5.06 Sketching polynomials

Shape of cubics

The cubic function belongs to the family of polynomial functions. The image below summarises the different shapes of cubics that you might encounter. The black circle is indicating the point of inflection on each photo. It is interesting to see that the cubics in photo A have turning points, called a local maximum or minimum, while the other cubics do not. While we don't need to use the terms written on the photos right now, they will reappear when we study Calculus later on.

Form of cubics

Just as with linear and quadratic functions, cubics can be represented in different forms. There are three main forms:

In this course we will focus on graphing polynomials in factored form.

 

Depending on the cubic involved, the function could be a one-to-one relation or a many-to-one relation. (Recall that a one-to-one relation is a relation where no two values of $x$x are mapped to the same value of $y$y, and a many-to-one relation is a relation where two or more values of $x$x are mapped to the same value of $y$y).

Factored form 

$y=k\left(x-a\right)\left(x-b\right)\left(x-c\right)$y=k(x−a)(x−b)(x−c)

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

• The $k$k value dilates (stretches) the graph and the larger the magnitude of $k$k the steeper the graph
• If $k>0$k>0 the graph is mostly increasing (with a positive slope from bottom left to top right)
• If $k<0$k<0 the graph is mostly decreasing (with a negative slope from top left to bottom right)
• For three distinct roots there will be three $x$x-intercepts at: $\left(a,0\right)$(a,0), $\left(b,0\right)$(b,0) and $\left(c,0\right)$(c,0)

Finding intercepts

The $y$y-intercept can be found by making $x=0$x=0. The $x$x-intercepts for the factored form can be determined directly from the equation as described above or solved in a similar way to a quadratic equation using the null factor law.

 

Worked example

Example 1

Consider the function $f(x)=2(x-1)(x-3)(x+2).$f(x)=2(x−1)(x−3)(x+2).

a) Find the $x$x-intercepts of the function.

b) Sketch the graph of $y=f(x).$y=f(x).

c) Sketch the graph of $y=|f(x)|.$y=|f(x)|.

 

a) Think: The $x$x-intercepts occur at $y=0$y=0. The equation is in factored form though so we can use the null factor law to read off the $x$x-intercepts.

Do: Set the value of $y$y to $0$0 and read off the $y$y-intercepts.

$2(x-1)(x-3)(x+2)$2(x−1)(x−3)(x+2)	$=$=	$0$0
$(x-1)(x-3)(x+2)$(x−1)(x−3)(x+2)	$=$=	$0$0
$x$x	$=$=	$-2,1,3$−2,1,3 using the null factor law

 

b)

Think: The curve will pass through the $x$x-axis at the each of the above intercepts since the power of each factor is $1$1. The $y$y-intercept of the curve can be found by substituting $x=0$x=0 into the function. 

Do: Substitute $x=0$x=0 into the equation:

$y$y	$=$=	$2(0-1)(0-3)(0+2)$2(0−1)(0−3)(0+2)
	$=$=	$12$12

 

Now that we have all the intercepts we can graph the function: 

c) To graph the absolute value function, we need to reflect any parts of the above curve that are below the $x$x-axis by the $x$x-axis, so that the whole function lies above the $x$x-axis. The parts of the curve that we need to reflect are highlighted below:

So by reflecting these parts about the x-axis we get the following graph of the function $y=|f(x)|$y=|f(x)|:

Practice questions

Question 1

Question 2

Question 3

Question 4