​​Cubic graphs

​​In a nutshell

Cubic functions are polynomial functions in the form $$f(x) = ax^3 + bx^2 + cx + d$$, where the coefficients are real numbers and $$a \ne 0$$. A graph of a cubic function may have up to three distinct roots where the curve crosses or touches the $$x$$-axis.

Sketching cubic graphs

To sketch a cubic graph, there are three important features that need to be identified from a given cubic function, $$f(x)$$: the $$y \text- intercept$$, the roots and the sign of the coefficient of $$x^3$$​.​

$$\boxed{f(x)=ax^3+bx^2+cx+d}$$​​

Procedure

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Note: Distinct roots cross the $$x$$-axis. Double repeated roots will touch the $$x$$-axis. Triple repeated roots will form points of inflection.

Example 1

Sketch the curve for the equation $$y= (x-1)(x+3)(x-2)$$.​

Identify the $$y \text- intercept$$ by multiplying all the constant values in the equation. ​

​$$-1 \times 3 \times -2 = 6$$​​

Mark the $$y \text- intercept$$ on a graph.

Find the roots by considering $$y=0$$​. 

$$\begin{aligned} x-1 &= 0 \\ x &= 1 \\ \\ x+3 &= 0\\ x &=-3 \\ \\ x-2 &= 0 \\ x &=2 \end{aligned}$$​​

Mark the roots with the $$y\text-intercept$$ on a graph.

Identify the coefficient ​of $$x^3$$ by multiplying all the $$x$$ terms together. 

$$x \times x \times x = x^3 = 1x^3$$​​

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The coefficient is positive, so the curve will start from the bottom left and end in the top right. 

Use this information to sketch the curve.  

Note: The notation $$y$$ and $$f(x)$$ can be used interchangeably for graphs which only have an $$x$$ and $$y$$-axis. 

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Example 2

Factorise and identify the roots of the function, $$f(x) = x^3 - 8x^2 + 12x$$.

Factorise out the common factor of $$x$$.

​$$f(x)=x(x^2 - 8x +12)$$​​

Factorise the quadratic equation using a method of your choice. 

$$x^2 - 8x +12$$

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​$$\begin{aligned} 12&= (-6) \times (-2)& \\ -8 &= (-6) + (-2) \end{aligned}$$

$$x^2 - 6x - 2x +12$$​​

$$x(x-6) - 2(x-6)$$​​

​$$(x-2)(x-6)$$​​

Therefore, $$\underline{ f(x) = x(x-2)(x-6).}$$

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Find the roots of $$f(x)$$​. ​

​$$\begin{aligned} f(x)&=0 \\ x(x-2)(x-6) &=0\\\\ x &= 0 \\ \\ x-2 &=0 \\ x&=2 \\ \\ x-6&=0 \\ x&=6\end{aligned}$$​​

Therefore, the roots of the equation occur when:  $$\underline{ x=0, 2 \ and \ 6.}$$​​