# 3.5 Transformation of functions  (Page 9/21)

 Page 9 / 21

Write a formula for the toolkit square root function horizontally stretched by a factor of 3.

$g\left(x\right)=f\left(\frac{1}{3}x\right)\text{\hspace{0.17em}}$ so using the square root function we get $\text{\hspace{0.17em}}g\left(x\right)=\sqrt{\frac{1}{3}x}$

## Performing a sequence of transformations

When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first.

When we see an expression such as $\text{\hspace{0.17em}}\text{\hspace{0.17em}}2f\left(x\right)+3,\text{\hspace{0.17em}}$ which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition.

Horizontal transformations are a little trickier to think about. When we write $\text{\hspace{0.17em}}g\left(x\right)=f\left(2x+3\right),\text{\hspace{0.17em}}$ for example, we have to think about how the inputs to the function $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ relate to the inputs to the function $\text{\hspace{0.17em}}\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ Suppose we know $\text{\hspace{0.17em}}f\left(7\right)=12.\text{\hspace{0.17em}}$ What input to $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ would produce that output? In other words, what value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ will allow $\text{\hspace{0.17em}}g\left(x\right)=f\left(2x+3\right)=12?\text{\hspace{0.17em}}$ We would need $\text{\hspace{0.17em}}2x+3=7.\text{\hspace{0.17em}}$ To solve for $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression.

This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function.

$f\left(bx+p\right)=f\left(b\left(x+\frac{p}{b}\right)\right)$

Let’s work through an example.

$f\left(x\right)={\left(2x+4\right)}^{2}$

We can factor out a 2.

$f\left(x\right)={\left(2\left(x+2\right)\right)}^{2}$

Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally.

## Combining transformations

When combining vertical transformations written in the form $\text{\hspace{0.17em}}af\left(x\right)+k,\text{\hspace{0.17em}}$ first vertically stretch by $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and then vertically shift by $\text{\hspace{0.17em}}k.$

When combining horizontal transformations written in the form $\text{\hspace{0.17em}}f\left(bx+h\right),\text{\hspace{0.17em}}$ first horizontally shift by $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ and then horizontally stretch by $\text{\hspace{0.17em}}\frac{1}{b}.$

When combining horizontal transformations written in the form $\text{\hspace{0.17em}}f\left(b\left(x+h\right)\right),\text{\hspace{0.17em}}$ first horizontally stretch by $\text{\hspace{0.17em}}\frac{1}{b}\text{\hspace{0.17em}}$ and then horizontally shift by $\text{\hspace{0.17em}}h.$

Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first.

## Finding a triple transformation of a tabular function

Given [link] for the function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ create a table of values for the function $\text{\hspace{0.17em}}g\left(x\right)=2f\left(3x\right)+1.$

 $x$ 6 12 18 24 $f\left(x\right)$ 10 14 15 17

There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal transformations, $\text{\hspace{0.17em}}f\left(3x\right)\text{\hspace{0.17em}}$ is a horizontal compression by $\text{\hspace{0.17em}}\frac{1}{3},\text{\hspace{0.17em}}$ which means we multiply each $\text{\hspace{0.17em}}x\text{-}$ value by $\text{\hspace{0.17em}}\frac{1}{3}.$ See [link] .

 $x$ 2 4 6 8 $f\left(3x\right)$ 10 14 15 17

Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. See [link] .

 $x$ 2 4 6 8 $2f\left(3x\right)$ 20 28 30 34

Finally, we can apply the vertical shift, which will add 1 to all the output values. See [link] .

 $x$ 2 4 6 8 $g\left(x\right)=2f\left(3x\right)+1$ 21 29 31 35

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