We can sketch a graph of this new function by adding 20 to each of the output values of the original function. This will have the effect of shifting the graph vertically up.

Notice that for each input value, the output value has increased by 20, so if we call the new function [latex]S\left(t\right)[/latex], we could write

[latex]S\left(t\right)=V\left(t\right)+20[/latex]

This notation tells us that, for any value of [latex]t,S\left(t\right)[/latex] can be found by evaluating the function [latex]V[/latex] at the same input and then adding 20 to the result. This defines [latex]S[/latex] as a transformation of the function [latex]V[/latex], in this case a vertical shift up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change.

[latex]t[/latex]	0	8	10	17	19	24
[latex]V\left(t\right)[/latex]	0	0	220	220	0	0
[latex]S\left(t\right)[/latex]	20	20	240	240	20	20

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:

[latex]\begin{cases}f\left(2\right)=1\hfill & \text{Given}\hfill \\ g\left(x\right)=f\left(x\right)-3\hfill & \text{Given transformation}\hfill \\ g\left(2\right)=f\left(2\right)-3\hfill & \hfill \\ =1 - 3\hfill & \hfill \\ =-2\hfill & \hfill \end{cases}[/latex]

Subtracting 3 from each [latex]f\left(x\right)[/latex] value, we can complete a table of values for [latex]g\left(x\right)[/latex].

[latex]x[/latex]	2	4	6	8
[latex]f\left(x\right)[/latex]	1	3	7	11
[latex]g\left(x\right)[/latex]	−2	0	4	8

Analysis of the Solution

As with the earlier vertical shift, notice the input values stay the same and only the output values change.

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.

[latex]\begin{cases}{c}V\left(t\right)=\text{ the original venting plan}\\ \text{F}\left(t\right)=\text{starting 2 hrs sooner}\end{cases}[/latex]

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]. 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 [latex]V\left(10\right)=F\left(8\right)[/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 [latex]F\left(t\right)=V\left(t-\left(-2\right)\right)=V\left(t+2\right)[/latex].

Analysis of the Solution

Note that [latex]V\left(t+2\right)[/latex] has the effect of shifting the graph to the left.

Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function [latex]F\left(t\right)[/latex] uses the same outputs as [latex]V\left(t\right)[/latex], but matches those outputs to inputs 2 hours earlier than those of [latex]V\left(t\right)[/latex]. Said another way, we must add 2 hours to the input of [latex]V[/latex] to find the corresponding output for [latex]F:F\left(t\right)=V\left(t+2\right)[/latex].

The formula [latex]g\left(x\right)=f\left(x - 3\right)[/latex] tells us that the output values of [latex]g[/latex] are the same as the output value of [latex]f[/latex] when the input value is 3 less than the original value. For example, we know that [latex]f\left(2\right)=1[/latex]. To get the same output from the function [latex]g[/latex], we will need an input value that is 3 larger. We input a value that is 3 larger for [latex]g\left(x\right)[/latex] because the function takes 3 away before evaluating the function [latex]f[/latex].

[latex]\begin{cases}g\left(5\right)=f\left(5 - 3\right)\hfill \\ =f\left(2\right)\hfill \\ =1\hfill \end{cases}[/latex]

We continue with the other values to create this table.

[latex]x[/latex]	5	7	9	11
[latex]x - 3[/latex]	2	4	6	8
[latex]f\left(x\right)[/latex]	1	3	7	11
[latex]g\left(x\right)[/latex]	1	3	7	11

The result is that the function [latex]g\left(x\right)[/latex] has been shifted to the right by 3. Notice the output values for [latex]g\left(x\right)[/latex] remain the same as the output values for [latex]f\left(x\right)[/latex], but the corresponding input values, [latex]x[/latex], have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11.

Analysis of the Solution

The graph below represents both of the functions. We can see the horizontal shift in each point.

Notice that the graph is identical in shape to the [latex]f\left(x\right)={x}^{2}[/latex] function, but the x-values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so

[latex]g\left(x\right)=f\left(x - 2\right)[/latex]

Notice how we must input the value [latex]x=2[/latex] to get the output value [latex]y=0[/latex]; the x-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the [latex]f\left(x\right)[/latex] function to write a formula for [latex]g\left(x\right)[/latex] by evaluating [latex]f\left(x - 2\right)[/latex].

[latex]\begin{cases}f\left(x\right)={x}^{2}\hfill \\ g\left(x\right)=f\left(x - 2\right)\hfill \\ g\left(x\right)=f\left(x - 2\right)={\left(x - 2\right)}^{2}\hfill \end{cases}[/latex]

Analysis of the Solution

To determine whether the shift is [latex]+2[/latex] or [latex]-2[/latex] , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, [latex]f\left(0\right)=0[/latex]. In our shifted function, [latex]g\left(2\right)=0[/latex]. To obtain the output value of 0 from the function [latex]f[/latex], we need to decide whether a plus or a minus sign will work to satisfy [latex]g\left(2\right)=f\left(x - 2\right)=f\left(0\right)=0[/latex]. For this to work, we will need to subtract 2 units from our input values.

[latex]G\left(m\right)+10[/latex] can be interpreted as adding 10 to the output, gallons. This is the gas required to drive [latex]m[/latex] miles, plus another 10 gallons of gas. The graph would indicate a vertical shift.

[latex]G\left(m+10\right)[/latex] can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than [latex]m[/latex] miles. The graph would indicate a horizontal shift.