Master the complete method for sketching cubic functions, from finding key points to understanding concavity and inflection points.
Graph sketching is one of the most important skills in Grade 12 calculus. Being able to accurately sketch cubic functions requires understanding how the function behaves, where it crosses the axes, where it turns, and how it curves.
In this lesson, we'll learn the complete systematic method for sketching cubic graphs. We'll explore:
By the end of this lesson, you'll be able to sketch any cubic function confidently and understand what the graph tells us about the function's behavior.
Why is this important?
Cubic functions appear everywhere in the real world - from modeling the motion of objects under changing forces, to describing economic trends, to understanding how structures bend and deform. Being able to visualize and sketch these functions helps you understand the mathematics behind real-world phenomena.
A cubic function is a polynomial function of degree 3. It has the general form:
where:
Understanding the degree
The degree of a polynomial is the highest power of $x$. For cubic functions, this is 3. The degree tells us several important things:
Try these presets:
Experiment!
Try setting $a = 0$. Notice that the function is no longer cubic! This is why we require $a \\neq 0$ for a function to be cubic.
The leading coefficient $a$ has a crucial effect on the overall shape and orientation of the cubic graph. Understanding this parameter helps you quickly sketch the general shape before finding specific points.
The magnitude of a
The absolute value of $a$ (written $|a|$) affects how "steep" or "stretched" the graph is:
Adjust the slider to see how the magnitude affects the steepness of both graphs
Key Observation:
Both graphs show the same function form $f(x) = ax^3 - 3ax$, but with opposite signs for $a$. Notice how:
Intercepts are the points where the graph crosses or touches the axes. They are essential reference points for sketching the graph accurately.
The y-intercept occurs where $x = 0$. To find it:
Method:
Substitute $x = 0$ into $f(x)$
Example:
For $f(x) = 2x^3 - 5x^2 + 3x - 7$:
$$\begin{align*} f(0) &= 2(0)^3 - 5(0)^2 + 3(0) - 7 \\ &= -7 \end{align*}$$So the y-intercept is $(0, -7)$
The y-intercept is always just the constant term $d$ in $f(x) = ax^3 + bx^2 + cx + d$
The x-intercepts occur where $f(x) = 0$. To find them:
Method:
Solve $f(x) = 0$ using factoring techniques
Common techniques:
A cubic function can have 1, 2, or 3 x-intercepts. If you can't find factors easily, one or more roots may be irrational or complex.
Factor Theorem:
If $f(k) = 0$, then $(x - k)$ is a factor of $f(x)$
How to use it:
Worked Example: Finding x-intercepts
Find the x-intercepts of $f(x) = x^3 - 6x^2 + 11x - 6$
Step 1: Use factor theorem
Try small values. Let's try x = 1:
$$\begin{align*} f(1) &= (1)^3 - 6(1)^2 + 11(1) - 6 \\ &= 1 - 6 + 11 - 6 \\ &= 0 \end{align*}$$Since $f(1) = 0$, we know $(x - 1)$ is a factor.
Step 2: Find other factor
Divide by $(x - 1)$ or use inspection:
$f(x) = (x - 1)(x^2 - 5x + 6)$Step 3: Factor the quadratic
$$\begin{align*} f(x) &= (x - 1)(x^2 - 5x + 6) \\ &= (x - 1)(x - 2)(x - 3) \end{align*}$$Step 4: Find x-intercepts
Set each factor to zero: $x = 1$, $x = 2$, and $x = 3$
X-intercepts: $(1, 0)$, $(2, 0)$, $(3, 0)$
Rational Root Theorem
For $f(x) = ax^3 + bx^2 + cx + d$, any rational root $\frac{p}{q}$ must have:
This helps narrow down which values to test! For example, in $f(x) = 2x^3 - x^2 - 13x - 6$, possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$
Stationary points (also called turning points or critical points) are points where the gradient of the function is zero. These are where the graph "turns" - either reaching a local maximum or minimum.
Differentiate the cubic function using the power rule
Stationary points occur where the gradient is zero
This gives you a quadratic equation (since the derivative of a cubic is quadratic)
Substitute each x-value back into the original function $f(x)$
Use the second derivative test or the first derivative test
Worked Example:
Find and classify the stationary points of $f(x) = x^3 - 3x^2 - 9x + 5$
Step 1: Find f'(x)
$f'(x) = 3x^2 - 6x - 9$Step 2: Set f'(x) = 0
$3x^2 - 6x - 9 = 0$Step 3: Solve for x
$$\begin{align*} 3x^2 - 6x - 9 &= 0 \\ 3(x^2 - 2x - 3) &= 0 \\ 3(x - 3)(x + 1) &= 0 \\ x &= 3 \text{ or } x = -1 \end{align*}$$Step 4: Find y-values
$$\begin{align*} f(3) &= (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22 \\ f(-1) &= (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10 \end{align*}$$Stationary points: $(3, -22)$ and $(-1, 10)$
Step 5: Classify using second derivative
$f''(x) = 6x - 6$At $x = 3$: $f''(3) = 6(3) - 6 = 12 > 0$ → Local minimum
At $x = -1$: $f''(-1) = 6(-1) - 6 = -12 < 0$ → Local maximum
After finding the x-values where $f'(x) = 0$, use the second derivative to classify each point:
If $f''(x) > 0$:
The point is a local minimum (concave up)
If $f''(x) < 0$:
The point is a local maximum (concave down)
If $f''(x) = 0$:
The test is inconclusive (use first derivative test instead)
Test the sign of $f'(x)$ on either side of the stationary point:
Local Minimum:
$f'(x)$ changes from negative to positive
(gradient goes: decreasing → flat → increasing)
Local Maximum:
$f'(x)$ changes from positive to negative
(gradient goes: increasing → flat → decreasing)
How many turning points?
A cubic function can have 0, 1, or 2 turning points. When you solve $f'(x) = 0$, you get a quadratic equation, which can have:
Stationary Points Found:
Local Maximum at $(0.00, 0.00)$
$f''(0.00) = -6.00 < 0$
Local Minimum at $(2.00, -4.00)$
$f''(2.00) = 6.00 > 0$