Quadratic Functions, Graphs and the Discriminant

Master quadratic equations, understand parabolas, and use the discriminant to determine the nature of roots

SPEC 2.3 • ALGEBRA AND FUNCTIONS

Quadratic Functions

Definition

A quadratic function has the general form:

y = a x^{2} + b x + c

where

a \neq = 0

The graph of a quadratic function is called a parabola.

Key Features

Vertex (turning point): The highest or lowest point

Axis of symmetry: Vertical line through the vertex

y-intercept: Where the graph crosses the y-axis (at

x = 0

, so

y = c

)

x-intercepts (roots): Where the graph crosses the x-axis (where

y = 0

)

Shape of the Parabola

The sign of the coefficient

a

determines the shape:

a

> 0: U-shape

a

< 0: ∩-shape

Maximum turning point

Remember

The larger the value of

∣ a ∣

, the narrower the parabola. The smaller

∣ a ∣

, the wider it opens.

Key Features of a Parabola

Formulas

Axis of symmetry:

x = - \frac{b}{2 a}

y-intercept:

(0, c)

x-intercepts: Solve

a x^{2} + b x + c = 0

Finding Roots by Factorising

When a quadratic can be factorised, we can find the roots (x-intercepts) easily:

Method

a x^{2} + b x + c = 0

Factorise into

(p x + q) (r x + s) = 0

Then either

p x + q = 0

or

r x + s = 0

Steps

Set the quadratic equal to zero

Factorise the expression

Set each factor equal to zero

Solve each linear equation

Important

Factorising only works when the roots are rational numbers. For irrational roots, other methods are needed (completing the square or the formula).

Worked Example01

Find the roots of $x^{2} - 5 x + 6 = 0$ by factorising.

1

Find two numbers that multiply to give +6 and add to give −5

2

The numbers are −2 and −3 (since (−2) × (−3) = 6 and (−2) + (−3) = −5)

3

Factorise:

x^{2} - 5 x + 6 = (x - 2) (x - 3) = 0

4

Set each factor to zero:

x - 2 = 0

or

x - 3 = 0

Answer: $x = 2$ or $x = 3$

Sketching Quadratic Graphs

5-Step Method

Determine shape: Check sign of

a

(positive = U, negative = ∩)

Find y-intercept: Set

x = 0

, giving point

(0, c)

Find x-intercepts: Solve

a x^{2} + b x + c = 0

(if roots exist)

Find vertex: Use

x = - \frac{b}{2 a}

, then substitute to find y

Sketch: Plot all key points and draw smooth parabola

Exam Tip

Always label: the vertex coordinates, axis of symmetry, y-intercept, and x-intercepts (roots) if they exist.

Worked Example02

Sketch the graph of $y = x^{2} - 4 x - 5$ , showing all key features.

1

Shape:

a = 1 > 0

, so U-shaped parabola (minimum)

2

y-intercept: When

x = 0

:

y = - 5

, so

(0, - 5)

3

x-intercepts:

x^{2} - 4 x - 5 = (x - 5) (x + 1) = 0

So

x = 5

or

x = - 1

, giving

(- 1, 0)

and

(5, 0)

4

Vertex:

x = - \frac{- 4}{2 ( 1 )} = 2

When

x = 2

:

y = 4 - 8 - 5 = - 9

Vertex:

(2, - 9)

The Discriminant

Definition

The discriminant of a quadratic

a x^{2} + b x + c

is:

det = b^{2} - 4 a c

The discriminant tells us about the nature of the roots without solving the equation:

DiscriminantNature of RootsGraph Behaviour

b^{2} - 4 a c > 0

Two distinct real rootsCrosses x-axis twice

b^{2} - 4 a c = 0

One repeated real rootTouches x-axis once

b^{2} - 4 a c < 0

No real rootsDoesn't cross x-axis

Visualising the Discriminant

When

a

> 0 (U-shaped parabola):

det > 0

det = 0

det < 0

No real roots

When

a

< 0 (∩-shaped parabola):

det > 0

det = 0

det < 0

No real roots

Key Insight

The discriminant determines the number of roots regardless of the sign of

a

.

Worked Example03

Determine the nature of roots of $3 x^{2} - 2 x + 1 = 0$ using the discriminant.

1

Identify:

a = 3

,

b = - 2

,

c = 1

2

Calculate discriminant:

det = b^{2} - 4 a c

3

det = (- 2)^{2} - 4 (3) (1) = 4 - 12 = - 8

4

Since

det = - 8 < 0

, the equation has no real roots.

Answer: No real roots (the parabola doesn't cross the x-axis)

Discriminant with Parameters

When a quadratic contains unknown constants, we use the discriminant to find conditions on those constants.

