Graphing real polynomials | Year 12

01 / LEARN

The idea, step by step

Build a sketch from information you can justify

The real roots locate horizontal-axis intercepts. The value $P (0)$ locates the vertical-axis intercept. Root multiplicity tells you whether the graph crosses the axis or touches and turns. These facts are stronger than an unsupported drawing from a few plotted points.

The leading term controls the far ends. A positive even-degree leading term sends both ends upwards; a negative even-degree term sends both downwards. A positive odd-degree leading term goes down to the left and up to the right; a negative odd-degree term reverses those directions.

Add enough detail without inventing features

Use a sign table across the real roots to check which intervals lie above or below the axis. You may plot extra points or use graphing technology to refine the sketch. If precise stationary points are requested, differentiate using the Methods prerequisite, solve $P^{'} (x) = 0$ , and substitute into $P$ .

A sketch is a reasoned representation, not a scale drawing unless the question asks for scale. Do not infer a missing root solely because the curve seems near the axis. Repeated roots, steep sections and a small viewing window can all mislead a graphing display.

A common trap

A root with multiplicity two crosses the axis twice.

Instead: A double root is one distinct intercept, and the sign stays the same across it.

02 / WORKED EXAMPLES

See the thinking

Example 1

A cubic with a double root

Sketch $P (x) = {(x + 1)}^{2} (x - 2)$ , showing roots, the y-intercept and end behaviour.

The roots are $- 1$ (double, touches) and $2$ (simple, crosses).
$P (0) = - 2$ , so the y-intercept is $(0, - 2)$ .
The leading term is $x^{3}$ : left end down, right end up. The sign is negative for $x < 2$ except at the zero $- 1$ , and positive for $x > 2$ .

Touch at

(- 1, 0)

; cross at

(2, 0)

; pass through

(0, - 2)

Example 2

Stationary points when required

Find the stationary points of $P (x) = x^{3} - 3 x$ and identify their type.

$P^{'} (x) = 3 x^{2} - 3 = 0$ gives $x = \pm 1$ .
$P (- 1) = 2$ and $P (1) = - 2$ .
$P^{'^{'}} (x) = 6 x$ is negative at $- 1$ and positive at $1$ , so the points are a local maximum and a local minimum respectively.

Local maximum

(- 1, 2)

; local minimum

(1, - 2)

Check your understanding

What happens at each end of $y = - 3 x^{4} + 2 x - 1$ ?

Show answer

Both ends go downwards.

The negative leading coefficient and even degree determine the end behaviour.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $P (x) = x {(x - 2)}^{2}$ , state roots, multiplicities and end behaviour.
2 marks
Find the y-intercept and end behaviour of $P (x) = - 2 (x + 1) (x - 3)$ .
2 marks
State the sign of ${(x + 1)}^{2} (x - 2)$ on the intervals separated by $- 1$ and $2$ .
2 marks
Write a monic quartic that touches the axis at $- 2$ and $1$ and has no other roots. Find its y-intercept.
2 marks
Find the stationary points of $P (x) = x^{3} - 6 x^{2} + 9 x$ .
4 marks
Explain why a positive-leading cubic must have at least one real zero.
3 marks

Read the worked solutions

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

For $P (x) = x {(x - 2)}^{2}$ , state roots, multiplicities and end behaviour.
2 marks
The leading term is $x^{3}$ and the factor powers give multiplicities.
$0$ simple/crosses; $2$ double/touches; left down, right up.
Find the y-intercept and end behaviour of $P (x) = - 2 (x + 1) (x - 3)$ .
2 marks
$P (0) = - 2 (1) (- 3) = 6$ ; leading term $- 2 x^{2}$ .
Y-intercept $(0, 6)$ ; both ends down.
State the sign of ${(x + 1)}^{2} (x - 2)$ on the intervals separated by $- 1$ and $2$ .
2 marks
The square is positive except at $- 1$ , so the other factor determines the sign.
Negative on $(- \infty, - 1)$ and $(- 1, 2)$ ; positive on $(2, \infty)$ .
Write a monic quartic that touches the axis at $- 2$ and $1$ and has no other roots. Find its y-intercept.
2 marks
Each touching root requires an even multiplicity; two double roots use degree four.
$P (x) = {(x + 2)}^{2} {(x - 1)}^{2}$ ; y-intercept $(0, 4)$ .
Find the stationary points of $P (x) = x^{3} - 6 x^{2} + 9 x$ .
4 marks
$P^{'} = 3 (x - 1) (x - 3)$ , so candidates are $1, 3$ . $P^{'^{'}} = 6 x - 12$ determines the types.
$(1, 4)$ local maximum and $(3, 0)$ local minimum.
Explain why a positive-leading cubic must have at least one real zero.
3 marks
Polynomial graphs are continuous, so a change from negative to positive cannot occur without taking the value zero.
Its far-left values are negative and far-right values positive; continuity forces a crossing.

Checking saved progress…

Continue learning

Solving real polynomial equations

Next lesson

Curriculum reference & lesson scope

SACE Board of South Australia

A recognisable real-polynomial chapter supporting Stage 2 Specialist polynomial factorisation and graph analysis; original lessons use current SACE scope and the school chapter structure requested by the user.

Official curriculum source ↗

Back to Real Polynomials