Learning Objectives
For the rational parent function [latex]f(x)=\dfrac{1}{x}[/latex],
- Perform vertical and horizontal shifts
- Perform vertical stretches and compressions
- Perform reflections across the [latex]x[/latex]-axis
- Perform reflections across the [latex]y[/latex]-axis
- Determine the transformations performed on the parent function [latex]f(x)=\dfrac{1}{x}[/latex] to get the rational function [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k[/latex]
Vertical Shifts
If we shift the graph of the rational function [latex]f(x)=\dfrac{1}{x}[/latex] up 5 units, all of the points on the graph increase their [latex]y[/latex]-coordinates by 5, but their [latex]x[/latex]-coordinates remain the same. Therefore, the equation of the function [latex]f(x)=\dfrac{1}{x}[/latex] after it has been shifted up 5 units transforms to [latex]f(x)=\dfrac{1}{x}+5[/latex]. Table 1 shows the changes to specific values of this function, which are replicated on the graph in figure 1.
| [latex]x[/latex] | [latex]\dfrac{1}{x}[/latex] | [latex]\dfrac{1}{x}+5[/latex] |
 Figure 1. Shifting the graph of the function [latex]f(x)=\frac{1}{x}[/latex] up 5 units. |
|---|---|---|---|
| [latex]–2[/latex] | [latex]-\dfrac{1}{2}[/latex] | [latex]\dfrac{9}{2}[/latex] | |
| [latex]–1[/latex] | [latex]–1[/latex] | [latex]4[/latex] | |
| [latex]-\dfrac{1}{2}[/latex] | [latex]–2[/latex] | [latex]3[/latex] | |
| [latex]\dfrac{1}{2}[/latex] | [latex]2[/latex] | [latex]7[/latex] | |
| [latex]1[/latex] | [latex]1[/latex] | [latex]6[/latex] | |
| [latex]2[/latex] | [latex]\dfrac{1}{2}[/latex] | [latex]\dfrac{11}{2}[/latex] | |
| [latex]3[/latex] | [latex]\dfrac{1}{3}[/latex] | [latex]\dfrac{16}{3}[/latex] | |
| Table 1. [latex]f(x)=\frac{1}{x}[/latex] is transformed to[latex]f(x)=\frac{1}{x}+5[/latex]. | |||
If we shift the graph of the function [latex]f(x)=\frac{1}{x}[/latex] down 8 units, all of the points on the graph decrease their [latex]y[/latex]-coordinates by 8, but their [latex]x[/latex]-coordinates remain the same. Therefore, the equation of the function [latex]f(x)=\frac{1}{x}[/latex] after it has been shifted down 8 units transforms to [latex]f(x)=\frac{1}{x}-8[/latex]. Table 2 shows the changes to specific values of this function, which are replicated on the graph in figure 2.
| [latex]f(x)[/latex] | [latex]\dfrac{1}{x}[/latex] | [latex]f(x)=\dfrac{1}{x}-8[/latex] |
 Figure 2. Shifting the graph of the function [latex]f(x)=\frac{1}{x}[/latex] down 8 units. |
|---|---|---|---|
| [latex]–2[/latex] | [latex]-\dfrac{1}{2}[/latex] | [latex]-\dfrac{17}{2}[/latex] | |
| [latex]–1[/latex] | [latex]–1[/latex] | [latex]–9[/latex] | |
| [latex]-\dfrac{1}{2}[/latex] | [latex]–2[/latex] | [latex]–10[/latex] | |
| [latex]\dfrac{1}{2}[/latex] | [latex]2[/latex] | [latex]–6[/latex] | |
| [latex]1[/latex] | [latex]1[/latex] | [latex]–7[/latex] | |
| [latex]2[/latex] | [latex]\frac{1}{2}[/latex] | [latex]-\frac{15}{2}[/latex] | |
| [latex]3[/latex] | [latex]\frac{1}{3}[/latex] | [latex]-\frac{23}{3}[/latex] | |
| Table 2. [latex]f(x)=\frac{1}{x}[/latex] is transformed to [latex]f(x)=\frac{1}{x}-8[/latex]. | |||
Vertical shifts
We can represent a vertical shift of the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] by adding a constant, [latex]k[/latex], to the function:
[latex]f(x)=\dfrac{1}{x}+k[/latex]
If [latex]k>0[/latex], the graph shifts upwards and if [latex]k<0[/latex] the graph shifts downwards.
