Sketching Graphs Revision Notes for Grade 12 NSC Matric Mathematics

Sketching Graphs

Understanding cubic functions

Cubic functions are polynomial functions of degree 3, written in the standard form:

$y = ax^3 + bx^2 + cx + d$

The coefficient 'a' determines the basic shape and orientation of the cubic graph. This is crucial for sketching.

infoNote

Understanding the effect of the leading coefficient is your first step in any cubic graph sketch. This fundamental concept guides the overall orientation of your curve.

When $a > 0$ : The graph rises from left to right with an S-shaped curve
When $a < 0$ : The graph falls from left to right with an inverted S-shaped curve

Understanding this fundamental shape helps you begin any cubic graph sketch correctly.

Finding intercepts

Intercepts are the points where the graph crosses the coordinate axes. These provide essential reference points for your sketch.

y-intercept

The y-intercept occurs where the graph crosses the y-axis (when $x = 0$ ).

Method: Substitute $x = 0$ into the function equation.

x-intercepts

The x-intercepts occur where the graph crosses the x-axis (when $y = 0$ ).

Method: Set the function equal to zero and solve for x: $f(x) = 0$

infoNote

Different polynomial functions have different numbers of intercepts:

Linear functions: exactly 1 x-intercept
Quadratic functions: 0, 1, or 2 x-intercepts
Cubic functions: 1, 2, or 3 x-intercepts

lightbulbExample

Worked Example: Finding intercepts

Question: Given $f(x) = -x^3 + 4x^2 + x - 4$ , find the x- and y-intercepts.

Solution:

Step 1: Find the y-intercept

Substitute $x = 0$ :

$y = -(0)^3 + 4(0)^2 + (0) - 4 = -4$

The y-intercept is $(0; -4)$ .

Step 2: Find the x-intercepts using factorisation

Set $f(x) = 0$ : $-x^3 + 4x^2 + x - 4 = 0$

Test $x = 1$ : $f(1) = -(1)^3 + 4(1)^2 + (1) - 4 = 0$

Therefore $(x - 1)$ is a factor.

Factor further: $f(x) = -(x - 1)(x^2 - 3x - 4) = -(x - 1)(x + 1)(x - 4)$

Setting $f(x) = 0$ : $x = -1$ , $x = 1$ , or $x = 4$

The x-intercepts are $(-1; 0)$ , $(1; 0)$ , and $(4; 0)$ .

Stationary points

Stationary points (also called turning points) are points where the graph changes from increasing to decreasing, or vice versa. At these points, the gradient equals zero.

chatImportant

At stationary points, the first derivative equals zero: $f'(x) = 0$ . This is the fundamental condition for finding all turning points on a curve.

Types of stationary points:

Local maximum: The highest point in a local region (like a hill)
Local minimum: The lowest point in a local region (like a valley)

Finding stationary points

Method:

Find the first derivative: $f'(x)$
Set $f'(x) = 0$ and solve for x-coordinates
Substitute x-values back into original function to find y-coordinates

lightbulbExample

Worked Example: Finding stationary points

Question: Calculate the stationary points of $p(x) = x^3 - 6x^2 + 9x - 4$ .

Solution:

Step 1: Find the derivative

$p'(x) = 3x^2 - 12x + 9$

Step 2: Set $p'(x) = 0$ and solve

$3x^2 - 12x + 9 = 0$
$x^2 - 4x + 3 = 0$
$(x - 3)(x - 1) = 0$

Therefore $x = 1$ or $x = 3$

Step 3: Find y-coordinates

$p(1) = (1)^3 - 6(1)^2 + 9(1) - 4 = 1 - 6 + 9 - 4 = 0$
$p(3) = (3)^3 - 6(3)^2 + 9(3) - 4 = 27 - 54 + 27 - 4 = -4$

The stationary points are $(1; 0)$ and $(3; -4)$ .

Concavity and points of inflexion

Concavity describes how the curve bends. Understanding concavity helps you draw smooth, accurate curves.

Types of concavity:

Concave up: Curve bends upward (like a smile ∪)
Concave down: Curve bends downward (like a frown ∩)

infoNote

Memory tip: Think "smile = concave up, frown = concave down" to remember the direction of bending.

