Cubic and quartic graphs and models | Year 11

01 / LEARN

Understand the idea.

Degree, sign and multiplicity

The leading term determines a polynomial’s end behaviour. A positive cubic falls to the left and rises to the right; a positive quartic rises at both ends. Negative leading coefficients reverse those directions, while lower-degree terms determine the middle of the graph.

At a root of odd multiplicity the sign changes, so the curve crosses the horizontal axis. At a root of even multiplicity it touches without crossing. Combine roots, multiplicities, the vertical-axis intercept and end behaviour before refining the sketch.

The graph of y equals the product of (x plus two) and (x minus one) squared, from x minus three to three and y minus five to ten. Gold points retain the original labels minus two, zero and one, zero.

Use structure and context

An equation containing only even powers can sometimes be reduced by $u = x^{2}$ . Solve the resulting equation in $u$ , keeping $u \geq 0$ for real x. Each positive $u$ gives two real x-values, zero gives one, and negative $u$ gives none.

A physical polynomial model may use only a small part of its full graph. For a box formed by cutting a sheet, positive cut sizes and remaining side lengths determine the feasible interval. Establish that domain before interpreting roots, negative outputs or turning points.

Graph of y=x fourth power-5x squared+4 on x=-2.5 to 2.5 and y=-4 to 8. The four simple roots at -2, -1, 1 and 2, y-intercept (0,4), and minima at plus or minus square root of five halves with height negative nine quarters are marked.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A repeated cubic root

Describe $y = (x + 2) {(x - 1)}^{2}$ .

Worked solution

1
Find the roots and their multiplicities.
$y = (x + 2) {(x - 1)}^{2}, x = - 2 (simple), x = 1 (double)$
The factor x plus two occurs once, while x minus one occurs twice.
2
Describe the behaviour at the intercepts.
$(- 2, 0) is a crossing, (1, 0) is a touch$
Odd multiplicity changes the sign across a root; even multiplicity keeps the same sign on both sides.
3
Calculate the vertical-axis intercept.
$y (0) = (0 + 2) {(0 - 1)}^{2} = 2$
The curve therefore passes through the point with coordinates zero and two.
4
Determine the direction of both ends.
$y = x^{3} - 3 x + 2$
The positive cubic leading term makes the left end fall and the right end rise; these directions supplement the intercept information.

Answer

Crosses (-2,0), touches (1,0), passes (0,2).

EXAMPLE 02

Quadratic in x squared

Solve $x^{4} - 5 x^{2} + 4 = 0$ .

Worked solution

1
Reduce the equation using a non-negative substitution.
$u = x^{2} \geq 0, u^{2} - 5 u + 4 = 0$
The fourth power is the square of the second power, so the new equation is quadratic.
2
Factor and solve for the new variable.
$\begin{aligned} (u - 1) (u - 4) & = 0 \\ u & = 1 or u = 4 \end{aligned}$
The constants negative one and negative four have sum negative five and product four.
3
Return to x using both square-root branches.
$\begin{aligned} x^{2} = 1 & ⟹ x = \pm 1 \\ x^{2} = 4 & ⟹ x = \pm 2 \end{aligned}$
Both u-values are positive, so each supplies a positive and a negative real root.
4
Confirm all roots in a real factorisation.
$x^{4} - 5 x^{2} + 4 = (x - 1) (x + 1) (x - 2) (x + 2)$
These four factors give exactly the four roots negative two, negative one, one and two.

Answer

$x = - 2, - 1, 1, 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

Does $y = {(x - 3)}^{2} {(x + 1)}^{2}$ cross at x=3?

Show answer & reasoning

No; it touches.

The factor ${(x - 3)}^{2}$ has even multiplicity, and the other factor is nonzero at $x = 3$ . The sign does not change there, so the curve touches the horizontal axis rather than crossing it.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Describe crossing/touching roots of $y = (x + 1) {(x - 2)}^{2}$ .
2 marks
State end behaviour of $y = - 2 x^{4} + 3 x - 1$ .
2 marks
Solve $x^{4} - 10 x^{2} + 9 = 0$ .
3 marks
Solve $x^{4} + x^{2} - 6 = 0$ for real $x$ .
3 marks
Cut squares of side $x$ cm from the corners of a 12 cm by 8 cm sheet and fold a box. Give volume and domain.
3 marks
A box has volume $V = x (12 - 2 x) (8 - 2 x)$ cubic centimetres. Find its volume when $x = 1$ and when $x = 2$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Describe crossing/touching roots of $y = (x + 1) {(x - 2)}^{2}$ .
2 marks
Use multiplicities one and two.
Crosses at -1; touches at 2.
State end behaviour of $y = - 2 x^{4} + 3 x - 1$ .
2 marks
The leading degree is even and coefficient negative.
Both ends fall.
Solve $x^{4} - 10 x^{2} + 9 = 0$ .
3 marks
Let $u = x^{2}$ : $(u - 1) (u - 9) = 0$ .
Use both roots of each positive $u$ .
$x = - 3, - 1, 1, 3$
Solve $x^{4} + x^{2} - 6 = 0$ for real $x$ .
3 marks
The u-equation is $(u + 3) (u - 2) = 0$ .
Reject $u = - 3$ for real $x$ .
$x = \pm \sqrt{2}$
Cut squares of side $x$ cm from the corners of a 12 cm by 8 cm sheet and fold a box. Give volume and domain.
3 marks
The three dimensions must be positive.
Multiply height, length and width.
$V = x (12 - 2 x) (8 - 2 x)$ cubic centimetres; $0 < x < 4$ .
A box has volume $V = x (12 - 2 x) (8 - 2 x)$ cubic centimetres. Find its volume when $x = 1$ and when $x = 2$ .
2 marks
At 1: $1 (10) (6) = 60$ .
At 2: $2 (8) (4) = 64$ .
60 cubic centimetres and 64 cubic centimetres.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

In this lesson

Understand the idea.

Degree, sign and multiplicity

Use structure and context

See the method in action.

A repeated cubic root

Find the roots and their multiplicities.

Describe the behaviour at the intercepts.

Calculate the vertical-axis intercept.

Determine the direction of both ends.

Quadratic in x squared

Reduce the equation using a non-negative substitution.

Factor and solve for the new variable.

Return to x using both square-root branches.

Confirm all roots in a real factorisation.

Try it yourself.

Check your understanding

Keep practising