Transformations of Cubic Functions

Translating, dilating and reflecting $y=ax^3$y=ax3

Recall that the cubic function is given by $y=ax^3+bx^2+cx+d$y=ax3+bx2+cx+d. 

Now if $b^2=3ac$b2=3ac, the function can be re-written in the form $y=a\left(x-h\right)^3+k$y=a(x−h)3+k.

This second form can be considered as the translated form of the cubic given by $y=ax^3$y=ax3. Specifically, if the curve of the function $y=ax^3$y=ax3 is translated (shifted in position without distorting the shape) $h$h units to the right and $k$k units up, then the translated function becomes $y=a\left(x-h\right)^3+k$y=a(x−h)3+k.

That is, if we wish to translate a cubic function $y=ax^3$y=ax3 so that its central point moves from the origin to any other point $\left(h,k\right)$(h,k) in the cartesian plane, then all we need to do is to replace $x$x with $\left(x-h\right)$(x−h) and $y$y with $\left(y-k\right)$(y−k). 

As an example, suppose we wish to translate $y=2x^3$y=2x3 so that its central point is $\left(-3,1\right)$(−3,1) then we simply change $x$x to $x-\left(-3\right)=x+3$x−(−3)=x+3 and $y$y to $y-\left(1\right)=y-1$y−(1)=y−1 so that the translated function becomes $y-1=2\left(x+3\right)^3$y−1=2(x+3)3, which when written explicitly becomes $y=2\left(x+3\right)^3+1$y=2(x+3)3+1. It is as simple as that!

The simpler function $y=ax^3$y=ax3 for $\left|a\right|\ne1$|a|≠1 can be thought of as the dilation (a stretch or compression in the $y$y direction) of the base cubic function $y=x^3$y=x3. 

When $a<0$a<0 the base function $y=ax^3$y=ax3 is a reflected version of the same base function $y=\left|a\right|x^3$y=|a|x3. As an example, the function $y=-2x^3$y=−2x3 is a reflected version (reflected in the $x$x axis) of $y=2x^3$y=2x3.  

Using the Applet 

Exercise 1:

Open up the applet below and you will see the graph of $y=x^3$y=x3, written as $y=1\left(x-0\right)^3+0$y=1(x−0)3+0. Click on the box marked "show base function", and then, without changing anything else, move the 'a' slider along to $2$2, so that the function then reads $y=2\left(x-0\right)^3+0$y=2(x−0)3+0. Notice how the curve narrows, simply because its rising at a faster rate.

Now move the a slider back to $-2$−2. Watch how the cubic flips, so that it becomes a reflected image.

Exercise 2:

First move the $a$a slider back to $1$1.

Then, for the translation, move the $h$h slider to $-3$−3 and the $k$k slider to $-1$−1, so that the function then reads $y=1\left(x--3\right)^3+-1$y=1(x−−3)3+−1 .What the function has changed to is, of course, $y=\left(x+3\right)^3-1$y=(x+3)3−1 so that the central point has shifted from the origin to the point $\left(-3,-1\right)$(−3,−1).

Finally move the a slider up and down from $2$2 to $-2$−2 to see the independent effect of the dilation. Note that the central point remains the same. This is a critical understanding.

Very importantly, note that the presence of a negative $a$a value shows a reflected cubic function in the elevated line $y=k$y=k (and not the $x$x - axis as it was for the base function).  

Exercise 3:

Work out what you must slide to show a cubic equation that has a central point at $\left(-3,2\right)$(−3,2) and a dilation factor $a=-\frac{1}{2}$a=−12​. Try to describe what you see. 

 

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3