4.10 Transformations of functions

A number of graphs and forms of many functions have been studied so far, including:

Function	Base form	Transformed
Quadratic	$y=x^2$y=x2	$y=a\left(x-h\right)^2+k$y=a(x−h)2+k
Cubic	$y=x^3$y=x3	$y=a\left(x-h\right)^3+k$y=a(x−h)3+k
Hyperbola	$y=\frac{1}{x}$y=1x​	$y=\frac{a}{x-h}+k$y=ax−h​+k
Square root	$y=\sqrt{x}$y=√x	$y=a\sqrt{x-h}+k$y=a√x−h+k

From experimenting with applets and summarising the effects of $a$a, $h$h and $k$k for each type of function, did you notice that each variable always had the same effect?

Recall that:

• $a$a dilates (stretches) the graph by a factor of $a$a away from the $x$x-axis
• When $a<0$a<0 the graph was reflected in the $x$x-axis
• $h$h translates the graph $h$h units horizontally
• $k$k translates the graph $k$k units vertically

These transformations work in the same way for a general function $f\left(x\right)$f(x). Let's go through them and summarise now.

 

Starting with a base function $y_1=f\left(x\right)$y1​=f(x), what effect does multiplying the function by $a$a have and why? In other words, how will the graph of the function $y_2=af\left(x\right)$y2​=af(x) compare to the graph of $y_1$y1​?

Take a quick look at the functions $y_1=x^2$y1​=x2 and $y_2=2x^2$y2​=2x2:

$x$x	$-3$−3	$-2$−2	$-1$−1	$0$0	$1$1	$2$2	$3$3
$y_1=x^2$y1​=x2	$9$9	$4$4	$1$1	$0$0	$1$1	$4$4	$9$9
$y_2=2x^2$y2​=2x2	$18$18	$8$8	$2$2	$0$0	$2$2	$8$8	$18$18

The $y$y-values for $y_2$y2​ were all double that of $y_1$y1​. This would have the effect of stretching the graph by a factor of $2$2 vertically (or away from the $x$x-axis). As we are only multiplying the $y$y-coordinate of each point, we are not changing the horizontal location of any key features.

More generally, if $y_1=f\left(x\right)$y1​=f(x) has points $\left(x,f\left(x\right)\right)$(x,f(x)) then $y_2=af\left(x\right)$y2​=af(x) will have coordinates $\left(x,af\left(x\right)\right)$(x,af(x)). Since this involves multiplying the $y$y-coordinate by $a$a, this will dilate (stretch) the graph by a factor of $a$a vertically.

 

Also notice from the coordinates of $y_1=f\left(x\right)$y1​=f(x) compared to $y_2=-f\left(x\right)$y2​=−f(x) that the $y$y-coordinates differ only by their sign (one is multiplied by $-1$−1 to give the other). This will make the point the same distance but the opposite side of the $x$x-axis. This will cause the graph to retain its shape but be reflected in the $x$x-axis.

 

How about $y_1=f\left(x\right)$y1​=f(x) compared to $y_2=f\left(x\right)+k$y2​=f(x)+k? Coordinates of $y_2$y2​ are $\left(x,f\left(x\right)+k\right)$(x,f(x)+k), so a constant value has been added to each $y$y-coordinate. This will translate (shift) the whole graph vertically by $k$k units.

 

Now, compare $y_1=f\left(x\right)$y1​=f(x) to $y_2=f\left(x-h\right)$y2​=f(x−h). Consider the case where $y_1=x^2$y1​=x2 and $y_2=\left(x-2\right)^2$y2​=(x−2)2. Compare their tables of values to understand the change:

$x$x	$-3$−3	$-2$−2	$-1$−1	$0$0	$1$1	$2$2	$3$3
$y_1=x^2$y1​=x2	$9$9	$4$4	$1$1	$0$0	$1$1	$4$4	$9$9

$x$x	$-1$−1	$0$0	$1$1	$2$2	$3$3	$4$4	$5$5
$y_2=\left(x-2\right)^2$y2​=(x−2)2	$9$9	$4$4	$1$1	$0$0	$1$1	$4$4	$9$9

Notice that the graphs have the same $y$y-coordinates but where these values occur is now shifted $2$2 to the right for the $x$x-coordinate.

In general, in comparison to the graph of $y_1=f\left(x\right)$y1​=f(x) the graph of $y_2=f\left(x-h\right)$y2​=f(x−h) will be translated $h$h units horizontally.

 

This can also be thought of as a reflection in the $y$y-axis and a dilation in the horizontal direction (away from the $y$y-axis). Below is a summary of the transformations for a general function $f\left(x\right)$f(x):

Transformation	Effect on graph	Effect on coordinates
$-f\left(x\right)$−f(x)	reflects $f\left(x\right)$f(x) in the $x$x-axis	multiplies $y$y-coordinate by $-1$−1
$af\left(x\right)$af(x)	dilates $f\left(x\right)$f(x) by a factor of $a$a away from the $x$x-axis	multiplies $y$y-coordinate by $a$a
$f\left(x\right)+k$f(x)+k	translates $f\left(x\right)$f(x) by $k$k units vertically	adds $k$k to each $y$y-coordinate
$f\left(x-h\right)$f(x−h)	translates $f\left(x\right)$f(x) by $h$h units horizontally	adds $h$h to each $x$x-coordinate
$f\left(-x\right)$f(−x)	reflects $f\left(x\right)$f(x) in the $y$y-axis	multiplies $x$x-coordinate by $-1$−1
$f\left(ax\right)$f(ax)	dilates $f\left(x\right)$f(x) by a factor of $\frac{1}{a}$1a​ away from the y-axis	divides $x$x-coordinates by $a$a

Consider some examples of functions that you are familiar with, and try to justify the last two lines in the summary table.

 

Transformation of relations

The circle centred on the origin with radius $2$2 has the equation $x^2+y^2=4$x2+y2=4.

Suppose that $x$x was replaced with $\left(x-15\right)$(x−15) and $y$y with $\left(y-9\right)$(y−9), so that the new equation becomes $\left(x-15\right)^2+\left(y-9\right)^2=4$(x−15)2+(y−9)2=4. The two relations are shown in the image below:

The change from $x$x to $\left(x-15\right)$(x−15) and from $y$y to $\left(y-9\right)$(y−9) caused the circle to shift to the right by $15$15 units and upward by $9$9 units. The centre moved from $\left(0,0\right)$(0,0) to $\left(15,9\right)$(15,9) and the radius (and thus the overall shape of the circle) remained unchanged.

More generally, for any relation in terms of $x$x and $y$y, the substitution of $\left(x-h\right)$(x−h) for $x$x and $\left(y-k\right)$(y−k) for $y$y will translate the relation $h$h units horizontally and $k$k units vertically.

 

Practice questions

Question 1

Question 2

Question 3

Question 4

Question 5