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.

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].

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{align}{c}V\left(t\right)&=\text{ the original venting plan}\\ F\left(t\right)&=\text{starting 2 hrs sooner}\end{align}[/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]f\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.

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 [latex]x[/latex]–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 [latex]x[/latex]-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.

a. Reflecting the graph vertically means that each output value will be reflected over the horizontal [latex]t[/latex]–axis as shown below.

Vertical reflection of the square root function

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.

b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown below.

Horizontal reflection of the square root function

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 [latex]x[/latex]-values stay the same and each output value will be the opposite of the original output value.
   

   [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.
   

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

Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Let’s begin with the rule for even functions.

[latex]f\left(-x\right)={\left(-x\right)}^{3}+2\left(-x\right)=-{x}^{3}-2x[/latex]

This does not return us to the original function, so this function is not even. We can now test the rule for odd functions.

[latex]-f\left(-x\right)=-\left(-{x}^{3}-2x\right)={x}^{3}+2x[/latex]

Because [latex]-f\left(-x\right)=f\left(x\right)[/latex], this is an odd function.

Analysis of the Solution

Consider the graph of [latex]f[/latex]. Notice that the graph is symmetric about the origin. For every point [latex]\left(x,y\right)[/latex] on the graph, the corresponding point [latex]\left(-x,-y\right)[/latex] is also on the graph. For example, (1, 3) is on the graph of [latex]f[/latex], and the corresponding point [latex]\left(-1,-3\right)[/latex] is also on the graph.

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]

Figure 16

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}\cdot{f}(4) =\frac{1}{2}\cdot\left(3\right)=\frac{3}{2}[/latex]

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

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

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

(a) Original population graph (b) Compressed population graph

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.

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.

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

[latex]h\left(x\right)=f\left(x - 1\right)+2[/latex]

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

[latex]h\left(x\right)=\sqrt{x - 1}+2[/latex]

Analysis of the Solution

Note that this transformation has changed the domain and range of the function. This new graph has domain [latex]\left[1,\infty \right)[/latex] and range [latex]\left[2,\infty \right)[/latex].

This equation combines three transformations into one equation.

• A horizontal reflection: [latex]f\left(-t\right)={2}^{-t}[/latex]
• A vertical reflection: [latex]-f\left(-t\right)=-{2}^{-t}[/latex]
• A vertical shift: [latex]-f\left(-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 [latex](0, 1)[/latex] and [latex](1, 2)[/latex].

1. First, we apply a horizontal reflection: [latex](0, 1) (–1, 2)[/latex].
2. Then, we apply a vertical reflection: [latex](0, −1) (1, –2)[/latex].
3. Finally, we apply a vertical shift: [latex](0, 0) (1, 1)[/latex].

This means that the original points, [latex](0,1)[/latex] and [latex](1,2)[/latex] become [latex](0,0)[/latex] and [latex](1,1)[/latex] after we apply the transformations.

In the graphs below, 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.

Analysis of the Solution

As a model for learning, this function would be limited to a domain of [latex]t\ge 0[/latex], with corresponding range [latex]\left[0,1\right)[/latex].

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].

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.

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

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

[latex]f\left(\frac{1}{2}x+1\right)-3=f\left(\frac{1}{2}\left(x+2\right)\right)-3[/latex]

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 [latex](0,2)[/latex] remains at [latex](0,2)[/latex] while the point [latex](2,0)[/latex] will stretch to [latex](4,0)[/latex].

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

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