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# Rational Functions

Do you ever think of that one person who is totally odd in thought and decision? Such people would pick hot drinks on a hot day rather than cold drinks on a scorching day. Well, not everyone is rational, hence hereafter, you would be learning about rational functions. Let's get rational!

## Definition of a Rational Function

A rational function is a function made up of two polynomials, one divided by the other. They look a little something like this:

$$f(x)=\frac{Q(x)}{P(x)}$$

In other words, a rational function is any function that can be expressed as a fraction, where the numerator and the denominator are both polynomials, where the denominator is not equal to 0.

The denominator $$Q(x)$$ needs to be at least of degree one. This means that $$Q(x)$$ cannot be just a constant; it must involve a variable.

$$f(x)=\dfrac{2+x^{2}}{x}$$- Is a Fraction

$$g(x)=\dfrac{2+x^{2}}{5}$$- Is not a Fraction

Remember that the degree of a polynomial in one variable is the highest exponent of the powers that comprise it. When there is no stated exponent on the variable, then the exponent is 1 and if a polynomial is equal to a constant, its degree is 0.

There may be values of $$x$$ for which $$Q(x) = 0$$. Because dividing by zero leads to an undefined expression, the rational function is not defined at that point.

$$f(x)=\dfrac{x^{2}}{x-3}$$

For this example, rational function, the denominator $$Q(x) = x-3$$ becomes $$0$$ when $$x = 3$$. Therefore, this rational function is not defined for $$x=3$$.

An important example of a rational function is $$y=1/x$$ which is the parent rational function. The graph of the parent rational function is called a hyperbola, and it comprises two symmetrical parts called branches, this is shown below. The straight lines represent the function's asymptotes.

Fig. 1 - The graph of the rational function $$y=\frac{1}{x}$$ - StudySmarter Originals

## The Rational Function Equation

The rational function equation is one of the simpler equations to recognize. The general form of rational functions is as follows:

$$f(x)=\frac{p(x)}{q(x)}$$

It's important to note that rational functions are just rational equations inside of a function, such that we can plot a graph of the equation. Any time you see the word rational function, you should realize that it will be a fraction and that the numerator and the denominator will each be polynomials that you will want to simplify.

Rational Equations can be solved by cross multiplying. More details can be found in the article Solving rational equations.

## Examples of How to simplify Rational Functions

As mentioned before, all rational functions can be written as a fraction of polynomial functions. And, just as with other fractions, a rational function can be simplified if the numerator and the denominator have common factors. Simplifying the rational function makes it more manageable, so it is recommended to simplify whenever it is possible. Let's take a look at a specific example to illustrate how we can simplify these functions.

$$f(x)=\frac{x^{2}-5x-24}{x^{2}-8x-33}$$

Solution:

In this case, we have $$P(x)=x^{2}-5x-24$$ and $$Q(x)=x^{2}-8x-33$$.

To see if we can simplify this rational function, we can factor $$x^{2}-5x-24$$ as $$(x+3)(x-8)$$.

Similarly $$Q(x)$$ can be rewritten as $$(x+3)(x-11)$$.

Putting that altogether gives us the following:

$$f(x)=\frac{(x+3)(x-8)}{(x+3)(x-11)}$$

Notice that the numerator and the denominator of this rational expression have as common factor $$(x+3)$$. Therefore, these factors cancel each other out, leaving us with only $$(x-8)$$in the numerator and $$(x-11)$$in the denominator.

$$f(x)=\left(\frac{x-8}{x-11}\right)$$

Simplify the following rational function:

$$\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}$$

Solution:

Let the two be functions as $$p(x)=x^{2}+10x+24$$ and $$q(x)=x^{3}-x^{2}-20x$$

We will simplify both of them individually and then take the ratio of them at the end.

Factorising $$p(x)$$:

$$\begin{split} p(x) &=x^2 +10x+24 \\ &=x^2 +6x+4x+24 \\ &=x(x+6)+4(x+6) \\ &=(x+4)(x+6) \end{split}$$

All we did is splitting the middle term and then taking out the common factor.

Now, factorising $$q(r)$$:

$$\begin{split} q(x) &=x^3 -x^2-20x \\ &=x(x^2 - x -20) \\ &=x(x^2 -5x+4x-20) \\ &=x(x-5)(x+4) \end{split}$$

Here we took out $$x$$ as a common factor and then split the middle term.

Therefore, we can evaluate $$\frac{x^2 +10x+24}{x^3 -x^2 -20}$$:

$$\begin{split} \frac{x^2 +10x+24}{x^3 -x^2 -20} &=\frac{(x+4)(x+6)}{x(x-5)(x+4)} \\ &=\frac{x+6}{x(x-5)} \end{split}$$

## How do we Find the Inverse of a Rational Function?

Before going through the steps for finding the inverse of a rational function, let's review what an inverse function is.

