Open Math›Year 11›Mathematics Advanced›Working with functions›Cubic and reciprocal functions

Cubic and reciprocal functions

Identify x-intercepts and end behaviour of cubic functions in factored form, sketch and analyse reciprocal functions of the form

f(x) = k/x

, and match equations to graphs by justifying key features.

Worked examples

Finding x-intercepts and end behaviour of a cubic

Straightforward

Problem

For

f(x) = (x + 2)(x - 1)(x - 4)

, state the x-intercepts and describe the end behaviour.

Find the x-intercepts by setting each factor equal to zero.

x + 2 = 0 \implies x = -2, \qquad x - 1 = 0 \implies x = 1, \qquad x - 4 = 0 \implies x = 4

Identify the leading term by multiplying the leading term from each factor.

x \cdot x \cdot x = x^3 \quad (\text{positive leading coefficient})

State the end behaviour. A positive leading coefficient means the graph behaves like

y = x^3

: it falls to the left and rises to the right.

\text{As } x \to +\infty,\; f(x) \to +\infty \qquad \text{As } x \to -\infty,\; f(x) \to -\infty

Answer

x-intercepts at

x = -2

x = 1

and

x = 4

. As

x \to +\infty

f(x) \to +\infty

; as

x \to -\infty

f(x) \to -\infty

Graphing a reciprocal function

Moderate

Problem

For

f(x) = \dfrac{-3}{x}

, state the asymptotes, domain and range, and which quadrants the graph occupies.

Find the vertical asymptote. This occurs where the denominator equals zero.

x = 0 \quad (\text{vertical asymptote})

Find the horizontal asymptote by considering what happens to

f(x)

|x|

becomes very large.

\text{As } |x| \to \infty,\; \frac{-3}{x} \to 0 \quad (\text{horizontal asymptote } y = 0)

State the domain and range. Neither

x

nor

y

can equal zero.

\text{Domain: } x \in \mathbb{R},\; x \neq 0 \qquad \text{Range: } y \in \mathbb{R},\; y \neq 0

Determine which quadrants the graph occupies by checking the sign of

f(x)

for positive and negative

x

x > 0:\; f(x) = \frac{-3}{x} < 0 \implies \text{Quadrant IV} \qquad x < 0:\; f(x) = \frac{-3}{x} > 0 \implies \text{Quadrant II}

Answer

Vertical asymptote

x = 0

; horizontal asymptote

y = 0

. Domain:

x \neq 0

. Range:

y \neq 0

. The graph lies in Quadrants II and IV.

Finding a cubic equation from graph features

Challenging

Problem

A cubic graph has x-intercepts at

x = -2

x = 0

and

x = 3

, and as

x \to +\infty

y \to -\infty

. Find the equation of the cubic in fully factored form.

Write the corresponding factor for each x-intercept.

x = -2 \implies (x + 2), \qquad x = 0 \implies x, \qquad x = 3 \implies (x - 3)

Write the general factored form with an unknown leading coefficient

a

f(x) = a \cdot x(x + 2)(x - 3)

Determine the sign of

a

from the end behaviour. As

x \to +\infty

y \to -\infty

means the leading term is negative, so

a < 0

. Take

a = -1

\text{Leading term: } a \cdot x^3 < 0 \text{ for large positive } x \implies a = -1

Write the final equation and verify the end behaviour.

f(x) = -x(x + 2)(x - 3) \qquad \text{As } x \to +\infty,\; -x^3 \to -\infty \checkmark

Answer

f(x) = -x(x + 2)(x - 3)

Practise

Q1·Straightforward

The function

f(x) = (x - 2)(x + 3)(x - 5)

has three x-intercepts. What is the largest x-intercept?

Explanation

Setting each factor to zero:

x - 2 = 0 \implies x = 2

x + 3 = 0 \implies x = -3

x - 5 = 0 \implies x = 5

. The x-intercepts are

-3

2

and

5

. The largest is

5

Q2·Straightforward

For

f(x) = (x + 1)(x - 2)(x - 4)

, which statement correctly describes the end behaviour?

As $x \to +\infty$, $f(x) \to +\infty$ and as $x \to -\infty$, $f(x) \to -\infty$As $x \to +\infty$, $f(x) \to -\infty$ and as $x \to -\infty$, $f(x) \to +\infty$As $x \to \pm\infty$, $f(x) \to +\infty$As $x \to \pm\infty$, $f(x) \to -\infty$

Explanation

Expanding

(x+1)(x-2)(x-4)

, the leading term is

x \cdot x \cdot x = x^3

, which is positive. So the graph rises to the right: as

x \to +\infty

f(x) \to +\infty

. It falls to the left: as

x \to -\infty

f(x) \to -\infty

Q3·Straightforward

The function

f(x) = (x + 1)(x - 3)(x + 5)

has three x-intercepts. What is their sum?