Manipulate the graph in figure 3 to shift the graph vertically. Pay attention to what happens with the function as [latex]k[/latex] changes value. Also, watch what happens with the asymptotes.
Figure 3. Vertical Transformations
Example 1
1. Use figure 3 to graph the function [latex]f(x)=\frac{1}{x}-6[/latex]. What is the horizontal asymptote of the function? What is the relationship between the value of [latex]k[/latex] and the horizontal asymptote? What is the vertical asymptote of the function?
2. Without graphing the function, what is the horizontal asymptote of the function [latex]g(x)=\frac{1}{x}+5[/latex]? What is the vertical asymptote?
Solution
1.
The horizontal asymptote is the line [latex]y=-6[/latex]. The value of [latex]k[/latex] is [latex]-6[/latex] so the horizontal asymptote mimics the value of [latex]k[/latex].
The vertical asymptote is the line [latex]x=0[/latex].
2. Since [latex]k=5[/latex] the graph of [latex]f(x)=\frac{1}{x}[/latex] gets shifted up by 5 units. This means that the horizontal asymptote is shifted up 5 units to [latex]y=5[/latex]. The vertical asymptote does not move so is still [latex]x=0[/latex].
Try It 1
1. Use figure 3 to graph the function [latex]f(x)=\frac{1}{x}+3[/latex]. What is the horizontal asymptote of the function? What is the relationship between the value of [latex]k[/latex] and the horizontal asymptote? What is the vertical asymptote of the function?
2. Without graphing the function, what is the horizontal asymptote of the function [latex]g(x)=\frac{1}{x}-7[/latex]? What is the vertical asymptote?
Example 2
The graph of the rational function [latex]y=f(x)[/latex] has a horizontal asymptote at [latex]y=9[/latex] and a vertical asymptote at [latex]x=0[/latex]. Determine a rational functional [latex]f(x)[/latex] whose graph meets these criteria.
Solution
Since the vertical asymptote is at [latex]x=0[/latex] there is no horizontal shift from the parent function [latex]f(x)=\dfrac{1}{x}[/latex]. Since the horizontal asymptote is at [latex]y=9[/latex], there has been a vertical shift of 9 units up from the parent function [latex]f(x)=\dfrac{1}{x}[/latex]. This means that [latex]k=9[/latex], and the function is [latex]f(x)=\dfrac{1}{x}+9[/latex].
Try It 2
The graph of the rational function [latex]y=f(x)[/latex] has a horizontal asymptote at [latex]y=-12[/latex] and a vertical asymptote at [latex]x=0[/latex]. Determine a rational functional [latex]f(x)[/latex] whose graph meets these criteria.
Horizontal Shifts
If we shift the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] right 7 units, all of the points on the graph increase their [latex]x[/latex]-coordinates by 7, but their [latex]y[/latex]-coordinates remain the same. The point (1, 1) in the original graph is moved to (8, 1). Any point [latex](x, y)[/latex] on the original graph is moved to [latex](x+7, y)[/latex](figure 4).
But what happens to the original function [latex]f(x)=\dfrac{1}{x}[/latex]? An automatic assumption may be that since [latex]x[/latex] moves to [latex]x+7[/latex] that the function will become [latex]f(x)=\dfrac{1}{x+7}[/latex]. But that is NOT the case. Remember that the [latex]x[/latex]-intercept is moved to (8, 1) and if we substitute [latex]x=8[/latex] into the function [latex]f(x)=\dfrac{1}{x+7}[/latex] we get [latex]f(x)=\dfrac{1}{15} \neq 1[/latex]!! The way to get a function value of 1 is for the transformed function to be [latex]f(x)=\dfrac{1}{x-7}[/latex]. Then[latex]f(8)=\dfrac{1}{8-7}=1[/latex]. So the function [latex]f(x)=\dfrac{1}{x}[/latex] transforms to [latex]f(x)=\dfrac{1}{x-7}[/latex] after being shifted 7 units to the right. The reason is that when we move the function 7 units to the right, the [latex]x[/latex]-value increases by 7 and to keep the corresponding [latex]y[/latex]-coordinate the same in the transformed function, the [latex]x[/latex]-coordinate of the transformed function needs to subtract 7 to get back to the original [latex]x[/latex] that is associated with the original [latex]y[/latex]-value. Table 3 shows the changes to specific values of this function, and the graph is shown in figure 4.
