IB Mathematics AA β Topic 2: Functions
Comprehensive Guide to Quadratic Functions
Introduction to Quadratic Functions
A quadratic function is a polynomial function of degree 2, forming a U-shaped curve called a parabola. Quadratics are among the most important functions in mathematics, appearing in physics (projectile motion), economics (profit optimization), engineering (bridge design), and countless other applications.
General form: \(f(x) = ax^2 + bx + c\) where \(a \neq 0\). The coefficient \(a\) determines whether the parabola opens upward (\(a > 0\)) or downward (\(a < 0\)).
Key features: Every quadratic has a vertex (maximum or minimum point), an axis of symmetry, and may have 0, 1, or 2 x-intercepts (roots). Understanding these features allows you to analyze and sketch the function accurately.
In this guide: We'll explore the three forms of quadratic functions (general, factored, vertex), master completing the square, understand the discriminant's role in determining roots, find vertices, solve quadratic equations, and develop sketching skills essential for IB exams.
1. Three Forms of Quadratic Functions
Understanding the Forms
Quadratic functions can be written in three different forms, each revealing different information about the parabola.
Three Standard Forms
Form 1: General (Standard) Form
\(f(x) = ax^2 + bx + c\)
Shows clearly: y-intercept at \((0, c)\)
Axis of symmetry: \(x = -\frac{b}{2a}\)
Use when: You need to find the y-intercept or use the quadratic formula
Form 2: Factored (Intercept) Form
\(f(x) = a(x - p)(x - q)\)
Shows clearly: x-intercepts (roots) at \((p, 0)\) and \((q, 0)\)
Axis of symmetry: \(x = \frac{p + q}{2}\) (midpoint of roots)
Use when: You know the roots or can factor easily
Form 3: Vertex Form
\(f(x) = a(x - h)^2 + k\)
Shows clearly: Vertex at \((h, k)\)
Axis of symmetry: \(x = h\)
Use when: You know the vertex or need to identify transformations
π‘ Form Selection Tips:
- Given roots β use factored form
- Given vertex β use vertex form
- Need to solve for x β convert to factored or use quadratic formula
- Completing the square converts general form to vertex form
2. Completing the Square
The Method
Completing the square is a technique for converting a quadratic from general form to vertex form. This reveals the vertex and makes solving equations easier.
Algorithm for \(ax^2 + bx + c\):
- If \(a \neq 1\), factor out \(a\) from the first two terms
- Take half the coefficient of \(x\), then square it: \(\left(\frac{b}{2a}\right)^2\)
- Add and subtract this value inside the brackets
- Factor the perfect square trinomial
- Simplify to get \(a(x - h)^2 + k\) form
Quick Formula (when \(a = 1\)):
\(x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\)
β Common Mistakes:
- Forgetting to factor out \(a\): Must have coefficient 1 on \(x^2\) inside brackets
- Sign errors: Be careful with negative coefficients
- Not balancing: What you add, you must also subtract
- Distributing incorrectly: When factoring out \(a\), divide correctly
Example 1: Completing the Square
Problem: Express \(f(x) = 2x^2 - 12x + 7\) in vertex form and find the vertex.
Solution:
Step 1: Factor out the coefficient of \(x^2\) from the first two terms
\(f(x) = 2(x^2 - 6x) + 7\)
Step 2: Complete the square inside brackets
Take half of \(-6\): \(\frac{-6}{2} = -3\)
Square it: \((-3)^2 = 9\)
Add and subtract 9 inside the brackets:
\(f(x) = 2(x^2 - 6x + 9 - 9) + 7\)
\(= 2((x^2 - 6x + 9) - 9) + 7\)
Step 3: Factor the perfect square
\(f(x) = 2((x - 3)^2 - 9) + 7\)
Step 4: Distribute and simplify
\(= 2(x - 3)^2 - 18 + 7\)
\(= 2(x - 3)^2 - 11\)
Vertex form: \(f(x) = 2(x - 3)^2 - 11\)
Vertex: \((3, -11)\)
Since \(a = 2 > 0\), parabola opens upward, so \((3, -11)\) is a minimum point.
3. The Discriminant
Definition and Significance
The discriminant is the expression under the square root in the quadratic formula. It determines the nature and number of roots without solving the equation completely.
