Quadratic Functions & Models — SL 2.3 Quadratic functions...

SECTION 1

Historical Context & Motivation

Long before anyone wrote down y = ax² + bx + c, ancient civilizations wrestled with problems that secretly involved quadratic relationships. Babylonian scribes around 2000 BCE solved area problems on clay tablets that amounted to finding the sides of rectangles given their area and perimeter — essentially solving quadratic equations without any algebraic notation. The Greeks studied the curves produced by slicing a cone at various angles, and the parabola was one of the shapes they discovered. Over the centuries, mathematicians from the Islamic Golden Age through the European Renaissance developed increasingly powerful methods for handling these equations, eventually giving us the formulas and graphing techniques you will learn in this lesson.

~2000 BCE

Babylonian Quadratic Problems

Babylonian mathematicians solved geometric area problems equivalent to quadratic equations using algorithmic recipes inscribed on clay tablets.

~300 BCE

Greek Conic Sections

Apollonius of Perga classified curves formed by slicing a cone, naming the parabola, ellipse, and hyperbola — linking geometry to what would later become algebraic functions.

~820 CE

Al-Khwarizmi's Algebra

Persian scholar al-Khwarizmi published systematic methods for solving quadratic equations by completing the square, giving us the word 'algebra' itself.

1637

Descartes & Coordinate Geometry

René Descartes merged algebra with geometry, making it possible to graph equations like y = x² on a coordinate plane and analyze their shapes visually.

1980s–Present

Graphing Calculators & CAS

Technology such as graphing calculators and computer algebra systems (CAS) transformed how students explore, solve, and model quadratic functions in real time.

Today, quadratic functions appear everywhere — in the arc of a basketball, the shape of a satellite dish, and the profit curve of a business. The central question this lesson addresses is: How do we write, graph, and solve quadratic functions so we can model curved, real-world relationships?

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 symmetric, U-shaped curve called a parabola. Understanding a few foundational ideas unlocks everything else about these functions.

Standard Form

Written as f(x) = ax² + bx + c. The coefficient a controls the opening direction and width; c is the y-intercept.

Vertex Form

Written as f(x) = a(x − h)² + k. The vertex (h, k) is immediately visible, making it easy to identify the maximum or minimum point of the parabola.

Axis of Symmetry

Every parabola has a vertical axis of symmetry passing through the vertex at x = h (or x = −b / 2a in standard form). The graph is a mirror image on either side of this line.

Roots / Zeros

The roots (also called zeros or x-intercepts) are the x-values where f(x) = 0. A quadratic can have two, one, or zero real roots depending on the discriminant.

Discriminant

The value Δ = b² − 4ac determines the number of real roots: Δ > 0 gives two roots, Δ = 0 gives one repeated root, and Δ < 0 gives no real roots.

✦ KEY TAKEAWAY

Think of a parabola like the path of a ball you toss to a friend. The highest point of the ball's arc is the vertex, the line running straight down from that peak is the axis of symmetry, and the two spots where the ball leaves and returns to your hand level are the roots. The coefficient a controls how steep or gentle that arc is.

SECTION 3

Anatomy of a Parabola

The diagram above shows the parabola f(x) = (x − 3)² − 4. The vertex at (3, −4) is the lowest point. The dashed vertical line through the vertex is the axis of symmetry (x = 3). The two cyan dots mark the roots at (1, 0) and (5, 0), and the amber dot shows the y-intercept at (0, 5).

Notice how every feature of the parabola connects to a specific part of the equation. In vertex form f(x) = a(x − h)² + k, the values h = 3 and k = −4 place the vertex directly at the point (3, −4). Because a = 1 (positive), the parabola opens upward, making the vertex a minimum. The roots occur where the curve crosses the x-axis, and they are equidistant from the axis of symmetry — each exactly 2 units away from x = 3. This symmetry is one of the most useful properties of quadratic functions.

SECTION 4

Mathematical Framework

There are three main forms of a quadratic function, and each reveals different information at a glance. In the IB Applications and Interpretation course, you are expected to move fluently between these forms and use technology (GDC or CAS) to solve equations and find key features.

STANDARD FORM

f(x) = ax² + bx + c

Where a ≠ 0 controls the opening direction and width, b shifts the axis of symmetry, and c = y-intercept. Axis of symmetry: x = −b / (2a).