| [latex]x[/latex] | [latex]x-7[/latex] | [latex]f(x)=\dfrac{1}{x-7}[/latex] | 
Figure 4. Shifting the graph of the function [latex]f(x)=\frac{1}{x}[/latex] right 7 units. |
|---|---|---|---|
| [latex]5[/latex] | [latex]-2[/latex] | [latex]-\frac{1}{2}[/latex] | |
| [latex]6[/latex] | [latex]-1[/latex] | [latex]-1[/latex] | |
| [latex]\frac{13}{2}[/latex] | [latex]-\frac{1}{2}[/latex] | [latex]-2[/latex] | |
| [latex]\frac{15}{2}[/latex] | [latex]\frac{1}{2}[/latex] | [latex]2[/latex] | |
| [latex]8[/latex] | [latex]1[/latex] | [latex]1[/latex] | |
| [latex]9[/latex] | [latex]2[/latex] | [latex]\frac{1}{2}[/latex] | |
| [latex]10[/latex] | [latex]3[/latex] | [latex]\frac{1}{3}[/latex] | |
| Table 3. Shifting the graph right by 7 units transforms [latex]f(x)=\frac{1}{x}[/latex] into [latex]f(x)=\frac{1}{x-7}[/latex] . | |||
On the other hand, if we shift the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] left by 11 units, all of the points on the graph decrease their [latex]x[/latex]-coordinates by 11, but their [latex]y[/latex]-coordinates remain the same. So any point [latex](x, y)[/latex] on the original graph moves to [latex](x-11, y)[/latex]. Consequently, to keep the same [latex]y[/latex]-values we need to increase the [latex]x[/latex]-value by 11 in the transformed function. The equation of the function after being shifted left 11 units is [latex]f(x)=\dfrac{1}{x+11}[/latex]. Table 4 shows the changes to specific values of this function, and the graph is shown in figure 5.
| [latex]x[/latex] | [latex]x+11[/latex] | [latex]f(x)=\dfrac{1}{x+11}[/latex] |
 Figure 5. Shifting the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] left 11 units. |
|---|---|---|---|
| [latex]–13[/latex] | [latex]-2[/latex] | [latex]-\frac{1}{2}[/latex] | |
| [latex]-12[/latex] | [latex]-1[/latex] | [latex]-1[/latex] | |
| [latex]-\frac{23}{2}[/latex] | [latex]-\frac{1}{2}[/latex] | [latex]-2[/latex] | |
| [latex]-\frac{21}{2}[/latex] | [latex]\frac{1}{2}[/latex] | [latex]2[/latex] | |
| [latex]-10[/latex] | [latex]1[/latex] | [latex]1[/latex] | |
| [latex]-9[/latex] | [latex]2[/latex] | [latex]\frac{1}{2}[/latex] | |
| [latex]-8[/latex] | [latex]3[/latex] | [latex]\frac{1}{3}[/latex] | |
| Table 4. Shifting the graph left by 11 units transforms [latex]f(x)=\dfrac{1}{x}[/latex] into [latex]f(x)=\dfrac{1}{x+11}[/latex]. | |||
horizontal shifts
We can represent a horizontal shift of the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] by subtracting a constant, [latex]h[/latex], from the variable [latex]x[/latex].
[latex]f(x)=\dfrac{1}{x-h}[/latex]
If [latex]h>0[/latex] the graph shifts toward the right and if [latex]h<0[/latex] the graph shifts to the left.
Manipulate the graph in figure 6 to shift the graph horizontally. Pay attention to what happens with the function as [latex]h[/latex] changes value. Also, watch what happens with the asymptotes.
Figure 6. Horizontal shifts
Example 3
1. Use figure 6 to graph the function [latex]f(x)=\frac{1}{x+6}[/latex]. What is the vertical asymptote of the function? What is the relationship between the value of [latex]h[/latex] and the vertical asymptote? What is the horizontal asymptote of the function?
