Quadratic Functions and Equations
Subject: Additional Mathematics
Topic: 2
Cambridge Code: 4037 / 0606
Quadratic Functions
Quadratic Function - Polynomial of degree 2
f(x)=ax2+bx+c,a=0
where:
- a is the coefficient of x2
- b is the coefficient of x
- c is the constant term
Features of Parabola
f(x)=a(x−h)2+k
where (h,k) is the vertex
Completing the Square
Convert f(x)=ax2+bx+c to vertex form:
f(x)=a(x2+abx)+c
f(x)=a[(x+2ab)2−4a2b2]+c
f(x)=a(x+2ab)2−4ab2+c
Example: Complete the square for f(x)=2x2−8x+5
f(x)=2(x2−4x)+5
f(x)=2[(x−2)2−4]+5
f(x)=2(x−2)2−8+5
f(x)=2(x−2)2−3
Vertex: (2,−3)
Direction of Parabola
- If a>0: parabola opens upward (minimum)
- If a<0: parabola opens downward (maximum)
Axis of Symmetry
x=−2ab=h
Vertex
y=f(−2ab)=k
Or from vertex form: (h,k)
Solving Quadratic Equations
Quadratic Equation - ax2+bx+c=0
Method 1: Factoring
If ax2+bx+c=(px+q)(rx+s)=0, then:
x=−pq or x=−rs
Example: x2+5x+6=0
(x+2)(x+3)=0
x=−2 or x=−3
Method 2: Complete the Square
Example: Solve x2−6x+5=0
x2−6x+5=0
(x−3)2−9+5=0
(x−3)2=4
x−3=±2
x=5 or x=1
x=2a−b±b2−4ac
Example: Solve 2x2−7x+3=0 where a=2,b=−7,c=3
x=4−(−7)±49−24=47±25=47±5
x=3 or x=21
The Discriminant
Discriminant - Expression Δ=b2−4ac that determines the nature of roots
Relationship to Roots
Δ>0:Two distinct real roots
Δ=0:One repeated real root (equal roots)
Δ<0:No real roots (complex roots)
Example Analysis
For x2−5x+6=0:
Δ=25−24=1>0
Two distinct real roots ✓
For x2−2x+1=0:
Δ=4−4=0
One repeated root: x=1 ✓
For x2+x+1=0:
Δ=1−4=−3<0
No real roots ✓
Sum and Product of Roots
For quadratic ax2+bx+c=0 with roots α and β:
Sum of roots: α+β=−ab
Product of roots: αβ=ac
Example
For 2x2−5x+2=0:
Sum=25=2.5
Product=22=1
Verify: roots are x=2 and x=0.5
Sum: 2+0.5=2.5 ✓
Product: 2×0.5=1 ✓
Graph Analysis
Finding Intercepts
x-intercepts (roots): Solve f(x)=0
y-intercept: f(0)=c
Sketching Parabola
- Find vertex using x=−2ab
- Calculate y-intercept: f(0)
- Determine direction: a positive/negative
- Find x-intercepts if they exist
- Sketch parabola through these points
Key Points to Remember
- Quadratic has form ax2+bx+c
- Vertex form shows vertex directly
- Discriminant tells number of real roots
- Quadratic formula always works
- Parabola symmetric about vertical line through vertex
- Sum and product of roots relate to coefficients
Worked Examples
Example 1: Complete the Square
Express f(x)=x2−8x−3 in vertex form
f(x)=(x2−8x)−3
f(x)=(x−4)2−16−3
f(x)=(x−4)2−19
Vertex: (4,−19), opens upward, minimum value −19
Solve 3x2+2x−1=0
x=6−2±4+12=6−2±4
x=31 or x=−1
Example 3: Using Discriminant
For what value of k does x2−4x+k=0 have equal roots?
Δ=0
16−4k=0
k=4
Practice Questions
-
Complete the square for:
- f(x)=x2−6x+5
- f(x)=2x2+8x−3
-
Solve using any method:
- x2−7x+12=0
- 2x2−x−3=0
-
Find values of m for which x2+mx+4=0 has:
- Two distinct real roots
- Equal roots
- No real roots
Revision Tips
- Completing the square reveals vertex form
- Discriminant quickly determines nature of roots
- Quadratic formula is most reliable method
- Sum and product of roots useful for checking