2.08 Graphs of cubics

The shape of a cubic

The cubic function belongs to the family of polynomial functions and has the general form: 

$y=ax^3+bx^2+cx+d$y=ax3+bx2+cx+d

Various combinations of choices of the four coefficients $a$a, $b$b, $c$c and $d$d will create variations in the shape and position of the curve. However, the curve will always exhibit certain properties and key features that we will outline here.

Before we look at these properties it might be beneficial to describe the shape using a rope analogy.

Take a piece of rope and lay it on the table. Put two turning points in it so that it looks like photo A in the diagram below. One of the turning points is known as a local maximum and the other is known as a local minimum. Photo A is one of the common shapes of a cubic function.

In between the two turning points (circled) is the point of inflection of the cubic function. A point of inflection is a point where the graph changes concavity. For each example above the graph changes from concave down to concave up at the circled point. All cubic functions undergo a single change in concavity (from concave down to concave up, or from concave up to concave down) and this change always occurs at the cubic's point of inflection. Cubic graphs have $180$180$^\circ$° point symmetry about the point of inflection.

For $y=ax^3+bx^2+cx+d$y=ax3+bx2+cx+d the point of inflection is always found at $x=\frac{-b}{3a}$x=−b3a​.

Now imagine that both ends of the rope are pulled - the left end pulled down and the right end pulled up. At some point in the pulling it will become something like that depicted in photo B. The turning points will have merged together to form a single point known as a horizontal point of inflection or, alternatively, a stationary point of inflection. 

Finally imagine pulling on the ends further. You will end up with something like that depicted in photo C.

With a real rope you would be able to pull it into a straight line, but with the cubic curve, a complete straightening is impossible. No matter how hard you 'pull' on the cubic function, there will always be a central point of inflection (no longer horizontal but nevertheless an inflection) with a concave down section on the left of the centre and a concave up section to the right.

Forms of a cubic function

Recall from Chapter 1 where we looked at factorising and solving cubic equations, there will always be one x-intercept but depending on the factorisation of the quadratic factor we may have one, two or three $x$x-intercepts. 

Let's explore these forms below and try to find the effect of the different parameters. 

 

Translated form

$y=a\left(x-h\right)^3+k$y=a(x−h)3+k 

This form looks very similar to turning point form  for quadratics. Unlike quadratics, not all cubics can be written in this form. Only cubics similar to diagram B with a horizontal point of inflection can be written in this form. Start the applet using $a=1$a=1, $h=0$h=0 and $k=0$k=0 to view the basic form of a cubic $y=x^3$y=x3. Then adjust $a$a, $h$h, and $k$k and try to summarise their impact on the graph.

Created with Geogebra

All the graphs of this form can be obtained by transformations of the graph $y=x^3$y=x3.

Summary:

• $a$a dilates(stretches) the graph by a factor of $a$a from the $x$x-axis, the larger the magnitude of $a$a the steeper the graph
• If $a>0$a>0, the graph is increasing
• If $a<0$a<0, the graph is decreasing, this is a reflection of the basic graph about the $x$x-axis
• $h$h translates the graph $h$h units horizontally
• $k$k translates the graph $k$k units vertically
• Hence, the horizontal point of inflection is at $\left(h,k\right)$(h,k)

Factored form (3 linear factors)

$y=a\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)$y=a(x−α)(x−β)(x−γ)

Just like factored form for quadratics, this form allows us to read off the $x$x-intercepts directly. Try changing the values in the applet below, try cases with all 3 values different and other cases where some or all of the values match.

Created with Geogebra

Summary:

• The $a$a value dilates(stretches) the graph and the larger the magnitude of $a$a the steeper the graph
• If $a>0$a>0 the graph is mostly increasing (with a positive slope from bottom left to top right)
• If $a<0$a<0 the graph is mostly decreasing (with a negative slope from top left to bottom right)
• For three distinct roots there will three $x$x-intercepts at:$\left(\alpha,0\right)$(α,0), $\left(\beta,0\right)$(β,0) and $\left(\gamma,0\right)$(γ,0)
• For two distinct roots, that is a graph of the form $y=a\left(x-\alpha\right)^2\left(x-\beta\right)$y=a(x−α)2(x−β), there are two $x$x-intercepts: a touch at$\left(\alpha,0\right)$(α,0) and a cut at $\left(\beta,0\right)$(β,0)
• For one distinct root, that is graphs of the form $y=a\left(x-\alpha\right)^3$y=a(x−α)3, there is a single $x$x-intercept and horizontal point of inflection at $\left(\alpha,0\right)$(α,0). This is the same as translated form with $k=0$k=0

An interesting fact, not easily spotted from the applet, is that the point of inflection will occur at the average of the roots. This is always true when the roots are real.

 

General form

The third applet allows you to play with all four constants of the form $y=ax^3+bx^2+cx+d$y=ax3+bx2+cx+d. See what you can discover.

Created with Geogebra

Summary:

• The $y$y-intercept occurs at$\left(0,d\right)$(0,d)
• The graph can have $1$1, $2$2 or $3$3 $x$x-intercepts
• The graph can be of one of the three shapes given by diagrams A, B and C.