An Inverse function is essentially the opposite of a given function. In other words, an inverse function $$f^{-1}(x)$$ reverses the inputs and outputs of the original function $$f(x)$$. This is true for any function and does not only apply to rational ones.

To find the inverse of a rational function $$f(x)$$ there are four steps to follow:

1. We replace $$f(x)$$ with $$y$$.
2. We swap every $$x$$ and every $$y$$ in the equation.
3. We solve the new equation for $$y$$ in terms of $$x$$.
4. Finally, we replace $$y$$ with $$f^{-1}(x)$$ which indicates this is the inverse function of the original $$f(x)$$.

$$f(x)=\frac{x-8}{x-11}$$

2. Next we need to replace the $$f(x)$$ with $$y$$:

$$y=\frac{x-8}{x-11}$$

3. Every $$x$$ and $$y$$ need to be swapped around:

$$x=\frac{y-8}{y-11}$$

4. We now solve this equation for $$y$$ in terms of $$x$$. This will take a few steps, as it is not something we can complete in one go. The first thing to do is multiply both sides by $$(y-11)$$:

$$y-8=x(y-11)$$

Then we distribute $$x$$ on the right-hand side:

$$y-8=xy-11x$$

Then, we subtract $$xy-8$$ from both sides:

$$y(1-x)=8-11x$$

Finally, we can divide both sides by $$1-x$$:

$$y=\frac{8-11x}{1-x}$$

5. The last thing we need to do is replace $$y$$ with $$f^{-1}(x)$$ which indicates the fact that it is the inverse of the original rational function $$f(x)$$.

$$f^{-1}(x)=\frac{8-11x}{1-x}$$

To check whether the obtained answer is correct or not, we can take a value of $$x$$ and evaluate it at $$f(x)$$:

$$f(0)=\frac{0-8}{0-11}=\frac{8}{11}$$

Now we know that $$f(0)=8/11$$, so we need to check if we get $$f^{-1}(8/11)=0$$ from our inverse equation:

$$f^{-1}\left(\frac{8}{11}\right)=\frac{8-11 \left( \frac{8}{11} \right)}{1-\frac{8}{11}}=0$$

Hence our answer is indeed correct.

## Asymptotes of rational functions

Details about Asymptotes can be found here. As a summary:

An asymptote is a straight line that approaches a curve, but the curve never actually crosses or touches it.

Rational functions can have three types of asymptotes: horizontal, vertical, and oblique. While the curve of a rational function will never come into contact with its asymptotes, they are useful in that they help us predict how a graph will behave at extreme values of $$x$$.

### Vertical Asymptotes

A rational function has a vertical asymptote at any point where its denominator is equal to $$0$$. Since the denominator is a polynomial $$Q(x)$$, a rational function can have multiple vertical asymptotes. This is also why we cannot have a rational function where the denominator is always equal to $$0$$, there would be no graph!

### Horizontal Asymptotes

Rational functions can only have either one or no horizontal asymptotes. There are three things to check to determine whether or not the rational function has a horizontal asymptote:

1. If the degree of the polynomial in the numerator $$P(x)$$ is greater than the degree of the polynomial in the denominator $$Q(x)$$ then the function does not have a horizontal asymptote.

2. If the degree of the numerator is lower than the degree of the denominator, then there is one horizontal asymptote at $$y=0$$. In other words, the x-axis is the horizontal asymptote.

3. If both polynomials' degrees are the same, then the rational function has one horizontal asymptote and its equation is determined by the leading coefficients of the polynomials.

$$y=\frac{\text{leading coefficient of }P(x)}{\text{leading coefficient of } Q(x)}$$

Remember that the leading coefficient refers to the number multiplying the variable x with the highest exponent of the terms that form the polynomial. If there is no stated coefficient, then the coefficient is 1.

### Oblique Asymptotes

An oblique asymptote refers to an asymptote that is slanted. If the degree of the numerator is greater than the degree of the denominator by one, then there is an oblique asymptote. Note that the conditions for having a horizontal and an oblique asymptote are mutually exclusive. Therefore, a rational function cannot have both a horizontal and an oblique asymptote, just one or the other. If the rational function has an oblique asymptote, we can find its equation by dividing the numerator by the denominator using polynomial division. Since the asymptote is a straight line, it will be of the form $$y=mx+b$$. We just need to identify the part of the quotient that has this form.

## How do we Graph Rational Functions?

The below 5 steps should be followed to graph the rational functions.