The Discriminant Formula
\(\Delta = b^2 - 4ac\)
For the quadratic \(ax^2 + bx + c = 0\)
Interpreting the Discriminant
Three Cases:
\(\Delta > 0\) (Positive)
Two distinct real roots
Graph crosses x-axis at two points
Example: \(\Delta = 25\) β roots are \(\frac{-b \pm 5}{2a}\)
\(\Delta = 0\) (Zero)
One repeated real root (or two equal roots)
Graph touches x-axis at exactly one point (the vertex)
Root is \(x = -\frac{b}{2a}\)
\(\Delta < 0\) (Negative)
No real roots (two complex roots)
Graph does not cross or touch the x-axis
Cannot take square root of negative number in real numbers
π‘ Discriminant Applications:
- Determine number of x-intercepts before graphing
- Find values of parameters that give specific numbers of roots
- Verify factorability: \(\Delta\) is perfect square β factorable over integers
- Quick check before attempting to factor
Example 2: Using the Discriminant
Problem: For the quadratic equation \(kx^2 - 4x + 1 = 0\):
(a) Find the value of \(k\) for which the equation has exactly one solution
(b) Find the range of values of \(k\) for which there are two distinct real solutions
Solution:
Identify: \(a = k\), \(b = -4\), \(c = 1\)
Discriminant: \(\Delta = b^2 - 4ac = (-4)^2 - 4(k)(1) = 16 - 4k\)
(a) Exactly one solution (repeated root):
Need \(\Delta = 0\)
\(16 - 4k = 0\)
\(16 = 4k\)
\(k = 4\)
Answer: \(k = 4\)
(b) Two distinct real solutions:
Need \(\Delta > 0\)
\(16 - 4k > 0\)
\(16 > 4k\)
\(4 > k\)
Answer: \(k < 4\)
Note: Also need \(k \neq 0\) for it to be a quadratic, so \(k \in (-\infty, 0) \cup (0, 4)\)
4. Vertex and Solving Quadratic Equations
Finding the Vertex
The vertex is the turning point of the parabola:
Method 1: Formula (from general form)
x-coordinate: \(h = -\frac{b}{2a}\)
y-coordinate: \(k = f(h) = f\left(-\frac{b}{2a}\right)\)
Vertex: \(\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)\)
Method 2: From vertex form
If \(f(x) = a(x - h)^2 + k\), vertex is immediately \((h, k)\)
Method 3: From roots (factored form)
If roots are \(p\) and \(q\), x-coordinate is \(h = \frac{p+q}{2}\)
Then find \(k = f(h)\)
Solving Quadratic Equations
Four Methods:
1. Factoring
If \(ax^2 + bx + c = a(x - p)(x - q)\), then \(x = p\) or \(x = q\)
Works when: Easily factorable
2. Quadratic Formula
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Works when: Always! (if real solutions exist)
Given in IB formula booklet
3. Completing the Square
Convert to vertex form, then solve \(a(x - h)^2 + k = 0\)
Works when: Useful for deriving quadratic formula or finding vertex
4. Graphing (GDC)
Graph \(y = ax^2 + bx + c\) and find x-intercepts
Works when: Calculator allowed, approximate solutions acceptable
β Common Errors:
- Sign errors in quadratic formula: Be careful with \(-b\) and \(\pm\)
- Forgetting Β± sign: Usually two solutions!
- Wrong vertex formula: It's \(-\frac{b}{2a}\), not \(\frac{b}{2a}\)
- Not checking discriminant first: Save time by checking if real solutions exist
Example 3: Complete Quadratic Analysis (IB-Style)
Problem: Consider the function \(f(x) = -x^2 + 6x - 5\)
(a) Find the vertex and state whether it's a maximum or minimum
(b) Find the x-intercepts
(c) Find the y-intercept
(d) Sketch the graph
Solution:
(a) Vertex:
From \(f(x) = -x^2 + 6x - 5\): \(a = -1\), \(b = 6\), \(c = -5\)
x-coordinate: \(h = -\frac{b}{2a} = -\frac{6}{2(-1)} = -\frac{6}{-2} = 3\)
y-coordinate: \(k = f(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4\)
Vertex: \((3, 4)\)
Since \(a = -1 < 0\), parabola opens downward β Maximum point
(b) x-intercepts: Solve \(f(x) = 0\)
\(-x^2 + 6x - 5 = 0\)
Multiply by \(-1\): \(x^2 - 6x + 5 = 0\)
Factor: \((x - 1)(x - 5) = 0\)
\(x = 1\) or \(x = 5\)
x-intercepts: \((1, 0)\) and \((5, 0)\)
(c) y-intercept: Set \(x = 0\)
\(f(0) = -(0)^2 + 6(0) - 5 = -5\)
y-intercept: \((0, -5)\)
(d) Sketch:
Key features to show:
- Parabola opens downward (\(a < 0\))
- Vertex (maximum) at \((3, 4)\)
- x-intercepts at \((1, 0)\) and \((5, 0)\)
- y-intercept at \((0, -5)\)
- Axis of symmetry: \(x = 3\)
Note: Vertex is midway between x-intercepts: \(\frac{1+5}{2} = 3\) β
π Quadratics Formula Summary
| Property | Formula | Notes |
|---|---|---|
| General Form | \(f(x) = ax^2 + bx + c\) | Shows y-intercept \((0,c)\) |
| Vertex | \(\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)\) | Turning point |
| Axis of Symmetry | \(x = -\frac{b}{2a}\) | Vertical line through vertex |
| Discriminant | \(\Delta = b^2 - 4ac\) | Determines number of roots |
| Quadratic Formula | \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) | In formula booklet |
| Vertex Form | \(f(x) = a(x-h)^2 + k\) | Vertex at \((h,k)\) |
π― IB Exam Strategy
Common Question Types:
- "Find the vertex": Use \(x = -\frac{b}{2a}\) or complete the square
- "Express in vertex form": Complete the square
- "Solve the equation": Factor, use formula, or GDC
- "Find values of k for which...": Use discriminant
- "Sketch the graph": Find vertex, intercepts, note direction
Calculator Tips:
- Use graph to verify your algebraic work
- Find maximum/minimum using CALC menu
- Find zeros (roots) with root finder
- Always show algebraic work even if using GDC
π Master Quadratic Functions!
Quadratic functions are foundational to all of mathematics. Understanding the three forms, mastering completing the square, interpreting the discriminant, and finding vertices gives you complete control over parabolasβessential for calculus, optimization, and modeling!
Key Success Factors:
- β Know which form reveals which information
- β Master completing the square technique
- β Use discriminant to predict number of roots
- β Vertex formula: \(x = -\frac{b}{2a}\) (negative!)
- β Quadratic formula is in your formula booklet
- β Check direction: \(a > 0\) opens up, \(a < 0\) opens down
Practice All Three Forms β’ Use the Discriminant β’ Verify with GDC
Master quadratics and unlock advanced mathematics! π