VERTEX FORM

f(x) = a(x − h)² + k

The vertex is at (h, k). If a > 0 the parabola opens upward (minimum at vertex). If a < 0 it opens downward (maximum at vertex).

QUADRATIC FORMULA

x = (−b ± √(b² − 4ac)) / (2a)

Solves ax² + bx + c = 0 for x. The expression under the square root, Δ = b² − 4ac, is the discriminant. It determines the nature and number of roots.

DISCRIMINANT CONDITIONS

Δ > 0 → two distinct real roots | Δ = 0 → one repeated root | Δ < 0 → no real roots

The discriminant tells you the number of x-intercepts the parabola has before you even graph it.

📱 IB TECHNOLOGY NOTE

On the IB exam, you are expected to use your GDC (graphing display calculator) to solve quadratic equations, find vertices, and determine roots. Practice entering equations into your calculator's equation solver or graphing mode. The TI-84, TI-Nspire, and Casio fx-CG50 all have built-in polynomial solvers — learn yours.

SECTION 5

Comparing Forms & the Role of the Discriminant

Each algebraic form of a quadratic function gives you quick access to different information. The table below summarizes what each form reveals and when you might prefer to use it.

Three forms of a quadratic function and their uses

Form	Equation	Reveals at a Glance	Best For
Standard	ax² + bx + c	y-intercept (c), coefficients for the quadratic formula	Applying the quadratic formula, identifying the y-intercept quickly
Vertex	a(x − h)² + k	Vertex (h, k), axis of symmetry (x = h), direction of opening	Graphing, optimization problems (finding max/min)
Factored	a(x − p)(x − q)	Roots / x-intercepts at x = p and x = q	Reading off roots, sketching graphs quickly

When Δ > 0, the parabola crosses the x-axis at two points (two real roots). When Δ = 0, the parabola just touches the x-axis at one point (a repeated root). When Δ < 0, the parabola does not intersect the x-axis at all (no real roots).

The discriminant is incredibly useful in modeling contexts. For example, if you model a projectile's height as a quadratic function of horizontal distance, the discriminant tells you whether the projectile ever reaches a certain target height. If Δ < 0 for the equation f(x) = target, the object never reaches that height.

SECTION 6

Worked Example — Modeling a Projectile

A ball is launched from a platform 2 metres above the ground. Its height h (in metres) after t seconds is modeled by h(t) = −4.9t² + 14t + 2. Find the maximum height of the ball, the time it takes to reach the maximum, and when it hits the ground.

Projectile Motion Problem

Step 1 — Identify the Form and Key ValuesThe function h(t) = −4.9t² + 14t + 2 is in standard form with a = −4.9, b = 14, and c = 2. Since a < 0, the parabola opens downward, so the vertex gives a maximum height.

Step 2 — Find the Time at Maximum HeightThe axis of symmetry gives the time at the vertex. Using t = −b / (2a): t = −14 / (2 × (−4.9)) = −14 / (−9.8) = 1.43 seconds (to 3 s.f.).

t = 1.43 s

Step 3 — Calculate the Maximum HeightSubstitute t = 1.43 into h(t): h(1.43) = −4.9(1.43)² + 14(1.43) + 2 = −4.9(2.0449) + 20.02 + 2 = −10.02 + 20.02 + 2 = 12.0 metres (to 3 s.f.).

Maximum height ≈ 12.0 m

Step 4 — Find When the Ball Hits the GroundSet h(t) = 0 and solve: −4.9t² + 14t + 2 = 0. Using the quadratic formula (or your GDC): t = (−14 ± √(14² − 4(−4.9)(2))) / (2(−4.9)) = (−14 ± √(196 + 39.2)) / (−9.8) = (−14 ± √235.2) / (−9.8). √235.2 ≈ 15.34. So t = (−14 + 15.34) / (−9.8) = −0.137 (rejected, negative time) or t = (−14 − 15.34) / (−9.8) = 2.99 seconds.

Ball hits ground at t ≈ 2.99 s

Step 5 — Verify with Technology (GDC)On your GDC, graph Y1 = −4.9X² + 14X + 2 and use the 'maximum' and 'zero' features to confirm: maximum at approximately (1.43, 12.0) and positive zero at approximately x = 2.99. This matches our algebraic work.

