The formula[latex]\,g\left(x\right)=f\left(x\right)-3\,[/latex] tells us that we can find the output values of [latex]\,g\,[/latex] by subtracting 3 from the output values of[latex]\,f.\,[/latex] For example:

We can set[latex]\,V\left(t\right)\,[/latex]to be the original program and [latex]\,F\left(t\right)\,[/latex] to be the revised program.

In the new graph, at each time, the airflow is the same as the original function [latex]\,V\,[/latex] was 2 hours later. For example, in the original function [latex]\,V,\,[/latex] the airflow starts to change at 8 a.m., whereas for the function[latex]\,F,\,[/latex] the airflow starts to change at 6 a.m. The comparable function values are[latex]\,V\left(8\right)=F\left(6\right).\,[/latex] See Figure 6. Notice also that the vents first opened to [latex]\,220{\text{ ft}}^{2}\,[/latex] at 10 a.m. under the original plan, while under the new plan, the vents reach [latex]\,220{\text{ ft}}^{\text{2}}\,[/latex] at 8 a.m., so

<p style=”text-align: center”>[latex] V(10) = F(8)[/latex].

In both cases, we see that, because[latex]\,F\left(t\right)\,[/latex]starts 2 hours sooner,[latex]\,h=-2.\,[/latex]That means that the same output values are reached when

<p style=”text-align: center”>[latex]\,F\left(t\right)=V\left(t-\left(-2\right)\right)=V\left(t+2\right).[/latex]

The function [latex]\,f\,[/latex] is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of [latex]\,h\,[/latex] has transformed [latex]\,f\,[/latex] in two ways: [latex]\,f\left(x+1\right)\,[/latex] is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in [latex]\,f\left(x+1\right)-3\,[/latex] is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 9.

Let us follow one point of the graph of [latex]\,f\left(x\right)=|x|.[/latex]

• The point [latex]\left(0,0\right)[/latex] is transformed first by shifting left 1 unit:[latex]\left(0,0\right)\to \left(-1,0\right)[/latex]
• The point[latex]\left(-1,0\right)[/latex] is transformed next by shifting down 3 units:[latex]\left(-1,0\right)\to \left(-1,-3\right)[/latex]

Figure 10 shows the graph of [latex]\,h.[/latex]

The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as

Using the formula for the square root function, we can write

1. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis, as shown in Figure 13.

   

   Because each output value is the opposite of the original output value, we can write

   [latex]V\left(t\right)=-s\left(t\right)\text{ or }V\left(t\right)=-\sqrt{t}[/latex]

   Notice that this is an outside change, or vertical shift, that affects the output [latex]\,s\left(t\right)\,[/latex] values, so the negative sign belongs outside of the function.

2. Reflecting horizontally means that each input value will be reflected over the vertical axis, as shown in Figure 14.

   

   Because each input value is the opposite of the original input value, we can write

   [latex]H\left(t\right)=s\left(-t\right)\text{ or }H\left(t\right)=\sqrt{-t}[/latex]

   Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function.

   Note that these transformations can affect the domain and range of the functions. While the original square root function has domain [latex]\,\left[0,\infty \right)\,[/latex] and range [latex]\,\left[0,\infty \right),\,[/latex] the vertical reflection gives the [latex]\,V\left(t\right)\,[/latex] function the range [latex]\left(-\infty ,\,0\right][/latex] and the horizontal reflection gives the[latex]\,H\left(t\right)\,[/latex] function the domain [latex]\left(-\infty ,\,0\right].[/latex]

1. For [latex]\,g\left(x\right),\,[/latex] the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value. See Table 7.

   Table 7

   [latex]x[/latex]	2	4	6	8
   [latex]\,g\left(x\right)\,[/latex]	–1	–3	–7	–11

2. For [latex]\,h\left(x\right),\,[/latex] the negative sign inside the function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the [latex]\,h\left(x\right)\,[/latex] values stay the same as the [latex]\,f\left(x\right)\,[/latex] values. See Table 8.

   Table 8

   [latex]x[/latex]	−2	−4	−6	−8
   [latex]h\left(x\right)[/latex]	1	3	7	11

This equation combines three transformations into one equation.

• A horizontal reflection:[latex]f\left(\text{−}t\right)={2}^{-t}[/latex]
• A vertical reflection:[latex]\,-f\left(\text{−}t\right)=-{2}^{-t}[/latex]
• A vertical shift:[latex]\,-f\left(\text{−}t\right)+1=-{2}^{-t}+1[/latex]

We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each of the three transformations. We will choose the points (0, 1) and (1, 2).

1. First, we apply a horizontal reflection: (0, 1) (–1, 2).
2. Then, we apply a vertical reflection: (0, -1) (-1, –2)
3. Finally, we apply a vertical shift: (0, 0) (-1, -1)).

This means that the original points, (0,1) and (1,2), become (0,0) and (-1,-1) after we apply the transformations.

In Figure 16, the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit.

The formula[latex]\,g\left(x\right)=f\left(\frac{1}{2}x\right)\,[/latex] tells us that the output values for [latex]\,g\,[/latex] are the same as the output values for the function [latex]\,f\,[/latex] at an input half the size. Notice that we do not have enough information to determine [latex]\,g\left(2\right)\,[/latex] because [latex]\,g\left(2\right)=f\left(\frac{1}{2}\cdot 2\right)=f\left(1\right),\,[/latex] and we do not have a value for [latex]\,f\left(1\right)\,[/latex] in our table. Our input values to [latex]\,g\,[/latex] will need to be twice as large to get inputs for [latex]\,f\,[/latex] that we can evaluate. For example, we can determine [latex]\,g\left(4\right)\text{.}[/latex]

We do the same for the other values to produce Table 14.

Figure 25 shows the graphs of both of these sets of points.

There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal transformations,[latex]\,f\left(3x\right)\,[/latex]is a horizontal compression by [latex]\,\frac{1}{3},\,[/latex] which means we multiply each [latex]\,x\text{-}[/latex] value by[latex]\,\frac{1}{3}.[/latex] See Table 16.

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 Table 17.

Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 18.

To simplify, let’s start by factoring out the inside of the function.

By factoring the inside, we can first horizontally stretch by 2, as indicated by the[latex]\,\frac{1}{2}\,[/latex]on the inside of the function. Remember that twice the size of 0 is still 0, so the point (0,2) remains at (0,2), while the point (2,0) will stretch to (4,0). See Figure 28.

Next, we horizontally shift left by 2 units, as indicated by [latex]\,x+2.\,[/latex]See Figure 29.

Last, we vertically shift down by 3 to complete our sketch, as indicated by the [latex]\,-3\,[/latex] on the outside of the function. See Figure 30.]