2. Without graphing the function, what is the vertical asymptote of the function [latex]g(x)=\frac{1}{x+3}[/latex]? What is the horizontal asymptote?
Solution
1. 
The vertical asymptote is the line [latex]x=-6[/latex]. The value of [latex]h[/latex] is [latex]-6[/latex] so the vertical asymptote mimics the value of [latex]h[/latex].
The horizontal asymptote is the line [latex]y=0[/latex].
2. Since [latex]h=-3[/latex] the graph of [latex]f(x)=\frac{1}{x}[/latex] gets shifted left by 3 units. This means that the vertical asymptote is shifted left 3 units to [latex]x=-3[/latex]. The horizontal asymptote does not move so is still [latex]y=0[/latex].
Try It 3
1. Use figure 6 to graph the function [latex]f(x)=\frac{1}{x+3}[/latex]. What is the vertical asymptote of the function? What is the relationship between the value of [latex]h[/latex] and the vertical asymptote? What is the horizontal asymptote of the function?
2. Without graphing the function, what is the vertical asymptote of the function [latex]g(x)=\frac{1}{x-7}[/latex]? What is the horizontal asymptote?
Example 4
The graph of the rational function [latex]y=f(x)[/latex] has a horizontal asymptote at [latex]y=0[/latex] and a vertical asymptote at [latex]x=5[/latex]. Determine a rational functional [latex]f(x)[/latex] whose graph meets these criteria.
Solution
Since the horizontal asymptote is at [latex]y=0[/latex] there is no vertical shift from the parent function [latex]f(x)=\dfrac{1}{x}[/latex]. Since the vertical asymptote is at [latex]x=5[/latex], there has been a horizontal shift of 5 units right from the parent function [latex]f(x)=\dfrac{1}{x}[/latex]. This means that [latex]h=5[/latex], and the function is [latex]f(x)=\dfrac{1}{x-5}[/latex].
Try It 4
The graph of the rational function [latex]y=f(x)[/latex] has a horizontal asymptote at [latex]y=0[/latex] and a vertical asymptote at [latex]x=-9[/latex]. Determine a rational functional [latex]f(x)[/latex] whose graph meets these criteria.
Combining Vertical and Horizontal Shifts
vertical and horizontal shifts
We can represent both a vertical and a horizontal shift of the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] by the transformed function
[latex]f(x)=\dfrac{1}{x-h}+k[/latex]
If [latex]h>0[/latex] the graph shifts toward the right and if [latex]h<0[/latex] the graph shifts to the left.
If [latex]k>0[/latex] the graph shifts upwards and if [latex]k<0[/latex] the graph shifts downwards.
When the parent function [latex]f(x)=\dfrac{1}{x}[/latex] is shifted both vertically and horizontally, there will be non-zero values for both [latex]h[/latex] and [latex]k[/latex]. The graph in figure 7 can be manipulated both vertically and horizontally. Pay attention to what happens to the function as the values for [latex]h[/latex] and [latex]k[/latex] change. Also notice what happens to the asymptotes.
Figure 7. Vertical and horizontal shifts
Example 5
The graph of the parent function [latex]f(x)=\dfrac{1}{x}[/latex] is shifted up by 4 units and left by 7 units.
1. Determine the equation of the transformed function.
2. Determine the vertical asymptote.
3. Determine the horizontal asymptote.
4. The point [latex](2, \frac{1}{2})[/latex] lies on the parent function. Where does it end up after the transformation?
Solution
- Shifted up by 4 units means [latex]k=4[/latex] and shifted left by 7 units means [latex]h=-7[/latex]. Therefore the function is [latex]f(x)=\dfrac{1}{x-h}+k=\dfrac{1}{x+7}+4[/latex].
- Since the graph is shifted left 7 units, the vertical asymptote is also shifted left 7 units: [latex]x=-7[/latex].
- Since the graph is shifted up 4 units, the horizontal asymptote is also shifted up 4 units: [latex]y=4[/latex].
- Since the graph is shifted left 7 units, the [latex]x[/latex]-coordinate is shifted left 7 units to 2 – 7 = –5. Since the graph is shifted up 4 units, the [latex]y[/latex]-coordinate is shifted up 4 units to [latex]\frac{1}{2}+4=\frac{9}{2}[/latex]. Consequently,[latex](2, \frac{1}{2})[/latex] moves to [latex](-5, \frac{9}{2})[/latex].
