Lesson

Recall that the cubic function is given by $y=ax^3+bx^2+cx+d$`y`=`a``x`3+`b``x`2+`c``x`+`d`.

Now if $b^2=3ac$`b`2=3`a``c`, 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`=`a``x`3. Specifically, if the curve of the function $y=ax^3$`y`=`a``x`3 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`=`a``x`3 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`=2`x`3 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`=`a``x`3 for $\left|a\right|\ne1$|`a`|≠1 can be thought of as the enlargement (a stretch or compression in the $y$`y` direction) of the base cubic function $y=x^3$`y`=`x`3.

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

Open up the applet below and you will see the graph of $y=x^3$`y`=`x`3, 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.

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 enlargement. 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).

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 enlargement factor $a=-\frac{1}{2}$`a`=−12. Try to describe what you see.

Consider the function $y=\frac{1}{2}\left(x-3\right)^3$`y`=12(`x`−3)3

Is the cubic increasing or decreasing from left to right?

Increasing

ADecreasing

BIncreasing

ADecreasing

BIs the function more or less steep than the function $y=x^3$

`y`=`x`3 ?More steep

ALess steep

BMore steep

ALess steep

BWhat are the coordinates of the point of inflection of the function?

Inflection ($\editable{}$, $\editable{}$)

Plot the graph $y=\frac{1}{2}\left(x-3\right)^3$

`y`=12(`x`−3)3Loading Graph...

Consider the function $y=-2\left(x-2\right)^3$`y`=−2(`x`−2)3

Is the cubic increasing or decreasing from left to right?

Increasing

ADecreasing

BIncreasing

ADecreasing

BIs the function more or less steep than the function $y=-x^3$

`y`=−`x`3 ?More steep

ALess steep

BMore steep

ALess steep

BWhat are the coordinates of the point of inflection of the function?

Inflection ($\editable{}$, $\editable{}$)

Plot the graph $y=-2\left(x-2\right)^3$

`y`=−2(`x`−2)3Loading Graph...

This is a graph of $y=x^3$`y`=`x`3.

Loading Graph...

How do we shift the graph of $y=x^3$

`y`=`x`3 to get the graph of $y=\left(x-2\right)^3-3$`y`=(`x`−2)3−3?Move the graph to the right by $2$2 units and down by $3$3 units.

AMove the graph to the left by $3$3 units and down by $2$2 units.

BMove the graph to the right by $3$3 units and up by $2$2 units.

CMove the graph to the left by $2$2 units and up by $3$3 units.

DMove the graph to the right by $2$2 units and down by $3$3 units.

AMove the graph to the left by $3$3 units and down by $2$2 units.

BMove the graph to the right by $3$3 units and up by $2$2 units.

CMove the graph to the left by $2$2 units and up by $3$3 units.

DHence plot $y=\left(x-2\right)^3-3$

`y`=(`x`−2)3−3 on the same graph as $y=x^3$`y`=`x`3.Loading Graph...