💡 GDC TIP

On the TI-84, press [2nd][TRACE] to access the CALC menu, then select '4:maximum' or '2:zero'. On the Casio fx-CG50, go to Graph → G-Solv and choose MAX or ROOT. Practice this before your exam!

SECTION 7

Solving Quadratics — Methods Compared

There are several ways to solve a quadratic equation, and each has its strengths and limitations. In the IB Applications and Interpretation course, using technology is explicitly encouraged, but understanding when each method works best will make you more efficient.

Comparison of methods for solving quadratic equations

Method	Strengths	Limitations
Factoring	Fast, exact; reveals roots directly; no technology needed.	Only works when roots are rational; not always obvious how to factor.
Quadratic Formula	Always works for any quadratic; gives exact answers including irrational roots.	Algebraically heavy; easy to make sign errors with complex coefficients.
Completing the Square	Converts to vertex form; excellent for optimization and deriving the formula.	Can be tedious; requires comfortable fraction manipulation.
GDC / Technology	Fast, visual; handles messy coefficients; confirms algebraic work.	Gives decimal approximations (not exact); need to set an appropriate window.

✦ KEY TAKEAWAY

Think of these methods like tools in a toolkit. Factoring is the screwdriver — quick when the screw fits perfectly. The quadratic formula is the universal wrench — it always works but takes more effort. Your GDC is the power drill — fast and powerful, but you should understand what it is doing. On the IB exam, choose the tool that fits the problem.

SECTION 8

Connection to Higher-Degree Polynomials & Advanced Models

Quadratic functions are the simplest type of polynomial function with a curve. Once you move beyond quadratics, you encounter cubic (degree 3), quartic (degree 4), and higher-degree polynomial functions. Many of the ideas you learn with quadratics — such as roots, turning points, and the role of coefficients — extend naturally to these more complex functions.

Quadratics vs. higher-degree polynomials

Feature	Quadratic (Degree 2)	Higher-Degree Polynomials
Shape	Single U-shape (parabola), one turning point	Multiple turning points; S-curves, W-shapes, etc.
Max real roots	2	Equal to the degree (e.g., cubic has up to 3)
Solving algebraically	Quadratic formula always works	Cubic/quartic formulas exist but are rarely used; technology is essential
IB context	SL 2.3 — explicit study of properties and modeling	SL 2.4/2.5 — studied with GDC support for graphing and regression

In the IB AI course, you will also encounter quadratic regression — fitting a parabola to a set of real-world data points using technology. This extends the modeling idea: instead of being given the equation, you collect data, enter it into your GDC, and let the calculator find the best-fitting quadratic. This is a powerful skill that connects directly to the IA (Internal Assessment) and Paper 2 contexts.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

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

PROBLEM 2 — BASIC CALCULATION

Find the vertex of the quadratic function f(x) = 2x² − 8x + 3 using the formula x = −b / (2a).

PROBLEM 3 — INTERMEDIATE

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

PROBLEM 4 — APPLIED

A company's daily profit P (in thousands of dollars) depends on the number of items x it produces, modeled by P(x) = −0.5x² + 20x − 50. Find the number of items that maximizes profit and the maximum daily profit. Also find the break-even points (where profit is zero).

PROBLEM 5 — CRITICAL THINKING

A parabola passes through the points (1, 6), (3, 2), and (5, 6). Use these three points to set up a system of equations and find the quadratic function in standard form. Then convert it to vertex form and identify the vertex. Verify your answer using your GDC's quadratic regression feature.

SUMMARY

Lesson Summary

A quadratic function has the general form f(x) = ax² + bx + c and produces a parabola when graphed. The three key algebraic forms — standard form (ax² + bx + c), vertex form (a(x − h)² + k), and factored form (a(x − p)(x − q)) — each highlight different features of the same function. The vertex represents the maximum or minimum, the axis of symmetry divides the parabola into mirror halves, and the roots are the x-intercepts where f(x) = 0.

The discriminant Δ = b² − 4ac reveals whether there are two, one, or zero real roots. You can solve quadratics by factoring, using the quadratic formula, completing the square, or using your GDC to find zeros and maxima/minima graphically. In the IB AI course, technology-supported methods are central: use your calculator to graph, solve, and perform quadratic regression on real-world data. These skills are essential for modeling contexts on Paper 2 and for your Internal Assessment.