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). 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

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]

Figure 6.

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]

We continue with the other values to create (Figure).

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.

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

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]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.

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).

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 9.

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

Figure 10.

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).

   

   Figure 13. 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.

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

   

   Figure 14. 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 x-values stay the same and each output value will be the opposite of the original output value. See (Figure).

   [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 (Figure).

   [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), 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.

Figure 16.