5.01 Rational parent functions

Identify the function family to which each function belongs.

Recall the shape and characteristics of the parent function of each function family. We have previously seen linear, radical, exponential, absolute value, rational, polynomial, and logarithhmic functions. Characteristics to consider include: domain, range, x-intercept, y-intercept, absolute and relative extrema, asymptotes, and increasing/decreasing intervals.

For f\left(x\right), both logarithmic and square root functions have a minimum or maximum for their domain. Square root functions also have a minimum or maximum for their range. Since f\left(x\right), has no restriction on the range, it must belong to the logarithmic function family.

For g\left(x\right), only rational functions have two intervals on either side of an asymptote. Since both intervals are above y = 0, it must belong to the rational function family with a parent function y = \dfrac{1}{x^{2}}.

Compare the asymptotes of f\left(x\right) and g\left(x\right).

Recall that a function can have both horizontal or vertical asymptotes.

In part (a), we identified f\left(x\right) as a logarithmic function. Since there is no restriction on the range, there is no horizontal asymptote. The vertical asymptote, drawn on the graph, is at x = 0.

In part (a), we identified g\left(x\right) as rational, so we must consider horizontal and veritcal asymptotes.

The vertical asymptote, drawn on the graph, is at x = -4. The horizontal asymptote is at y = 0.

Describe the end behavior of each function.

We want to determine what happens to the y-values when the x-values are very small and very large. Consider how the asymptotes affect end behavior.

First, let's consider f\left(x\right).

Since f\left(x\right) is the logarithmic parent function, as x \to \infty, y \to \infty. For f\left(x\right), the domain is restricted to x\gt 0, so as x \to 0^{+}, y \to -\infty.

Next, let's consider the end behavior of g\left(x\right). As x \to \infty, y \to 0. As x \to -\infty, y \to 0.

When determining end behavior, it is important to consider the domain and the direction. For f\left(x\right), we cannot discuss the behavior of x \to -\infty because the domain is restricted to x\gt 0. We also cannot discuss the behavior of the function as x \to 0^{-} because the function is restricted to values of x greater than 0.

Complete the table.

We substitute each value of R to the inverse variation equation C\left(R\right) = \dfrac {200}{R} to find the value of C.

Now,

\qquadIf R=5, then C\left(5\right)= \dfrac {200}{5}=40.

\qquadIf R=10, then C\left(10\right)= \dfrac {200}{10}=20.

\qquadIf R=20, then C\left(20\right)= \dfrac {200}{20}=10.

\qquadIf R=25, then C\left(25\right)= \dfrac {200}{25}=8.

\qquadIf R=40, then C\left(40\right)= \dfrac {200}{40}=5.

Therefore, the complete table of values is:

Sketch the relationship between the current and resistance.

We assign the horizontal axis for variable R and the vertical axis for variable C.

We will plot the coordinates \left(R,C\right) on the coordinate plane. The points are: \left(5,40\right),\,\left(10,20\right),\,\left(20,10\right),\,\left(25,8\right) \text{ and }\left(40,5\right)

Then, we can graph the function C\left(R\right) = \dfrac {200}{R}

It's important to consider what representation is best for the types of questions we want to answer. The table clearly shows us exact values of the function. In the graph, we can see a few notable things that were less clear in the table of values, such as possible asymptotes or end behavior. Depending on what information we are looking for, one representation may be preferred over another.

Specify any asymptotes and the restrictions on the domain of the function.

Use the graph of the relationship between the current and resistance to determine the asymptotes, then use the asymptotes, equation of the function, and context to determine the domain.

We can see that the values of the function approach \infty as the value of R \to 0. There is a vertical asymptote at R=0.

The values of the function approach 0 as the value of R \to + \infty, and they will never reach 0. There is a horizontal asymptote at C\left(R\right)=0.

At the vertical asymptote, R=0, the function is undefined since we cannot evaluate C\left(0\right)=\dfrac{200}{0}, so we will exclude R=0 from the domain of the function.

Identify the domain and range.

The domain represents the x-values of the function, and the range represents the y-values of the function. Recall that the graph of a rational functions can have asymptotes and discontinuities.

From the graph, we can see that x is defined for all values on either side of the vertical asymptote at x=-6. Therefore, the domain is \left(-\infty, -6 \right) \cup \left(-6, \infty \right).

