A quadratic function is a second-degree polynomial function of the form:

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

Unlike linear functions, which exhibit a constant rate of change and graph as straight lines, quadratic functions graph as symmetric curves called parabolas. The rate of change along a parabola varies continuously, reaching a singular turning point known as the vertex (representing either a global maximum or minimum) and changing direction.

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Algebraic Forms of a Quadratic Function

Examiners expect students to transition fluently between three algebraic representations, choosing the best form based on the question's demands.

A. Standard Form: f(x)=ax2+bx+cf(x) = ax^2 + bx + c

  • y-intercept: Occurs at (0, c).
  • Orientation: If a > 0, the parabola opens upwards (∪\cup) with a minimum turning point; if a < 0, it opens downwards (∩\cap) with a maximum turning point.
  • Axis of Symmetry: The vertical line passing through x=−b2ax = -\dfrac{b}{2a}.

B. Vertex Form: f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k

  • Vertex Coordinates: The point (h, k).
  • Axis of Symmetry: The line x = h.
  • Extrema: If a > 0, the minimum value is k; if a < 0, the maximum value is k.
  • Geometric Transformations: Compared to the parent function y=x2y = x^2, the curve is translated horizontally by h units, scaled vertically by factor a, and translated vertically by k units.

C. Factored (Root) Form: f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q)

  • x-intercepts (Roots/Zeros): Occurs at (p, 0) and (q, 0).
  • Symmetry Property: The x-coordinate of the vertex lies precisely halfway between the roots:

h=p+q2h = \dfrac{p + q}{2}

Converting Standard Form to Vertex Form (Completing the Square)

Completing the square is the standard algebraic method for identifying the vertex and extreme values without using calculus.

Step-by-Step Procedure for f(x)=ax2+bx+cf(x) = ax^2 + bx + c

Step 1: Factor out the coefficient a from the first two terms:

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

Step 2: Halve the linear coefficient inside the bracket:

b2a\dfrac{b}{2a}

Step 3: Form the squared bracket and subtract the square of this halved term:

x2+bax=(x+b2a)2−(b2a)2x^2 + \dfrac{b}{a}x = \left(x + \dfrac{b}{2a}\right)^2 - \left(\dfrac{b}{2a}\right)^2

Step 4: Distribute a back across the terms and simplify the constant:

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

Worked Example

Factorisation and Root Determination

Solve the quadratic equation:

3x2+9x−12=03x^2 + 9x - 12 = 0

(a) Factorise the quadratic expression completely.

(b) Find the roots of the function.

(c) State the coordinates of the turning point.

Step 1: Factor out the highest common factor

3(x2+3x−4)=03(x^2 + 3x - 4) = 0

Step 2: Factorise the quadratic bracket

We need two numbers that multiply to give -4 and add to give +3. These are +4 and -1:

3(x+4)(x−1)=03(x + 4)(x - 1) = 0

Step 3: Solve for the roots

Set each factor to zero:

x+4=0⟹x=−4x + 4 = 0 \implies x = -4

x−1=0⟹x=1x - 1 = 0 \implies x = 1

The roots are x = -4 and x = 1.

Step 4: Find the turning point (vertex)

The x-coordinate of the vertex lies halfway between the roots:

h=−4+12=−32=−1.5h = \dfrac{-4 + 1}{2} = -\dfrac{3}{2} = -1.5

Substitute x = -1.5 into the original function:

f(−1.5)=3(−1.5)2+9(−1.5)−12=3(2.25)−13.5−12=6.75−25.5=−18.75f(-1.5) = 3(-1.5)^2 + 9(-1.5) - 12 = 3(2.25) - 13.5 - 12 = 6.75 - 25.5 = -18.75

The turning point is a minimum at (-1.5, -18.75).

Practice Questions & Solutions

1

A toy rocket is launched from a launchpad 2m above the ground with an initial velocity of 24m/s. Its height h in meters after t seconds is given by:

(a) Find the time taken to reach its maximum height.

(b) Calculate the maximum height reached by the rocket.

(c) Find the total flight time until the rocket impacts the ground, giving your answer to two decimal places.

Solution

(a)

The rocket's height equation is

.

The time to reach maximum height occurs at the vertex:

(b)

Evaluate h(2.4):

The maximum height reached is 30.8m

(c)

Set h(t) = 0 to find the impact time:

Apply the quadratic formula with , , :

Since :

(Rejecting the negative root ).

Total flight time is 4.88 seconds

2

A farmer wishes to construct a rectangular paddock using an existing straight riverbank as one boundary. The farmer has 80m of fencing for the remaining three sides.

(a) If x is the width of the paddock perpendicular to the riverbank, show that the enclosed area is

.

(b) By completing the square, express A(x) in vertex form.

(c) State the dimensions of the paddock that maximize the enclosed area, and calculate this maximum area.

Solution

(a)

Let the two sides perpendicular to the river have width , and the side parallel to the river have length y.

The fencing constraint is:

The enclosed area is:

(b)

Complete the square on

:

Halve -40 to get -20:

Substitute back:

(c)

From the vertex form

:

Optimal width:


Optimal length:


Maximum area:

3

A travel agency organizes coach tours. A tour requires a minimum of  20 passengers, charging  £60 per ticket. For every additional passenger beyond 20, the agency discounts the ticket price for all passengers by £1.50.

(a) If x represents the number of additional passengers beyond 20, show that total revenue is given by

.

(b) Determine the number of passengers that maximizes total tour revenue.

(c) Calculate the ticket price and maximum revenue at this group size.

Solution

(a)

Number of passengers:


Price per ticket:

Revenue

:


(b)

The revenue function is a downward-opening parabola. The maximum occurs at:

Total passengers

.

(c)

Ticket price:


Maximum revenue:

Alternatively:

4

The hypotenuse of a right-angled triangle is (2x + 1) cm. The other two sides have lengths of x cm and (2x - 1) cm.

(a) Use Pythagoras' Theorem to show that:

.

(b) Find the non-zero value of and state the lengths of all three sides.

(c) Determine the exact area of the triangle.

Solution

(a)

By Pythagoras' Theorem, the sum of the squares of the two shorter sides equals the square of the hypotenuse:

Expand both sides:

Subtract from both sides:

(b)

Factorise:

Since length must be non-zero,

.

Substitute x = 8 into each side:

Base: x = 8 cm

Perpendicular Height:

Hypotenuse:

(This forms the classic Pythagorean triple 8 - 15 - 17).


(c)

The area of a right-angled triangle is:

5

The length of a rectangular picture frame is 3 cm longer than its width. The area enclosed by the frame is .

(a) If x represents the width of the frame in centimetres, form an equation in terms of x and show that it simplifies to:

(b) Solve this equation by factorisation to find the width and length of the frame.

 

Solution

Let the width of the frame be .

The length is longer than the width:

The formula for the area of a rectangle is:

Substitute the given area of :

Expand the brackets:

Subtract from both sides to form a standard quadratic equation:

(b) Solve by Factorisation
Find two numbers that multiply to give and add to give .

These numbers are and :

Set each factor equal to zero:

Since a physical length cannot be negative, discard .

Width:
Length:

Summarise with AI:

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Gianpiero Placidi

UK-based Chemistry graduate with a passion for education, providing clear explanations and thoughtful guidance to inspire student success.