# 1.5 Transformation of functions  (Page 2/22)

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Given a tabular function, create a new row to represent a vertical shift.

1. Identify the output row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down.

## Shifting a tabular function vertically

A function $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is given in [link] . Create a table for the function $\text{\hspace{0.17em}}g\left(x\right)=f\left(x\right)-3.$

 $x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11

The formula $\text{\hspace{0.17em}}g\left(x\right)=f\left(x\right)-3\text{\hspace{0.17em}}$ tells us that we can find the output values of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ by subtracting 3 from the output values of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ For example:

Subtracting 3 from each $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ value, we can complete a table of values for $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ as shown in [link] .

 $x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11 $g\left(x\right)$ −2 0 4 8

The function $\text{\hspace{0.17em}}h\left(t\right)=-4.9{t}^{2}+30t\text{\hspace{0.17em}}$ gives the height $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ of a ball (in meters) thrown upward from the ground after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function $\text{\hspace{0.17em}}b\left(t\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}h\left(t\right),\text{\hspace{0.17em}}$ and then find a formula for $\text{\hspace{0.17em}}b\left(t\right).$

$b\left(t\right)=h\left(t\right)+10=-4.9{t}^{2}+30t+10$

## Identifying horizontal shifts

We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift , shown in [link] .

For example, if $\text{\hspace{0.17em}}f\left(x\right)={x}^{2},\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}g\left(x\right)={\left(x-2\right)}^{2}\text{\hspace{0.17em}}$ is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in $\text{\hspace{0.17em}}f.$

## Horizontal shift

Given a function $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ a new function $\text{\hspace{0.17em}}g\left(x\right)=f\left(x-h\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ is a constant, is a horizontal shift    of the function $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ is positive, the graph will shift right. If $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ is negative, the graph will shift left.

## Adding a constant to an input

Returning to our building airflow example from [link] , suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function.

We can set $\text{\hspace{0.17em}}V\left(t\right)\text{\hspace{0.17em}}$ to be the original program and $\text{\hspace{0.17em}}F\left(t\right)\text{\hspace{0.17em}}$ to be the revised program.

In the new graph, at each time, the airflow is the same as the original function $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ was 2 hours later. For example, in the original function $\text{\hspace{0.17em}}V,\text{\hspace{0.17em}}$ the airflow starts to change at 8 a.m., whereas for the function $\text{\hspace{0.17em}}F,\text{\hspace{0.17em}}$ the airflow starts to change at 6 a.m. The comparable function values are $\text{\hspace{0.17em}}V\left(8\right)=F\left(6\right).\text{\hspace{0.17em}}$ See [link] . Notice also that the vents first opened to at 10 a.m. under the original plan, while under the new plan the vents reach at 8 a.m., so $\text{\hspace{0.17em}}V\left(10\right)=F\left(8\right).$

In both cases, we see that, because $\text{\hspace{0.17em}}F\left(t\right)\text{\hspace{0.17em}}$ starts 2 hours sooner, $\text{\hspace{0.17em}}h=-2.\text{\hspace{0.17em}}$ That means that the same output values are reached when $\text{\hspace{0.17em}}F\left(t\right)=V\left(t-\left(-2\right)\right)=V\left(t+2\right).$

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