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{align}f\left(2\right)&=1 && \text{Given} \\[1.5mm] g\left(x\right)&=f\left(x\right)-3 && \text{Given transformation} \\[1.5mm] g\left(2\right)&=f\left(2\right)-3 \\& =1 - 3 \\ &=-2 \end{align}[/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].

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]{c}V\left(t\right)=\text{ the original venting plan}[/latex]

[latex]\text{F}\left(t\right)=\text{starting 2 hrs sooner}[/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].

Figure 6

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{align}g\left(5\right)&=f\left(5 - 3\right) \\ &=f\left(2\right) \\ &=1 \end{align}[/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 in Figure 7 represents both of the functions. We can see the horizontal shift in each point.

Figure 7

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{align}&f\left(x\right)={x}^{2} \\ &g\left(x\right)=f\left(x - 2\right) ={\left(x - 2\right)}^{2} \end{align}[/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 t-axis as shown in Figure 10.

Figure 10. Vertical reflection of the square root function

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

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 in Figure 11.

Figure 11. Horizontal reflection of the square root function

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

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.

   [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.

Figure 13