1. First, we have to find all the asymptotes of the rational function that has been given to us.

2. Next, we can draw the asymptotes as dotted or dashed lines. It's best to use a ruler, as asymptotes are straight lines!

3. We need to find any x or y-intercepts of the rational function. In other words, we need to find out if the graph crosses the x or y axes at any point. To find a y-intercept we set $$x=0$$, and to find any x-intercepts we set $$y=0$$.

4. Before we can plot the graph, we need to calculate the x and y values for several different values of x, and then plot these points.

5. Finally, we can plot the graph by connecting each point that we plotted already.

As you can see, Graphing Rational Functions is similar to graphing any other function on a graph. The only extra thing we need is to figure out any asymptotes, as they contain important information about the rational function.

Draw the graph the following function.

$$f(x)=\frac{2x^{2}+x+1}{x+1}$$

Solution:

1. Because $$Q(x)=x+1$$ becomes $$0$$ at $$x=-1$$, we can tell that there will be a vertical asymptote at that x-value. Furthermore, since the degree of the numerator is greater than the degree of the denominator by exactly one. Therefore, there is an oblique asymptote, and consequently, there are no horizontal asymptotes.

 Polynomial Degree $$P(x)=2x^{2}+x+1$$ $$2$$ $$Q(x)=x^{1}+1$$ $$1$$ Degree Difference $$2-1=1$$

To find the equation of the oblique asymptote, we need to do the polynomial division:

$$\frac{2x^{2}+x+1}{x+1}=2x-1+\frac{2}{x+1}$$

Since the equation of the asymptote must be of the form $$y=mx+b$$, we can identify the asymptote equation to be $$2x-1$$.

2. We draw these asymptotes on our paper. In this case, the vertical line is our vertical asymptote of $$x=-1$$ and our slant line is our oblique asymptote of $$2x-1$$:

Fig. 2 - A graph showing the asymptotes of a rational function - StudySmarter Originals

3. To find any intercepts we need to set $$x$$ to $$0$$ and then $$y$$ to $$0$$. First, we will set $$x$$ to $$0$$:

$$\frac{2(0)^{2}+0+1}{0+1}$$

which simplifies down to $$\frac{0+0+1}{0+1}$$. Therefore there is an intercept with the y axis at $$y = 1$$. Now, setting $$y=0$$, we obtain the equation

$$0=\frac{2(x)^{2}+x+1}{x+1}$$

where we can ignore the denominator, leaving us with $$2(x)^{2}+x+1=0$$. The reason why we can eliminate the denominator is that if we multiply both sides with it, multiplying with $$0$$ will vanish it anyways. This equation has no real solution. This implies that there is no point where the parabola of $$2(x)^{2}+x+1=0$$ crosses the x-axis. Therefore, there are no x-intercepts. Make sure to plot these points on the graph.

Fig. 3 - The graph involving the asymptotes and the y intercept - StudySmarter Originals

4. We now plot several values for $$x$$ against $$y$$, substituting an $$x$$ value into the equation and plotting the point on the graph:

Fig. 4 - The graph after plotting some more points - StudySmarter Originals

5. Finally, we can connect these points to draw our final graph:

Fig. 5 - The curve after joining all the points and asymptotes addressed - StudySmarter Originals

Notice that the graph is split up into two different pieces. This will happen with every rational function.

More examples on Graphing Rational Functions can be found here!

## Rational Functions - Key takeaways

• A rational function is a function that can be represented as the ratio of two polynomial functions, where the one in the denominator is not constant.
• The first thing to do when being faced with a rational function is to simplify the function as much as possible.
• An asymptote is a straight line that is approached by a curve.
• Rational functions can have three types of asymptotes: horizontal, vertical, or oblique.
• A rational function can only have a horizontal asymptote or an oblique asymptote, but not both.
• A rational function can have at most one non-vertical asymptote.
• To draw the graph of a rational function, it is important to know the equations of the asymptotes first.
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How to find zeros of a rational function?

Zeroes of a rational function are the same as its x-intercepts. Substitute for y=0 and find the value of x, which will be the zeroes of the rational function.

What is rational function with example?

A rational function is any function that can be written as a fraction where both the numerator and the denominator are polynomials.

A good example would be the ratio of any quadratic and any linear function.

How to find the range of rational functions?

The range of a rational function is the same as the domain of its inverse. So first find the inverse of the rational function and then determine its domain.

How to find x-intercept of a rational function?

For the function, plug-in y=0 and find the value of x, the value determined is the x-intercept.

How to solve rational functions?

For any rational function, first, find the common denominator.

Now multiply everything by the common denominator and simplify.

The obtained equation will now be much more simplified and the solutions can be determined explicitly.

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