We can also see that for each form $a$a has the same effect and can help us sketch the overall shape:

• $a$a dilates(stretches) the graph by a factor of $a$a from the $x$x-axis, the larger the magnitude of $a$a the steeper the graph
• If $a>0$a>0, the graph is mostly increasing. If there is a local maximum and minimum, the maximum will appear first.
• If $a<0$a<0, the graph is mostly decreasing. If there is a local maximum and minimum, the minimum will appear first.

In changing the values of $a$a, $b$b, $c$c and $d$d we can get any of the three shapes of cubic functions. So how do we graph from the general form? 

For all forms, we can find the $y$y-intercept by substituting $x=0$x=0 into the equation. For the $x$x-intercept we can substitute $y=0$y=0 into the equation and then for graphs in general form we will need to use our skills from Chapter 1 on factorising and solving cubic equations.

Finding factored forms

Reminder, some strategies for factorising cubics include:

• Looking for special patterns, such as the difference of two cubics, sum of two cubics
• Factorise by grouping in pairs
• Given one linear factor, find the quadratic factor by expansion and equating coefficients
• Given one linear factor, find the quadratic factor by polynomial division
• Find a linear factor using the remainder theorem or technology and then find the quadratic factor using one of the previous two methods above

 

Worked example

Example 1

Let's consider $y=2x^3-14x+12$y=2x3−14x+12. 

Shape: $a$a is positive, so the graphs will be mostly increasing and if it has turning points it will have a maximum followed by a minimum.

y-intercept: We can directly read off the $y$y-intercept: $\left(0,12\right)$(0,12)

Point of inflection: We can find the location of the point of inflection using the equation: $x=\frac{-b}{3a}$x=−b3a​

$x$x	$=$=	$\frac{-b}{3a}$−b3a​
	$=$=	$0/6$0/6
	$=$=	$0$0

Hence, the point of inflection is at the $y$y-intercept $\left(0,12\right)$(0,12)

x-intercepts: We need to solve $2x^3-14x+12=0$2x3−14x+12=0, if we can factorise the left-hand side we can then solve with the null factor law. Trying to find one factor we could guess factors of $d$d divided by factors of $a$a. Luckily we don't need to check too many since: $P\left(1\right)=2-14+12=0$P(1)=2−14+12=0 and so $\left(x-1\right)$(x−1) is a factor. Then using polynomial division, we see that $y=\left(x-1\right)\left(2x^2+2x-12\right)$y=(x−1)(2x2+2x−12) and lastly factorising the quadratic we reveal that: $y=2\left(x-1\right)\left(x+3\right)\left(x-2\right)$y=2(x−1)(x+3)(x−2). 

So we have $x$x-intercepts of $\left(-3,0\right)$(−3,0), $\left(1,0\right)$(1,0) and $\left(2,0\right)$(2,0)

Location of turning points: Later in this course, we will use calculus to find the exact location of the turning points. For this example, we have four points we can plot exactly and then put rough locations for the turning points. The local maximum is in between the first two $x$x-intercepts and the minimum being between the second two intercepts. We could also find these using technology to help us sketch the function accurately.

Graph of $y=2x^3-14x+12$y=2x3−14x+12

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

Applications and using technology

We will look more closely at applications of cubic functions later in the course when we can find the turning points algebraically. We can still set up and then solve problems involving cubics with the help of technology. We may need to use knowledge about the function such as axes intercepts to and practical restrictions to set up an appropriate view window. Just as common problems for quadratics involved areas, common problems for cubics often involve volumes.

Worked example

Example 2

An open chip box is made from a sheet of cardboard measuring $30$30 $cm$cm by $20$20 $cm$cm with four squares cut away from each corner as depicted in the diagram.

Here is the constructed box:

The store owner wishes to know the volume of the box as a function of the side $x$x of the cut-away squares. She suspects that the quantity of chips that can be contained varies according to the size of $x$x, but is not sure.

To answer this question, the length of the box, when constructed, will be $l=30-2x$l=30−2x and the width will likewise be $w=20-2x$w=20−2x.  With the height of the box as $x$x, the volume is given by $V=\left(30-2x\right)\left(20-2x\right)x$V=(30−2x)(20−2x)x or when expanded $V=4x^3-100x^2+600x$V=4x3−100x2+600x. 

This situation only makes sense for positive side lengths and positive volume. So when trying to create the domain restriction let's look at the sides of the box.

Length: $30-2x$30−2x, width: $20-2x$20−2x and height: $x$x. We can see the smallest $x$x can be is zero, creating a height of zero and no volume. What is the largest it can be? For the length to be positive, $x$x must be no more than $15$15 $cm$cm and for the width to be positive, $x$x must be no more than $10$10 $cm$cm. Since both must be positive we choose the smaller bound. Hence, the practical domain of this function is $0\le x\le10$0≤x≤10 .

Using technology, the graph of $V$V as a function of $x$x is shown here.

You can see from the graph that the maximum volume occurs when $x=4$x=4, and this volume is determined as $V=\left(30-2\times4\right)\left(20-2\times4\right)\times4=1056$V=(30−2×4)(20−2×4)×4=1056 $cm^3$cm3. 

Reflect: Notice that the value of our function $V(x)$V(x) is zero at $x=0$x=0, and $x=10$x=10. In other words, the volume of the box is zero in these edge cases. Some might argue that the volume of the box should be strictly positive, but in any case, we're concerned about the maximum at $x=4$x=4.

 

Practice questions

QUESTION 4

QUESTION 5