Try It 5
The graph of the parent function [latex]f(x)=\dfrac{1}{x}[/latex] is shifted down by 3 units and right by 2 units.
1. Determine the equation of the transformed function.
2. Determine the vertical asymptote.
3. Determine the horizontal asymptote.
4. The point [latex](2, \frac{1}{2})[/latex] lies on the parent function. Where does it end up after the transformation?
Stretching and Compressing
If we vertically stretch the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] by a factor of 7, all of the[latex]y[/latex]-coordinates of the points on the graph are multiplied by 7, but their [latex]x[/latex]-coordinates remain the same. The equation of the function after the graph is stretched is [latex]f(x)=7\times\dfrac{1}{x}=\dfrac{7}{x}[/latex]. The reason for multiplying [latex]f(x)=\dfrac{1}{x}[/latex] by 7 is that each [latex]y[/latex]-coordinate is made 7 times larger, and since [latex]y=\dfrac{1}{x}[/latex], [latex]\dfrac{1}{x}[/latex] is also made 7 times larger. Table 5 shows this change and the graph is shown in figure 8.
| [latex]x[/latex] | [latex]\dfrac{1}{x}[/latex] | [latex]f(x)=\dfrac{7}{x}[/latex] |
 Figure 8. Stretching the graph vertically. |
|---|---|---|---|
| [latex]-2[/latex] | [latex]-\frac{1}{2}[/latex] | [latex]-\frac{7}{2}[/latex] | |
| [latex]-1[/latex] | [latex]-1[/latex] | [latex]-7[/latex] | |
| [latex]-\frac{1}{2}[/latex] | [latex]-2[/latex] | [latex]-14[/latex] | |
| [latex]\frac{1}{2}[/latex] | [latex]2[/latex] | [latex]14[/latex] | |
| [latex]1[/latex] | [latex]1[/latex] | [latex]7[/latex] | |
| [latex]2[/latex] | [latex]\dfrac{1}{2}[/latex] | [latex]\frac{7}{2}[/latex] | |
| [latex]3[/latex] | [latex]\dfrac{1}{3}[/latex] | [latex]\frac{7}{3}[/latex] | |
| Table 5. Stretching the graph vertically 7 times transforms [latex]\dfrac{1}{x}[/latex] into [latex]\dfrac{7}{x}[/latex]. | |||
On the other hand, if we vertically compress the graph of the function [latex]\dfrac{1}{x}[/latex] into [latex]\dfrac{1}{5}[/latex] of its original height, all of the [latex]y[/latex]-coordinates of the points on the graph are divided by 5, but their [latex]x[/latex]-coordinates remain the same. This means the [latex]y[/latex]-coordinates are multiplied by [latex]\dfrac{1}{5}[/latex]. The equation of the function after being compressed is [latex]f(x)=\dfrac{1}{5}\times\dfrac{1}{x}=\dfrac{1}{5x}[/latex]. The reason for multiplying [latex]f(x)=\dfrac{1}{x}[/latex] by [latex]\dfrac{1}{5}[/latex] is that each [latex]y[/latex]-coordinate becomes [latex]\dfrac{1}{5}[/latex] of the original value. Table 6 shows this change and the graph is shown in figure 9.
| [latex]x[/latex] | [latex]\dfrac{1}{x}[/latex] | [latex]\dfrac{1}{5x}[/latex] |
 Figure 9. Compress the graph into [latex]\frac{1}{5}[/latex] of the original height. |
|---|---|---|---|
| [latex]-2[/latex] | [latex]-\dfrac{1}{2}[/latex] | [latex]-\dfrac{1}{10}[/latex] | |
| [latex]-1[/latex] | [latex]-1[/latex] | [latex]-\dfrac{1}{5}[/latex] | |
| [latex]-\dfrac{1}{2}[/latex] | [latex]-2[/latex] | [latex]-\dfrac{2}{5}[/latex] | |
| [latex]\dfrac{1}{2}[/latex] | [latex]2[/latex] | [latex]\dfrac{2}{5}[/latex] | |
| [latex]1[/latex] | [latex]1[/latex] | [latex]\dfrac{1}{5}[/latex] | |
| [latex]2[/latex] | [latex]\dfrac{1}{2}[/latex] | [latex]\dfrac{1}{10}[/latex] | |
| [latex]3[/latex] | [latex]\dfrac{1}{3}[/latex] | [latex]\dfrac{1}{15}[/latex] | |
| Table 6. Compressing the graph vertically into [latex]\frac{1}{5}[/latex] of the original height transforms [latex]f(x)=\frac{1}{x}[/latex] into [latex]f(x)=\frac{1}{5x}[/latex]. | |||
vertical stretching and compressing
A vertical stretch or compression of the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] can be represented by multiplying the function by a constant, [latex]a>0[/latex].