Using the second derivative:

When $f''(x) > 0$ : The graph is concave up
When $f''(x) < 0$ : The graph is concave down
When $f''(x) = 0$ : Possible point of inflexion

Points of inflexion

A point of inflexion is where the concavity changes from concave up to concave down (or vice versa).

Method to find points of inflection:

Find the second derivative: $f''(x)$
Set $f''(x) = 0$ and solve for x
Check that concavity actually changes at this point

General method for sketching cubic graphs

Follow this systematic 6-step approach for sketching any cubic function:

chatImportant

The Systematic 6-Step Method:

Consider the sign of 'a' and determine the general shape
Find the y-intercept by setting $x = 0$
Find the x-intercepts by factoring and solving $f(x) = 0$
Find stationary points by setting $f'(x) = 0$
Find y-coordinates of stationary points by substitution
Plot points and draw a smooth curve

lightbulbExample

Worked Example: Complete sketching process

Question: Sketch the graph of $g(x) = x^3 - 3x^2 - 4x$ .

Solution:

Step 1: Determine the shape

Coefficient of $x^3$ is $+1 > 0$ , so the graph rises from left to right.

Step 2: Find intercepts

y-intercept: $g(0) = 0$ , giving point $(0; 0)$

x-intercepts: $x^3 - 3x^2 - 4x = 0$

$x(x^2 - 3x - 4) = 0$
$x(x - 4)(x + 1) = 0$
$x = -1$ , $x = 0$ , or $x = 4$

Points: $(-1; 0)$ , $(0; 0)$ , $(4; 0)$

Step 3: Find stationary points\

$g'(x) = 3x^2 - 6x - 4$

Set $g'(x) = 0$ : $3x^2 - 6x - 4 = 0$

Using the quadratic formula:

$x = \frac{6 \pm \sqrt{36 + 48}}{6} = \frac{6 \pm \sqrt{84}}{6} = \frac{6 \pm 2\sqrt{21}}{6}$
$x \approx 2.53$ or $x \approx -0.53$

Step 4: Find y-coordinates

$g(2.53) \approx -13.13$ and $g(-0.53) \approx 1.13$

Stationary points: $(2.53; -13.13)$ and $(-0.53; 1.13)$

Step 5: Draw the sketch Plot all points and connect with a smooth S-shaped curve.

Interpreting derivative graphs

Sometimes you're given the graph of $f'(x)$ and asked to interpret information about $f(x)$ .

infoNote

Key relationships between a function and its derivative:

Where $f'(x) > 0$ : $f(x)$ is increasing
Where $f'(x) \< 0$ : $f(x)$ is decreasing
Where $f'(x) = 0$ : $f(x)$ has stationary points

lightbulbExample

Worked Example: Interpreting graphs

Question: Consider the graph of the derivative $g'(x)$ . For which values of x is $g(x)$ decreasing?

Solution: $g(x)$ is decreasing when $g'(x) < 0$ . From the graph, this occurs when $x \in (-2; 1)$ .

Understanding the relationship between functions and derivatives

The relationship between a function and its derivatives provides crucial information for sketching graphs.

infoNote

Key connections:

Turning points of the original function correspond to x-intercepts of the first derivative
Points of inflexion of the original function correspond to turning points of the first derivative
The first derivative shows where the function is increasing or decreasing
The second derivative shows the concavity of the function

Common exam tips

bookmarkSummary

Key Points to Remember:

Always start with the sign of the coefficient of $x^3$
Check your intercepts by substitution
Use factorisation systematically for x-intercepts
Show all working for stationary point calculations
Draw smooth curves - cubic graphs don't have sharp corners
Label key points clearly on your sketch

Summary

bookmarkSummary

Essential Concepts for Sketching Cubic Graphs:

Cubic functions have the form $f(x) = ax^3 + bx^2 + cx + d$ , where the sign of 'a' determines the basic orientation
Intercepts are found by setting $x = 0$ (y-intercept) or $y = 0$ (x-intercepts), providing essential reference points
Stationary points occur where $f'(x) = 0$ and represent local maxima or minima where the graph changes direction
Concavity is determined by the second derivative $f''(x)$ , with points of inflexion occurring where $f''(x) = 0$
The systematic 6-step method ensures you include all essential elements: shape, intercepts, stationary points, and smooth curve drawing

Sketching Graphs (Grade 12 NSC Matric Mathematics): Revision Notes