Similarly, we can see that y is defined for all values on either side of the horizontal asymptote at y=-3. Therefore, the range is \left(-\infty, -3 \right) \cup \left(-3, \infty \right).

We can also represent domain and range using set notation:

Domain: \{x \vert x \neq -6 \} or \{x \vert x \lt -6\text{ or }x \gt -6\}

Range: \{y \vert y \neq -3 \} or \{y \vert y \lt -3\text{ or } y \gt -3\}

We can also represent domain and range using inequalities:

Domain: All x such that x \neq -6

Range: All y such that y \neq -3

These different representations of domain and range can be used interchangeably.

Identify the increasing and decreasing intervals.

We want to determine what happens to the y-values for different values of x. We must consider points where the direction of the function could change.

Notice our graph is separated into two intervals. Starting from the left side of the first interval, we can see y is decreasing until we get to the vertical asymptote at x=-6.

Continuing from left to right on the second interval, we can see y is decreasing for the remainder of the domain.

Therefore, f\left(x\right) has no increasing intervals and is decreasing on the intervals \left( -\infty, -6 \right) \cup \left( -6, \infty \right).

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.

Any rational function of the form f\left(x\right) = \dfrac{a}{x-h} +k will only increase or decrease, not both, depending on the sign of a.

Identify the zero(s) of the function.

The zeros of a function occur when f\left(x\right) = 0. These are also the location of the x-intercepts.

There is only one zero at \left( -4, 0 \right) since this is the only point where f\left(x\right) touches the x-axis.

Identify the increasing and decreasing intervals.

This graph has two separate intervals. For each interval, move from left to right across the graph and determine whether the y-values are increasing or decreasing.

Starting from the left side of the graph for the first interval \left( - \infty, 0 \right), we can see y is increasing until we get to the vertical asymptote at x=0.

Continuing from left to right on the second interval \left( 0, \infty \right), we can see y is decreasing for the remainder of the domain.

Therefore, f\left(x\right) is increasing on the interval \left( - \infty, 0 \right) and decreasing on the interval \left( 0, \infty \right).

The behavior of increasing and decreasing intervals can help us determine whether a rational function has a parent function f\left(x\right) = \dfrac{1}{x} or f\left(x\right) = \dfrac{1}{x^2}.

From the previous example, notice that functions of the form f\left(x\right) = \dfrac{a}{x-h}+k have only increasing or decreasing intervals, but functions of the form f\left(x\right) = \dfrac{a}{\left(x-h\right)^{2}} +k have both increasing and decreasing intervals.

Identify the intercepts of the function.

The x-intercept is when the line touches or goes through the x-axis and the y-intercept is when the line touches or goes through the y-axis.

This function has no intercepts. It crossed neither the x-axis nor the y-axis.

Given the vertical asymptote at x=0, the function gets increasingly close to 0 but never actually reaches the y-axis. The horizontal asymptote at y=2 means that the function can never reach y-values below 2.

Describe the end behavior of the function as x\to-\infty and as x\to\infty.

We want to determine what happens to the y-values when the x-values get increasingly small or increasingly large.

As x gets very small, y approaches the horizontal asymptote y=2. So, as x\to -\infty, y\to 2.

As x gets very large, y again approaches the horizontal asymptote y=2. So, as x\to \infty, y\to 2.

Understanding the end behavior helps in sketching the graph's "tails" and gives a sense of the function's overall end shape.

What is the transformation of the parent function f\left(x\right) = \dfrac{1}{x}?

We can use the equation of the function to identify the transformation of the function, comparing the equation to f\left(x\right)=\dfrac{a}{x-h} +k.

We know there is no stretch, compression, or reflection because the numerator for g\left(x\right) remains the same as f\left(x\right)=\dfrac{1}{x}. The value of h=1 shifts the function horizontally, so we can state a horizontal translation 1 unit to the right. There was no vertical translation as k=0.

Therefore, g\left(x\right) is has been translated 1 unit to the right.

Complete the table of values.

Substitute each of the values of x in the equation to solve for the values of g\left(x\right).

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

Sketch a graph of the function.

We can translate the parent function f\left(x\right)=\dfrac{1}{x} one unit to the right, using the points from the table of values to help draw the curve.

Based on the equation, we can see there will be a vertical asymptote at x=1, where the function is undefined. The function can also be written as g\left(x\right)=\dfrac{1}{x-1} + 0, so the horizontal asymptote is at g\left(x\right)=0.

