Evaluating the expression \dfrac{1}{x - 1} at each value of x we get:

We can plot the points from the table of values and use this to help draw the curve.

The points in the table of values indicate that the vertical asymptote is between x = 0.5 and x = 1.5. We can confirm that the asymptote is the line x = 1 by looking at the equation of the function: y = \dfrac{1}{x - 1} is undefined at x = 1.

Looking at the graph, and in particular the location of the vertical asymptote, we can see that the function has been translated 1 unit to the right.

For the parent function f\left(x\right) = \dfrac{1}{x}, this function can be expressed as y = f\left(x\right) + 2, which is a vertical translation of 2 units upwards.

The graph of f\left(x\right) = \dfrac{1}{x} is shown on the same coordinate plane:

We know that this function is a vertical translation of the parent function by 2 units upwards. So the vertical asymptote hasn't changed, while the horizontal asymptote has translated 2 units upwards, which we can see in the graph.

The vertical asymptote has equation x = 0 and the horizontal asymptote has equation y = 2.

We can see that the domain of this rational function will be all values of x except for the value at the vertical asymptote, where the function is undefined.

Similarly, the range of this function will be all values of y except for the value at the horizontal asymptote, which is the one value the function does not obtain.

The domain of this function is "all real values of x except for 0", and the range is "all real values of y except for 2".

We can express this using interval notation as

• Domain: \left(-\infty, 0\right) \cup \left(0, \infty\right)
• Range: \left(-\infty, 2\right) \cup \left(2, \infty\right)

It is clearest to see how the function changes by looking at its graph. In this case, we can see that both branches of the function are decreasing curves - to the left of the asymptote, the function decreases at an increasing rate, while to the right of the asymptote the function decreases at a decreasing rate.

So the function has two decreasing intervals, \left(-\infty, 0\right) and \left(0, \infty\right), and no increasing intervals.

Note that even though the function is decreasing at every point in its domain, the domain is formed from two disconnected intervals (which are separated by the asymptote). So we cannot say that it has only one decreasing region.

We should check for each type of transformation: translations (vertical and/or horizontal), vertical stretch or compression, and reflection.

It may help to add the graph of y = \dfrac{1}{x} to the same coordinate plane:

We can see that the curve lies in the upper-right and lower-left sections, relative to the asymptotes, which is the same as that of y = \dfrac{1}{x}, so there has been no reflections.

The vertical asymptote of the graph is x = -3, which is 3 units to the left of the vertical asymptote of y = \dfrac{1}{x}. Their horizontal asymptotes are both the same (along the x-axis). So there has been a horizontal translation of 3 units to the left, and no vertical translation.

We can also see that the rate of change of the two graphs are different. Translating the point \left(1, 1\right) on the parent function to the left 3 units would result in the point \left(-2, 1\right), but the corresponding point on the graph is above it, at \left(-2, 2\right):

So there has been a vertical stretch by a factor of 2.

Reflections and translations of rational functions are relatively straightforward to see by looking at the graphs and the asymptotes. Vertical stretcha and compression can be less obvious, however, so make sure to check multiple points to confirm the vertical stretch.

For instance, we can also check the point \left(-0.5, -2\right) on the parent function:

We can confirm that a horizontal translation of 3 units to the left followed by a vertical stretch by a factor of 2 also matches for this point.

We can use the transformations that we described in part (a) and apply them to the function y = \dfrac{1}{x} to get an equation for the function shown.

Starting with y = \dfrac{1}{x}, we can apply the transformations one at a time:

So the equation of the function is y = \dfrac{2}{x + 3}.

It is important to note that the order of transformations matters in some cases. For example, applying a vertical translation and then a reflection across the x-axis will be different to reflecting first and then applying the same vertical translation.

In this case, the transformations of a vertical stretch and a horizontal translation are independant and can be applied in any order.

To identify the transformations involved, we will first aim to locate the horizontal and vertical asymptotes. This will directly tell us about any horizontal or vertical translations, and will also help to determine if there has been a reflection or stretch/compression.

Looking at the table of values, we can immediately see that the function is undefined at x = 0. So the vertical asymptote of the function is at x = 0, and this means there has been no horizontal translation.

Looking at the values on either side, we can see that they are symmetrically above and below the value y = -4. So the horizontal asymptote will be the line y = -4, and this means there has been a vertical translation of 4 units downwards.

With this in mind, we can also see that for values of x less than 0 (to the left of the vertical asymptote), the function values are greater than -4, i.e. they lie above the horizontal asymptote. For values of x greater than 0 (to the right of the horizontal asymptote), the function values are less than -4, i.e. they lie below the horizontal asymptote. This is opposite behaviour to that of y = \dfrac{1}{x}, and so there has been a reflection across the x-axis.

Finally, we can see that when x = 1, the function value is -7, which is 3 units below the horizontal asymptote. Similarly, when x = -1 the function value is -1, which is 3 units above the horizontal asymptote. So there has been a vertical stretch by a factor of 3.

We can confirm these transformations by creating a graph using the points in the table of values, and comparing it to the graph of y = \dfrac{1}{x}:

We can use the transformations that we described in part (a) and apply them to the function y = \dfrac{1}{x} to get an equation for the function shown.

Note that it is important to apply these transformations in the correct order. In particular, we will perform the reflection and stretch first, before translating the center of the function away from the origin.

Starting with y = \dfrac{1}{x}, we can apply the transformations one at a time:

So the equation of the function is y = -\dfrac{3}{x} - 4.