First-derivative graph analysis | Year 11

01 / LEARN

The idea, step by step

Build the sketch from features

Start with the domain and intercepts. Factor f to solve f=0, then differentiate to solve f′=0. These are different equations: intercepts need not be stationary, and stationary points need not be on an axis.

Plot the known points, then connect them using the sign chart and end behaviour from the leading term. A sketch records the qualitative shape and exact key points; it is not necessarily drawn to scale.

Check what is determined

A derivative fixes gradient behaviour but not a vertical position. Adding a constant shifts all heights without changing any derivative. A known point may supply the missing vertical information.

Technology can check the resulting shape, but the exact intercepts and turns should be supported by calculations. Compare any apparent extra turn with the derivative equation rather than trusting a coarse screen window.

A common trap

Treating the roots of the derivative as graph intercepts.

Instead: Solve f=0 for axis intercepts and f′=0 for stationary inputs.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Analyse $f (x) = x^{3} - 3 x^{2}$ .

$f = x^{2} (x - 3)$ gives intercepts (0,0) and (3,0).
$f^{'} = 3 x (x - 2)$ gives stationary points (0,0) and (2,−4).
Signs +, −, + show a maximum then minimum. The graph falls left and rises right.

A cubic touching the axis at its local maximum (0,0), dipping to (2,−4), then crossing at (3,0).

Example 2

Worked example 2

Analyse $g (x) = - x^{3} + 3 x$ .

Roots are $0, \pm \sqrt{3}$ .
$g^{'} = 3 - 3 x^{2}$ is zero at −1 and 1.
Signs −, +, − show a minimum (−1,−2) and maximum (1,2).

A negative-leading cubic increasing on (−1,1), decreasing outside.

Check your understanding

f and g have the same derivative and g(0)=f(0)+4. How are their graphs related?

Show answer

g is shifted vertically upwards by 4.

Adding a constant changes heights but not slopes.

03 / WORKSHEET

Your turn

6 questions · 24 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

List intercepts, stationary point and monotonic intervals of $f (x) = x^{2} - 4 x + 3$ .
5 marks
Analyse $f (x) = x^{3} - 3 x^{2}$ .
5 marks
A differentiable function has derivative zero at x=−2 and x=1. Its derivative is positive for x<−2, negative for −2<x<1, and positive for x>1. State the stationary types.
3 marks
Find roots and stationary points of $g (x) = x^{3} - 3 x$ .
5 marks
A straight-line function has $f' (x) = 2$ and $f (0) = - 3$ . Find f.
3 marks
Why can the derivative alone not determine vertical position?
3 marks

Read the worked solutions

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

List intercepts, stationary point and monotonic intervals of $f (x) = x^{2} - 4 x + 3$ .
5 marks
$f = (x - 1) (x - 3)$ ; $f' = 2 x - 4$ .
Evaluate f at 0 and 2 and use derivative signs.
Roots 1,3; y-intercept 3; minimum (2,−1); decreasing x<2, increasing x>2.
Analyse $f (x) = x^{3} - 3 x^{2}$ .
5 marks
$f = x^{2} (x - 3)$ and $f' = 3 x (x - 2)$ .
The derivative signs are +, −, +.
Intercepts (0,0),(3,0); maximum (0,0); minimum (2,−4).
A differentiable function has derivative zero at x=−2 and x=1. Its derivative is positive for x<−2, negative for −2<x<1, and positive for x>1. State the stationary types.
3 marks
The signs change + to − at −2 and − to + at 1.
No heights can be found from this information alone.
Maximum at x=−2; minimum at x=1.
Find roots and stationary points of $g (x) = x^{3} - 3 x$ .
5 marks
$g = x (x^{2} - 3)$ and $g' = 3 (x - 1) (x + 1)$ .
Use signs +, −, + and evaluate g at ±1.
Roots 0,±√3; maximum (−1,2), minimum (1,−2).
A straight-line function has $f' (x) = 2$ and $f (0) = - 3$ . Find f.
3 marks
A line of gradient 2 has form 2x+c.
The intercept condition makes c=−3.
$f (x) = 2 x - 3$
Why can the derivative alone not determine vertical position?
3 marks
For example x² and x²+5 both have derivative 2x but different heights.
Adding a constant leaves the derivative unchanged.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

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