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Quadratic Functions and Equations

Subject: Additional Mathematics
Topic: 2
Cambridge Code: 4037 / 0606

Quadratic Functions​

Quadratic Function - Polynomial of degree 2

General Form​

f(x)=ax2+bx+c,a≠0f(x) = ax^2 + bx + c, \quad a \neq 0

where:

  • aa is the coefficient of x2x^2
  • bb is the coefficient of xx
  • cc is the constant term

Features of Parabola​

Vertex Form​

f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k

where (h,k)(h, k) is the vertex

Completing the Square​

Convert f(x)=ax2+bx+cf(x) = ax^2 + bx + c to vertex form:

f(x)=a(x2+bax)+cf(x) = a\left(x^2 + \frac{b}{a}x\right) + c

f(x)=a[(x+b2a)2−b24a2]+cf(x) = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c

f(x)=a(x+b2a)2−b24a+cf(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c

Example: Complete the square for f(x)=2x2−8x+5f(x) = 2x^2 - 8x + 5

f(x)=2(x2−4x)+5f(x) = 2(x^2 - 4x) + 5 f(x)=2[(x−2)2−4]+5f(x) = 2[(x-2)^2 - 4] + 5 f(x)=2(x−2)2−8+5f(x) = 2(x-2)^2 - 8 + 5 f(x)=2(x−2)2−3f(x) = 2(x-2)^2 - 3

Vertex: (2,−3)(2, -3)

Direction of Parabola​

  • If a>0a > 0: parabola opens upward (minimum)
  • If a<0a < 0: parabola opens downward (maximum)

Axis of Symmetry​

x=−b2a=hx = -\frac{b}{2a} = h

Vertex​

y=f(−b2a)=ky = f\left(-\frac{b}{2a}\right) = k

Or from vertex form: (h,k)(h, k)

Solving Quadratic Equations​

Quadratic Equation - ax2+bx+c=0ax^2 + bx + c = 0

Method 1: Factoring​

If ax2+bx+c=(px+q)(rx+s)=0ax^2 + bx + c = (px + q)(rx + s) = 0, then: x=−qp or x=−srx = -\frac{q}{p} \text{ or } x = -\frac{s}{r}

Example: x2+5x+6=0x^2 + 5x + 6 = 0 (x+2)(x+3)=0(x+2)(x+3) = 0 x=−2 or x=−3x = -2 \text{ or } x = -3

Method 2: Complete the Square​

Example: Solve x2−6x+5=0x^2 - 6x + 5 = 0

x2−6x+5=0x^2 - 6x + 5 = 0 (x−3)2−9+5=0(x-3)^2 - 9 + 5 = 0 (x−3)2=4(x-3)^2 = 4 x−3=±2x - 3 = \pm 2 x=5 or x=1x = 5 \text{ or } x = 1

Method 3: Quadratic Formula​

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Example: Solve 2x2−7x+3=02x^2 - 7x + 3 = 0 where a=2,b=−7,c=3a=2, b=-7, c=3

x=−(−7)±49−244=7±254=7±54x = \frac{-(-7) \pm \sqrt{49-24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}

x=3 or x=12x = 3 \text{ or } x = \frac{1}{2}

The Discriminant​

Discriminant - Expression Δ=b2−4ac\Delta = b^2 - 4ac that determines the nature of roots

Relationship to Roots​

Δ>0:Two distinct real roots\Delta > 0: \text{Two distinct real roots} Δ=0:One repeated real root (equal roots)\Delta = 0: \text{One repeated real root (equal roots)} Δ<0:No real roots (complex roots)\Delta < 0: \text{No real roots (complex roots)}

Example Analysis​

For x2−5x+6=0x^2 - 5x + 6 = 0: Δ=25−24=1>0\Delta = 25 - 24 = 1 > 0 Two distinct real roots ✓

For x2−2x+1=0x^2 - 2x + 1 = 0: Δ=4−4=0\Delta = 4 - 4 = 0 One repeated root: x=1x = 1 ✓

For x2+x+1=0x^2 + x + 1 = 0: Δ=1−4=−3<0\Delta = 1 - 4 = -3 < 0 No real roots ✓

Sum and Product of Roots​

For quadratic ax2+bx+c=0ax^2 + bx + c = 0 with roots α\alpha and β\beta:

Sum of roots: α+β=−ba\text{Sum of roots: } \alpha + \beta = -\frac{b}{a}

Product of roots: αβ=ca\text{Product of roots: } \alpha \beta = \frac{c}{a}

Example​

For 2x2−5x+2=02x^2 - 5x + 2 = 0: Sum=52=2.5\text{Sum} = \frac{5}{2} = 2.5 Product=22=1\text{Product} = \frac{2}{2} = 1

Verify: roots are x=2x = 2 and x=0.5x = 0.5 Sum: 2+0.5=2.52 + 0.5 = 2.5 ✓ Product: 2×0.5=12 \times 0.5 = 1 ✓

Graph Analysis​

Finding Intercepts​

x-intercepts (roots): Solve f(x)=0f(x) = 0

y-intercept: f(0)=cf(0) = c

Sketching Parabola​

  1. Find vertex using x=−b2ax = -\frac{b}{2a}
  2. Calculate y-intercept: f(0)f(0)
  3. Determine direction: aa positive/negative
  4. Find x-intercepts if they exist
  5. Sketch parabola through these points

Key Points to Remember​

  1. Quadratic has form ax2+bx+cax^2 + bx + c
  2. Vertex form shows vertex directly
  3. Discriminant tells number of real roots
  4. Quadratic formula always works
  5. Parabola symmetric about vertical line through vertex
  6. Sum and product of roots relate to coefficients

Worked Examples​

Example 1: Complete the Square​

Express f(x)=x2−8x−3f(x) = x^2 - 8x - 3 in vertex form

f(x)=(x2−8x)−3f(x) = (x^2 - 8x) - 3 f(x)=(x−4)2−16−3f(x) = (x-4)^2 - 16 - 3 f(x)=(x−4)2−19f(x) = (x-4)^2 - 19

Vertex: (4,−19)(4, -19), opens upward, minimum value −19-19

Example 2: Solve Using Quadratic Formula​

Solve 3x2+2x−1=03x^2 + 2x - 1 = 0

x=−2±4+126=−2±46x = \frac{-2 \pm \sqrt{4+12}}{6} = \frac{-2 \pm 4}{6}

x=13 or x=−1x = \frac{1}{3} \text{ or } x = -1

Example 3: Using Discriminant​

For what value of kk does x2−4x+k=0x^2 - 4x + k = 0 have equal roots?

Δ=0\Delta = 0 16−4k=016 - 4k = 0 k=4k = 4

Practice Questions​

  1. Complete the square for:

    • f(x)=x2−6x+5f(x) = x^2 - 6x + 5
    • f(x)=2x2+8x−3f(x) = 2x^2 + 8x - 3

  2. Solve using any method:

    • x2−7x+12=0x^2 - 7x + 12 = 0
    • 2x2−x−3=02x^2 - x - 3 = 0

  3. Find values of mm for which x2+mx+4=0x^2 + 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