[latex]f(x)=a\dfrac{1}{x}[/latex]
The magnitude of [latex]a[/latex] indicates the stretch/compression of the graph. If [latex]a>1[/latex], the graph is stretched. If [latex]0<a<1[/latex], the graph is compressed.
Move the red dot in figure 10 to change the value of [latex]a[/latex]. Notice whether the graph is stretching or compressing depending on the value of [latex]a[/latex]. Notice also what happens to the function.
FIgure 10. Stretching and compressing
Example 6
1. The parent function [latex]f(x)=\dfrac{1}{x}[/latex] is stretched by a factor of 5. Determine the equation of the transformed function. The point (1, 1) lies on the parent function. What happens to this point after the transformation?
2. The parent function [latex]f(x)=\dfrac{1}{x}[/latex] is compressed to [latex]\frac{1}{2}[/latex] its size. Determine the equation of the transformed function. The point (1, 1) lies on the parent function. What happens to this point after the transformation?
Solution
1. Stretched by a factor of 5 means [latex]a=5[/latex], therefore, the transformed function is [latex]f(x)=5\left(\dfrac{1}{x}\right)[/latex]. This can also be written as [latex]f(x)=\dfrac{5}{x}[/latex].
When a function is stretched its [latex]x[/latex]-value stays the same while the [latex]y[/latex]-value is multiplied by the stretch factor. So, (1, 1) moves to (1, 5).
2. Compressed to [latex]\frac{1}{2}[/latex] its size means [latex]a=\dfrac{1}{2}[/latex], therefore, the transformed function is [latex]f(x)=\dfrac{1}{2}\left(\dfrac{1}{x}\right)[/latex]. This can also be written as [latex]f(x)=\dfrac{1}{2x}[/latex].
When a function is compressed its [latex]x[/latex]-value stays the same while the [latex]y[/latex]-value is multiplied by the compression. So, (1, 1) moves to [latex](1, \frac{1}{5})[/latex].
Try It 6
1. The parent function [latex]f(x)=\dfrac{1}{x}[/latex] is stretched by a factor of 4. Determine the equation of the transformed function. The point (1, 1) lies on the parent function. What happens to this point after the transformation?
2. The parent function [latex]f(x)=\dfrac{1}{x}[/latex] is compressed to [latex]\frac{1}{8}[/latex]th its size. Determine the equation of the transformed function. The point (1, 1) lies on the parent function. What happens to this point after the transformation?
Reflections
across the [latex]x[/latex]-axis
When the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] is reflected across the [latex]x[/latex]-axis, the [latex]y[/latex]-coordinates of all of the points on the graph change their signs, from positive to negative values or from negative to positive values, while the [latex]x[/latex]-coordinates remain the same. The equation of the function after [latex]f(x)=\dfrac{1}{x}[/latex] is reflected across the [latex]x[/latex]-axis is [latex]f(x)=-\dfrac{1}{x}[/latex]. Table 7 shows the effect of such a reflection on the functions values and the graph is shown in figure 11.
