solve h(t) = 0: −5t² + 14t + 2 = 0. Using the quadratic formula with a = −5, b = 14, c = 2: Δ = 196 + 40 = 236, so t = (−14 ± √236)/(−10). Taking the positive root: t = (−14 − √236)/(−10) ≈ (−14 − 15.36)/(−10) ≈ 2.94 seconds. The ball reaches a maximum height of 11.8 m at t = 1.4 s and hits the ground at approximately t = 2.94 s.

Question 5: CRITICAL THINKING

For what values of k does the equation 2x² + kx + 8 = 0 have exactly one real solution? Explain your reasoning and describe what happens graphically at this value of k.

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For exactly one real solution, the discriminant must equal zero: Δ = k² − 4(2)(8) = 0, so k² − 64 = 0, which gives k² = 64, meaning k = 8 or k = −8. At these values, the parabola y = 2x² + kx + 8 just touches the x-axis at exactly one point — the vertex lies on the x-axis. When k = 8, the equation becomes 2x² + 8x + 8 = 0, or 2(x + 2)² = 0, giving x = −2. When k = −8, the equation becomes 2x² − 8x + 8 = 0, or 2(x − 2)² = 0, giving x = 2. Both parabolas open upward (a = 2 > 0) with their vertex touching the x-axis, creating a tangent point rather than a crossing.

SUMMARY

Lesson Summary

Summary

A quadratic function f(x) = ax² + bx + c produces a parabola that opens upward when a > 0 and downward when a < 0. The vertex at (h, k), found using h = −b/(2a), represents the maximum or minimum value. The axis of symmetry x = h divides the parabola into two mirror-image halves. The y-intercept is always at (0, c), and the x-intercepts (roots) are found by solving f(x) = 0.

The discriminant Δ = b² − 4ac determines the nature of the roots: two distinct real roots when Δ > 0, one repeated root when Δ = 0, and no real roots when Δ < 0. You can solve quadratics by factoring (when possible), completing the square (which also yields vertex form), or applying the quadratic formula (which always works). Being able to switch between standard form, vertex form, and factored form is an essential skill for the IB exam and forms the foundation for studying polynomials, optimization, and calculus.