Explanation

Setting each factor to zero:

x + 1 = 0 \implies x = -1

x - 3 = 0 \implies x = 3

x + 5 = 0 \implies x = -5

. Their sum is

-1 + 3 + (-5) = -3

Q4·Straightforward

x \to -\infty

, for which function does

f(x) \to +\infty

$f(x) = (x - 1)(x + 2)(x - 3)$$f(x) = -(x - 1)(x + 2)(x - 3)$$f(x) = x^2(x - 1)$$f(x) = (x + 3)^3$

Explanation

For

f(x) = -(x-1)(x+2)(x-3)

, the leading term is

-x^3

. As

x \to -\infty

-x^3 \to -(-\infty)^3 = +\infty

. The other three functions all have leading term

+x^3

, so they tend to

-\infty

x \to -\infty

Q5·Moderate

For

f(x) = \dfrac{3}{x}

, what are the equations of the asymptotes?

$x = 0$ and $y = 0$$x = 3$ and $y = 3$$x = 0$ and $y = 3$$x = 3$ and $y = 0$

Explanation

The denominator equals zero at

x = 0

, giving vertical asymptote

x = 0

. As

|x| \to \infty

3/x \to 0

, giving horizontal asymptote

y = 0

. The constant numerator

3

does not shift the asymptotes.

Q6·Moderate

For

f(x) = \dfrac{-2}{x}

, in which quadrants does the graph lie?

Quadrants I and IIIQuadrants II and IVQuadrants I and IIQuadrants III and IV

Explanation

When

x > 0

f(x) = -2/x < 0

, so

(x, f(x))

lies in Quadrant IV. When

x < 0

f(x) = -2/x > 0

(negative divided by negative), so

(x, f(x))

lies in Quadrant II. The graph lies entirely in Quadrants II and IV.

Q7·Moderate

What is the domain of

f(x) = \dfrac{5}{x}

All real numbersAll real numbers except $x = 0$$x > 0$$x \geq 5$

Explanation

The denominator

x

equals zero when

x = 0

, making

5/x

undefined there. For all other values,

5/x

is defined. The domain is all real numbers except

x = 0

, written

x \in \mathbb{R},\, x \neq 0

. The range is also all real numbers except

y = 0

Q8·Moderate

The graph of

f(x) = \dfrac{k}{x}

passes through the point

(2, -6)

. What is the value of

k

Explanation

Substituting

x = 2

and

f(x) = -6

-6 = \frac{k}{2} \implies k = -6 \times 2 = -12.

Q9·Challenging

A cubic function has x-intercepts at

x = -3

x = 0

and

x = 4

, and its graph rises from bottom-left to top-right. Which equation matches?

$f(x) = x(x + 3)(x - 4)$$f(x) = x(x - 3)(x + 4)$$f(x) = -x(x + 3)(x - 4)$$f(x) = x(x + 3)(x + 4)$

Explanation

Roots

x = -3

x = 0

and

x = 4

give factors

(x + 3)

x

and

(x - 4)

. Rising from bottom-left to top-right means positive leading coefficient, so no overall negative factor. This gives

f(x) = x(x + 3)(x - 4)

. Option B has roots

0

3

-4

; Option C has a negative leading coefficient; Option D has roots

0

-3

-4

Q10·Challenging

A reciprocal function has its graph in Quadrants II and IV and passes through

(-1, 5)

. Which equation fits?

$f(x) = \dfrac{5}{x}$$f(x) = \dfrac{-5}{x}$$f(x) = \dfrac{1}{5x}$$f(x) = \dfrac{-1}{5x}$

Explanation

Quadrants II and IV require

k < 0

, ruling out Option A (

k = 5 > 0

). Checking Option B:

\frac{-5}{-1} = 5 \checkmark.

For Option C:

\dfrac{1}{5(-1)} = -\dfrac{1}{5} \neq 5

. For Option D:

\dfrac{-1}{5(-1)} = \dfrac{1}{5} \neq 5

. The answer is

f(x) = \dfrac{-5}{x}

Q11·Challenging

The graph of

y = \dfrac{k}{x}

lies in Quadrants II and IV and passes through

(-3, 7)

. What is the value of

k

Explanation

Substituting

x = -3

and

y = 7

7 = \frac{k}{-3} \implies k = 7 \times (-3) = -21.

Since

k = -21 < 0

, the graph lies in Quadrants II and IV, consistent with the given information.

Q12·Challenging

Which cubic equation has a negative leading coefficient and x-intercepts at

x = -1

x = 2

and

x = 5

$f(x) = (x + 1)(x - 2)(x - 5)$$f(x) = -(x + 1)(x - 2)(x - 5)$$f(x) = -(x - 1)(x + 2)(x + 5)$$f(x) = (x - 1)(x + 2)(x + 5)$

Explanation

Roots

x = -1

2

and

5

give factors

(x + 1)

(x - 2)

and

(x - 5)

. A negative leading coefficient requires multiplying by

-1

, giving

f(x) = -(x + 1)(x - 2)(x - 5)

. Option A has a positive leading coefficient; Options C and D have roots

1

-2

-5

(wrong signs on intercepts).