| [latex]x[/latex] | [latex]\dfrac{1}{x}[/latex] | [latex]-\dfrac{1}{x}[/latex] |
 Figure 11. Reflecting the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] across the [latex]x[/latex]-axis. |
|---|---|---|---|
| –2 | [latex]-\dfrac{1}{2}[/latex] | [latex]\dfrac{1}{2}[/latex] | |
| –1 | –1 | 1 | |
| [latex]-\dfrac{1}{2}[/latex] | –2 | 2 | |
| [latex]\dfrac{1}{2}[/latex] | 2 | –2 | |
| 1 | 1 | –1 | |
| 2 | [latex]\dfrac{1}{2}[/latex] | [latex]-\dfrac{1}{2}[/latex] | |
| 3 | [latex]\dfrac{1}{3}[/latex] | [latex]-\dfrac{1}{3}[/latex] | |
| Table 7. Reflecting the graph of [latex]f(x)=\dfrac{1}{x}[/latex] across the [latex]x[/latex]-axis transforms [latex]f(x)=\dfrac{1}{x}[/latex] into [latex]f(x)=-\dfrac{1}{x}[/latex]. | |||
across the [latex]y[/latex]-axis
When the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] is reflected across the [latex]y[/latex]-axis, the [latex]x[/latex]-coordinates of all of the points on the graph change their signs, from positive to negative values, or from negative to positive values, while the [latex]y[/latex]-coordinates remain the same. The equation of the function after [latex]f(x)=\dfrac{1}{x}[/latex] is reflected across the [latex]y[/latex]-axis is [latex]f(x)=\dfrac{1}{-x}=-\dfrac{1}{x}[/latex]. This equation is the same as the equation after being reflected across the [latex]x[/latex]-axis. Therefore, its graph is the same as the graph after being reflected across the [latex]x[/latex]-axis, even though it got there by a different route (note the differences in figures 11 and 13). Table 8 shows the effect of such a reflection on the functions values and the graph is shown in figure 13.
| [latex]x[/latex] | [latex]\dfrac{1}{x}[/latex] | [latex]\dfrac{1}{-x}[/latex] |
 Figure 13. Reflecting the graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] across the [latex]y[/latex]-axis. |
|---|---|---|---|
| –2 | [latex]-\frac{1}{2}[/latex] | [latex]\frac{1}{2}[/latex] | |
| –1 | –1 | 1 | |
| [latex]-\frac{1}{2}[/latex] | –2 | 2 | |
| [latex]\frac{1}{2}[/latex] | 2 | –2 | |
| 1 | 1 | –1 | |
| 2 | [latex]\frac{1}{2}[/latex] | [latex]-\frac{1}{2}[/latex] | |
| 3 | [latex]\frac{1}{3}[/latex] | [latex]-\frac{1}{3}[/latex] | |
| Table 8. Reflecting the graph of [latex]f(x)=\dfrac{1}{x}[/latex] across the [latex]y[/latex]-axis transforms [latex]f(x)=\dfrac{1}{x}[/latex] into [latex]f(x)=\dfrac{1}{-x}[/latex]. | |||
Example 7
Explain why the parent function [latex]f(x)=\dfrac{1}{x}[/latex] is transformed to [latex]f(x)=-\dfrac{1}{x}[/latex] when it is reflected across the [latex]x[/latex]-axis.
Solution
With reflection across the [latex]x[/latex]-axis, the [latex]x[/latex] values stay the same and the [latex]y[/latex]-values change sign. So, [latex]y=\dfrac{1}{x}[/latex] transforms to [latex]-y=\dfrac{1}{x}[/latex]. If we multiply (or divide) both sides of the equation by –1 we get, [latex]y=-\dfrac{1}{x}[/latex]. Consequently, the transformed function is [latex]f(x)=-\dfrac{1}{x}[/latex].
Try It 7
Explain why the parent function [latex]f(x)=\dfrac{1}{x}[/latex] is transformed to [latex]f(x)=-\dfrac{1}{x}[/latex] when it is reflected across the [latex]y[/latex]-axis.
Combining Transformations
Now that we have learned all the transformations for the function [latex]f(x)=\dfrac{1}{x}[/latex], we should be able to write the transformed function equation given specific transformations, and determine what transformations have been performed on the function [latex]f(x)=\dfrac{1}{x}[/latex], given an arbitrary transformed function [latex]f(x)=a\dfrac{1}{x-h}+k[/latex].
Combining Transformations
The graph of the function [latex]f(x)=\dfrac{1}{x}[/latex] can be shifted horizontally ([latex]h[/latex]-value), shifted vertically ([latex]k[/latex]-value), stretched or compressed ([latex]a[/latex]-value), reflected across the [latex]x[/latex]-axis (negative function value} or reflected across the [latex]y[/latex]-axis (negative [latex]x[/latex]-value) and can be represented by the function
[latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k[/latex]
Example 8
The parent function [latex]f(x)=\dfrac{1}{x}[/latex] is transformed. Write the equation of the transformed function.