Problem Types

"Equal roots" → Set

b^{2} - 4 a c = 0

and solve

"Two distinct real roots" → Set

b^{2} - 4 a c > 0

and solve inequality

"No real roots" → Set

b^{2} - 4 a c < 0

and solve inequality

"Real roots" → Set

b^{2} - 4 a c \geq 0

Common Exam Questions

Finding values of

k

,

p

,

m

etc. for which an equation has a specific number of roots is a very common A-Level question type.

Worked Example04

Find the values of $k$ for which $x^{2} + k x + 9 = 0$ has equal roots.

1

For equal roots, we need

det = 0

2

Identify:

a = 1

,

b = k

,

c = 9

3

Set up equation:

b^{2} - 4 a c = 0

k^{2} - 4 (1) (9) = 0

4

Solve:

k^{2} - 36 = 0

k^{2} = 36

k = \pm 6

Answer: $k = 6$ or $k = - 6$

Worked Example05

Find the range of values of $p$ for which $2 x^{2} - 3 x + p = 0$ has two distinct real roots.

1

For two distinct real roots, we need

det > 0

2

Identify:

a = 2

,

b = - 3

,

c = p

3

Set up inequality:

b^{2} - 4 a c > 0

(- 3)^{2} - 4 (2) (p) > 0

4

Solve:

9 - 8 p > 0

9 > 8 p

p < \frac{9}{8}

Answer: $p < \frac{9}{8}$

Worked Example06

The equation $x^{2} + (k - 2) x + (2 k + 1) = 0$ has no real roots. Find the range of values of $k$ .

1

For no real roots, we need

det < 0

2

Identify:

a = 1

,

b = (k - 2)

,

c = (2 k + 1)

3

Set up:

(k - 2)^{2} - 4 (1) (2 k + 1) < 0

4

Expand:

k^{2} - 4 k + 4 - 8 k - 4 < 0

k^{2} - 12 k < 0

k (k - 12) < 0

5

Critical values:

k = 0

and

k = 12

For product negative, we need

0 < k < 12

Answer: $0 < k < 12$

Practice

Question 1: Sketch

y = - x^{2} + 2 x + 3

showing roots, vertex, and y-intercept.

Step 1:

a = - 1 < 0

, so ∩-shaped (maximum)

Step 2: y-intercept:

(0, 3)

Step 3: Roots:

- x^{2} + 2 x + 3 = 0 \Rightarrow x^{2} - 2 x - 3 = 0 \Rightarrow (x - 3) (x + 1) = 0

Roots at

x = - 1

and

x = 3

Step 4: Vertex:

x = - \frac{2}{2 ( - 1 )} = 1

When

x = 1

:

y = - 1 + 2 + 3 = 4

Vertex:

(1, 4)

— this is a maximum

Practice

Question 2: Use the discriminant to find the values of

m

for which

m x^{2} - 4 x + m = 0

has real roots.

Step 1: For real roots,

det \geq 0

Step 2:

a = m

,

b = - 4

,

c = m

Step 3:

(- 4)^{2} - 4 (m) (m) \geq 0

16 - 4 m^{2} \geq 0

4 (4 - m^{2}) \geq 0

4 - m^{2} \geq 0

Step 4:

m^{2} \leq 4

- 2 \leq m \leq 2

Step 5: But

m \neq = 0

(otherwise not quadratic)

Answer: $- 2 \leq m \leq 2, m \neq = 0$

Practice

Question 3: Given

f (x) = 2 x^{2} + p x - 6

has roots at

x = - 3

and

x = q

, find

p

and

q

.

Method: Using factor form

If

x = - 3

is a root, then

(x + 3)

is a factor.

So

2 x^{2} + p x - 6 = 2 (x + 3) (x - q^{'})

for some

q^{'}

Expanding:

2 (x^{2} + 3 x - q^{'} x - 3 q^{'}) = 2 x^{2} + (6 - 2 q^{'}) x - 6 q^{'}

Comparing constant terms:

- 6 = - 6 q^{'}

, so

q^{'} = 1

So the other root is

q = 1

Comparing

x

terms:

p = 6 - 2 (1) = 4

Check:

2 x^{2} + 4 x - 6 = 2 (x^{2} + 2 x - 3) = 2 (x + 3) (x - 1)

✓

Answer: $p = 4$ and $q = 1$

Summary

Key Formulas

ConceptFormula

General Form

y = a x^{2} + b x + c

Discriminant

det = b^{2} - 4 a c

Axis of Symmetry

x = - \frac{b}{2 a}

y-intercept

(0, c)

Discriminant Summary

det > 0

→ Two distinct real roots

det = 0

→ One repeated real root

det < 0

→ No real roots

Exam Skills

Always check the sign of

a

to determine the shape

For parameter problems, set up the discriminant condition first

Remember

a \neq = 0

for a quadratic