Looking at the graph, and in particular the location of the vertical asymptote, we can see that the function g\left(x\right) has been translated 1 unit to the right from the parent function f\left(x\right)=\dfrac{1}{x}.

The points in the table of values indicate that the vertical asymptote is between x = 0.5 and x=1.5, as the values change from decreasing negative values to decreasing positive values. We can confirm that the asymptote is the line x = 1 by looking at the equation of the function, g\left(x\right) = \dfrac{1}{x - 1}, which is undefined at x=1.

Describe the transformation(s) used to get from the graph of y = \dfrac{1}{x} to the graph of this function.

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 no reflections have occurred.

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.

Shifting the parent function to the left 3 units, we can compare our graph to determine if any dilation has occurred:

Notice each y-value has been doubled to get our function. So the function has been dilated by a scale factor of 2.

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

Determine an equation for the function shown in the graph.

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:

• Vertical stretch by a factor of 2

• Horizontal translation of 3 units to the left

We can also identify the values of a,h, and k in the transformation form f\left(x\right) = \dfrac{a}{x-h} + k.

A dilation by a scale factor of 2 means a=2. A horizontal translation 3 units left means h=-3. Since we do not have any vertical translation, k=0. Substituting those into the equation, we have:

f\left(x\right) = \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 general, we apply reflections and dilations first, then translations.

Sketch a graph of the function.

For the parent function f\left(x\right) = \dfrac{1}{x^2}, this function can be expressed as y = -5 \cdot f\left(x - 3\right), which is a horizontal translation of 3 units right, a dilation with a scale factor of 5, and a reflection across the x-axis.

The graph of f\left(x\right) = \dfrac{1}{x^2} is shown on the same coordinate plane:

What are the equations of the asymptotes of the function?

We know that this function has been dilated with a scale factor of 5, reflected across the x-axis, and horizontally translated 3 units right. The dilation and reflection do not affect the asymptotes; only translations can affect asymptotes.

Since there was no vertical translation, the horizontal asymptote is the same as the parent function. The vertical asymptote has translated 3 units right, which we can see in the graph.

The vertical asymptote has the equation x = 3 and the horizontal asymptote has the equation y=0.

We can use the equation to find that asymptotes as well. Given that there is no x-value that makes the equation equal to 0, the horizontal asymptote occurs at y=k and since k=0, we can confirm that the horizontal asymptote occurs at y=0.

Since x=3 is a zero of the denominator, not the numerator, we can also confirm that the vertical asymptote occurs at x=3.

Using interval notation, what is the domain and range of the function?

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.

The function has been reflected, so the range will be all negative values of y.

The domain of this function is all real values of x except for 3, and the range is all real values of y less than 0.

We can express this using interval notation as

• Domain: \left(-\infty, 3\right) \cup \left(3, \infty\right)

• Range: \left(-\infty, 0\right)

Graph f\left(x\right) using transformations of the parent function.

Let's begin by making a table of values for the parent function, y=\dfrac{1}{x}:

Then, we will identify and apply the transformations that have occurred to get the given function.

The transformations from the parent function are a vertical stretch with a scale factor of 3 and a vertical shift 2 units down.

The scale factor multiplies to the y-values and the translation subtracts from the y-values, in that order. We can apply these transformations to create a new table of values for f\left(x\right):

As there has been no horizontal translation, the vertical asymptote will remain at x=0. Due to the vertical translation, the horizontal asymptote will be at y=-2.

Now, we can graph the function:

Graph the inverse function f^{-1}\left(x\right).

To graph the inverse of f\left(x\right), reflect the graph of f\left(x\right) over the line y=x. We can use the table of values from part (a) and interchange the inputs and outputs to create a table of values for the inverse.

We can set up a table where f^{-1}\left(x\right) is reflected across the line y=x. When a coordinate pair is reflected across the line y=x, the x and y values are switched.

Use the table of values to graph f^{-1}\left(x\right).

There are some special relationships between certain functions and their inverse.

The inverse of \dfrac{1}{x} is also \dfrac{1}{x}.

If we attempt to find the inverse of \dfrac{1}{x^2}, we will see it is not a function. Consider the table of values for f\left(x\right) = \dfrac{1}{x^{2}}:

Switching the coordinates, we can create a table of values for f^{-1}\left(x\right):

However, now the same x-values map to different y-values, and f^{-1}\left(x\right) is not a function.