1. Stretched by a factor of 2; shifted vertically up 3 units; shifted horizontally 5 units right.
2. Shifted horizontally left by 7 units; reflected across the [latex]x[/latex]-axis.
3. Compressed to [latex]\frac{1}{3}[/latex]rd its height; reflected across the [latex]y[/latex]-axis; shifted vertically down by 4 units.
Solution
- [latex]a=2,\;k=3,\;h=5[/latex] so function is [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k=2\left(\dfrac{1}{x-5}\right)+3[/latex]
- [latex]a=-1,\;k=0,\;h=-7[/latex] so function is [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k=-\dfrac{1}{x+7}[/latex]
- [latex]a=\frac{1}{3},\;x\rightarrow -x,\;k=-4[/latex] so function is [latex]f(x)=\dfrac{1}{3}\left(\dfrac{1}{-x}\right)-4=-\dfrac{1}{3}\left(\dfrac{1}{x}\right)-4[/latex]. This can also be written as [latex]f(x)=-\dfrac{1}{3x}-4[/latex].
Try It 8
The parent function [latex]f(x)=\dfrac{1}{x}[/latex] is transformed. Write the equation of the transformed function.
1. Stretched by a factor of 4; shifted vertically down by 7 units; shifted horizontally 3 units left.
2. Shifted horizontally left by 8 units; reflected across the [latex]x[/latex]-axis.
3. Compressed to [latex]\frac{1}{5}[/latex]th its height; reflected across the [latex]x[/latex]-axis; shifted vertically down by 4 units; shifted right by 2 units.
Example 9
Determine the transformations made to the parent function [latex]f(x)=\dfrac{1}{x}[/latex] to get the transformed function [latex]f(x)=-\dfrac{7}{x-4}+3[/latex].
Solution
[latex]f(x)=-\dfrac{7}{x-4}+3[/latex] can be written as [latex]f(x)=-7\left(\dfrac{1}{x-4}\right)+3[/latex].
Comparing [latex]f(x)=-7\left(\dfrac{1}{x-4}\right)+3[/latex] to [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k[/latex], we get [latex]a=-7,\;h=4,\;k=3[/latex], which translated to a reflection across the [latex]x[/latex]-axis, a stretch by a factor of 7, a horizontal shift to the right by 4 units, and a vertical shift up by 3 units.
Try It 9
Determine the transformations made to the parent function [latex]f(x)=\dfrac{1}{x}[/latex] to get the transformed function [latex]f(x)=-\dfrac{5}{3x}-7[/latex].
Formats of Rational Functions
You may be wondering why we defined a rational function as [latex]f(x)=\dfrac{P(x)}{Q(x)}[/latex] when we write a rational function using transformations in the form [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k[/latex].
It is simply that rational functions in the form [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k[/latex] are based on the parent rational function [latex]f(x)=\dfrac{1}{x}[/latex], and a special case of [latex]f(x)=\dfrac{P(x)}{Q(x)}[/latex].
We could take the function [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k[/latex] and write it as a single fraction by using a common denominator:
[latex]\begin{aligned}f(x)&=a\left(\dfrac{1}{x-h}\right)+k\\\\&=a\left(\dfrac{1}{x-h}\right)+k\color{blue}{\cdot \dfrac{x-h}{x-h}}\\\\&=\dfrac{a+k(x-h)}{x-h}\\\\&=\dfrac{kx+(a-hk)}{x-h}\end{aligned}[/latex]
Now the function is written in standard form with [latex]P(x)=kx+(a-hk)[/latex] and [latex]Q(x)=x-h[/latex]. Both [latex]P(x)[/latex] and [latex]Q(x)[/latex] have degree 1.
Consequently, a rational function in the transformed form [latex]f(x)=a\left(\dfrac{1}{x-h}\right)+k[/latex] is a special case of the standard form [latex]f(x)=\dfrac{P(x)}{Q(x)}[/latex] with [latex]P(x)[/latex] and [latex]Q(x)[/latex] having degree 1.
We leave the more in depth study of rational functions in the form [latex]f(x)=\dfrac{P(x)}{Q(x)}[/latex] to College Algebra.
