Because the population is always twice as large, the new population’s output values are always twice the original function’s output values.

If we choose four reference points, (0, 1), (3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2.

The following shows where the new points for the new graph will be located.

[latex]\begin{cases}\left(0,\text{ }1\right)\to \left(0,\text{ }2\right)\hfill \\ \left(3,\text{ }3\right)\to \left(3,\text{ }6\right)\hfill \\ \left(6,\text{ }2\right)\to \left(6,\text{ }4\right)\hfill \\ \left(7,\text{ }0\right)\to \left(7,\text{ }0\right)\hfill \end{cases}[/latex]

Symbolically, the relationship is written as

[latex]Q\left(t\right)=2P\left(t\right)[/latex]

This means that for any input [latex]t[/latex], the value of the function [latex]Q[/latex] is twice the value of the function [latex]P[/latex]. Notice that the effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal axis. The input values, [latex]t[/latex], stay the same while the output values are twice as large as before.

The formula [latex]g\left(x\right)=\frac{1}{2}f\left(x\right)[/latex] tells us that the output values of [latex]g[/latex] are half of the output values of [latex]f[/latex] with the same inputs. For example, we know that [latex]f\left(4\right)=3[/latex]. Then

[latex]g\left(4\right)=\frac{1}{2}f\left(4\right)=\frac{1}{2}\left(3\right)=\frac{3}{2}[/latex]
We do the same for the other values to produce this table.

[latex]x[/latex]	[latex]2[/latex]	[latex]4[/latex]	[latex]6[/latex]	[latex]8[/latex]
[latex]g\left(x\right)[/latex]	[latex]\frac{1}{2}[/latex]	[latex]\frac{3}{2}[/latex]	[latex]\frac{7}{2}[/latex]	[latex]\frac{11}{2}[/latex]

Analysis of the Solution

The result is that the function [latex]g\left(x\right)[/latex] has been compressed vertically by [latex]\frac{1}{2}[/latex]. Each output value is divided in half, so the graph is half the original height.

When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that [latex]g\left(2\right)=2[/latex]. With the basic cubic function at the same input, [latex]f\left(2\right)={2}^{3}=8[/latex]. Based on that, it appears that the outputs of [latex]g[/latex] are [latex]\frac{1}{4}[/latex] the outputs of the function [latex]f[/latex] because [latex]g\left(2\right)=\frac{1}{4}f\left(2\right)[/latex]. From this we can fairly safely conclude that [latex]g\left(x\right)=\frac{1}{4}f\left(x\right)[/latex].

We can write a formula for [latex]g[/latex] by using the definition of the function [latex]f[/latex].

[latex]g\left(x\right)=\frac{1}{4}f\left(x\right)=\frac{1}{4}{x}^{3}[/latex]

Symbolically, we could write

[latex]\begin{cases}R\left(1\right)=P\left(2\right),\hfill \\ R\left(2\right)=P\left(4\right),\text{ and in general,}\hfill \\ R\left(t\right)=P\left(2t\right).\hfill \end{cases}[/latex]

See below for a graphical comparison of the original population and the compressed population.

Analysis of the Solution

Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis.

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]

[latex]g\left(4\right)=f\left(\frac{1}{2}\cdot 4\right)=f\left(2\right)=1[/latex]

We do the same for the other values to produce the table below.

[latex]x[/latex]	4	8	12	16
[latex]g\left(x\right)[/latex]	1	3	7	11

This figure shows the graphs of both of these sets of points.

Analysis of the Solution

Because each input value has been doubled, the result is that the function [latex]g\left(x\right)[/latex] has been stretched horizontally by a factor of 2.

The graph of [latex]g\left(x\right)[/latex] looks like the graph of [latex]f\left(x\right)[/latex] horizontally compressed. Because [latex]f\left(x\right)[/latex] ends at [latex]\left(6,4\right)[/latex] and [latex]g\left(x\right)[/latex] ends at [latex]\left(2,4\right)[/latex], we can see that the [latex]x\text{-}[/latex] values have been compressed by [latex]\frac{1}{3}[/latex], because [latex]6\left(\frac{1}{3}\right)=2[/latex]. We might also notice that [latex]g\left(2\right)=f\left(6\right)[/latex] and [latex]g\left(1\right)=f\left(3\right)[/latex]. Either way, we can describe this relationship as [latex]g\left(x\right)=f\left(3x\right)[/latex]. This is a horizontal compression by [latex]\frac{1}{3}[/latex].

Analysis of the Solution

Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of [latex]\frac{1}{4}[/latex] in our function: [latex]f\left(\frac{1}{4}x\right)[/latex]. This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching.