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Rational Functions

Use arrow notation.

Rational Functions

Determine the values for which a rational expression is undefined (IA 7.1.1)
Find x- and y-intercepts (IA 3.1.4)

Here are some examples of rational expressions:

\[\begin{array}{llllllllll}-\frac{24}{56} & & & \ \frac{5x}{12y} & & & \ \frac{4x+1}{{x}^{2}-9} & & & \ \frac{4{x}^{2}+3x-1}{2x-8}\end{array}\]

Evaluate the following expression for the given values \(\frac{5x-10}{x+3}\)

Try it.

\(x=3\)

Try it.

\(x=2\)

Try it.

\(x=-3\)

Try it.

Why do we have a problem with evaluating this expression for \(x=-3\) but not for \(x=2\) ?

We say that this rational expression is undefined because its denominator equals 0.

Example

Try it.

Determine the value for which the rational expression \(\frac{x+3}{2{x}^{2}+9x-5}\) is undefined

Solution

Set the denominator equal to zero	\(2{x}^{2}+9x-5=0\)
Solve the equation	We will solve this quadratic equation by factoring \((2x-1)(x+5)=0\) \(\begin{array}{lll}2x-1=0 & & x+5=0 \\ 2x=1 & & x=-5 \\ x=\frac{1}{2} & & \end{array}\)
Answer:	Rational expression is undefined when \(x=\frac{1}{2}\) and \(x=-5\)

Determine the value for which each rational expression is undefined.

Try it.

\(\frac{3{y}^{2}}{5x}\)

Try it.

\(\frac{3x+5}{2x-3}\)

Try it.

\(\frac{m-5}{{m}^{2}+m-6}\)

Try it.

\(\frac{2a}{{a}^{2}-9}\)

Try it.

Determine the value for which the function \(f(x)=\frac{3}{x-2}\) is undefined.

What is the connection between the value you found and the graph of this function?

Because function is not defined at \(x=2\) , we say that the Domain of this function is \(\left(-\infty ,2\right)\left(2,\infty \right)\)

The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.

Try it.

Determine the domain of the function and express using interval notation.

ⓐ \(f(x)=\frac{x}{3x-1}\)
ⓑ \(f(x)=\frac{3}{{x}^{2}-5x+6}\)

Condensed — the full section is in OpenStax College Algebra 2e.

Using Arrow Notation

We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in , and notice some of their features.

Several things are apparent if we examine the graph of \(f(x)=\frac{1}{x}.\)

On the left branch of the graph, the curve approaches the x-axis \((y=0)\text{as}x\to -\infty .\)
As the graph approaches \(x=0\) from the left, the curve drops, but as we approach zero from the right, the curve rises.
Finally, on the right branch of the graph, the curves approaches the x-axis \((y=0)\text{as}x\to \infty .\)

To summarize, we use arrow notation to show that \(x\) or \(f(x)\) is approaching a particular value. See .

Symbol	Meaning
\(x\to {a}^{-}\)	\(x\) approaches \(a\) from the left ( \(x
\(x\to {a}^{+}\)	\(x\) approaches \(a\) from the right ( \(x>a\) but close to \(a\) )
\(x\to \infty\)	\(x\) approaches infinity ( \(x\) increases without bound)
\(x\to -\infty\)	\(x\) approaches negative infinity ( \(x\) decreases without bound)
\(f(x)\to \infty\)	the output approaches infinity (the output increases without bound)
\(f(x)\to -\infty\)	the output approaches negative infinity (the output decreases without bound)
\(f(x)\to a\)	the output approaches \(a\)

Let’s begin by looking at the reciprocal function, \(f(x)=\frac{1}{x}.\) We cannot divide by zero, which means the function is undefined at \(x=0;\) so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in .

\(x\)	–0.1	–0.01	–0.001	–0.0001
\(f(x)=\frac{1}{x}\)	–10	–100	–1000	–10,000

We write in arrow notation

\[\text{as }x\to {0}^{-},f(x)\to -\infty\]

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in .

\(x\)	0.1	0.01	0.001	0.0001
\(f(x)=\frac{1}{x}\)	10	100	1000	10,000

We write in arrow notation

\[\text{As }x\to {0}^{+},f(x)\to \infty .\]

See .

This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line \(x=0\) as the input becomes close to zero. See .

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Applied Problems Involving Rational Functions

In , we shifted a toolkit function in a way that resulted in the function \(f(x)=\frac{3x+7}{x+2}.\) This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

Example

Try it.

After running out of pre-packaged supplies, a nurse in a refugee camp is preparing an intravenous sugar solution for patients in the camp hospital. A large mixing tank currently contains 100 gallons of distilled water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of distilled water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the ratio of sugar to water, in pounds per gallon in the tank after 12 minutes. Is that a greater ratio of sugar to water, in pounds per gallon than at the beginning?

Solution

Let \(t\) be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:

\[\begin{array}{lll}\text{water: }W(t) & = & 100+10t\ \text{in gallons} \\ \text{sugar: }S(t) & = & 5+1t\ \text{in pounds}\end{array}\]

The ratio of sugar to water, in pounds per gallon, \(C\) , will be the ratio of pounds of sugar to gallons of water

\[C(t)=\frac{5+t}{100+10t}\]

The ratio of sugar to water, in pounds per gallon after 12 minutes is given by evaluating \(C(t)\) at \(t=\ 12.\)

\[\begin{array}{lll}C(12) & = & \frac{5+12}{100+10(12)} \\ & = & \frac{17}{220}\end{array}\]

This means the ratio of sugar to water, in pounds per gallon is 17 pounds of sugar to 220 gallons of water.

At the beginning, the ratio of sugar to water, in pounds per gallon is

\[\begin{array}{lll}C(0) & = & \frac{5+0}{100+10(0)} \\ & = & \frac{1}{20}\end{array}\]

Since \(\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05,\) the ratio of sugar to water, in pounds per gallon is greater after 12 minutes than at the beginning.

Condensed — the full section is in OpenStax College Algebra 2e.

Finding the Domains of Rational Functions

A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.

Example

Try it.

Find the domain of \(f(x)=\frac{x+3}{{x}^{2}-9}.\)

Solution

Begin by setting the denominator equal to zero and solving.

\[\begin{array}{lll}{x}^{2}-9 & = & 0 \\ {x}^{2} & = & 9 \\ x & = & \pm 3\end{array}\]

The denominator is equal to zero when \(x=\pm 3.\) The domain of the function is all real numbers except \(x=\pm 3.\)

Identifying Vertical Asymptotes of Rational Functions

By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

Example

Try it.

Find the vertical asymptotes of the graph of \(k(x)=\frac{5+2{x}^{2}}{2-x-{x}^{2}}.\)

Solution

First, factor the numerator and denominator.

\[\begin{array}{lll}k(x) & = & \frac{5+2{x}^{2}}{2-x-{x}^{2}} \\ & = & \frac{5+2{x}^{2}}{(2+x)(1-x)}\end{array}\]

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

\[\begin{array}{lll}(2+x)(1-x) & = & 0 \\ x & = & -2,1\end{array}\]

Neither \(x=-2\) nor \(x=1\) are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in confirms the location of the two vertical asymptotes.

Condensed — the full section is in OpenStax College Algebra 2e.

Identifying Horizontal Asymptotes of Rational Functions

While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.

There are three distinct outcomes when checking for horizontal asymptotes:

Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at \(y=0.\)

\[\text{Example: }f(x)=\frac{4x+2}{{x}^{2}+4x-5}\]

In this case, the end behavior is \(f(x)\approx \frac{4x}{{x}^{2}}=\frac{4}{x}.\) This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function \(g(x)=\frac{4}{x},\) and the outputs will approach zero, resulting in a horizontal asymptote at \(y=0.\) See . Note that this graph crosses the horizontal asymptote.

Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.

\[\text{Example: }f(x)=\frac{3{x}^{2}-2x+1}{x-1}\]

In this case, the end behavior is \(f(x)\approx \frac{3{x}^{2}}{x}=3x.\) This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function \(g(x)=3x.\) As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of \(g(x)=3x\) looks like a diagonal line, and since \(f\) will behave similarly to \(g,\) it will approach a line close to \(y=3x.\) This line is a slant asymptote.

To find the equation of the slant asymptote, divide \(\frac{3{x}^{2}-2x+1}{x-1}.\) The quotient is \(3x+1,\) and the remainder is 2. The slant asymptote is the graph of the line \(g(x)=3x+1.\) See .

\[\text{Example: }f(x)=\frac{3{x}^{2}+2}{{x}^{2}+4x-5}\]\[f(x)=\frac{3{x}^{5}-{x}^{2}}{x+3}\]\[f(x)\approx \frac{3{x}^{5}}{x}=3{x}^{4},\]\[x\to \pm \infty ,f(x)\to \infty\]

Condensed — the full section is in OpenStax College Algebra 2e.

Graphing Rational Functions

In , we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.

The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See .

When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See .

For example, the graph of \(f(x)=\frac{{(x+1)}^{2}(x-3)}{{(x+3)}^{2}(x-2)}\) is shown in .

At the x-intercept \(x=-1\) corresponding to the \({(x+1)}^{2}\) factor of the numerator, the graph "bounces", consistent with the quadratic nature of the factor.
At the x-intercept \(x=3\) corresponding to the \((x-3)\) factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
At the vertical asymptote \(x=-3\) corresponding to the \({(x+3)}^{2}\) factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function \(f(x)=\frac{1}{{x}^{2}}.\)
At the vertical asymptote \(x=2,\) corresponding to the \((x-2)\) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side.

Condensed — the full section is in OpenStax College Algebra 2e.

Writing Rational Functions

Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.

Condensed — the full section is in OpenStax College Algebra 2e.

Key Equations

Rational Function

\(f(x)=\frac{P(x)}{Q(x)}=\frac{{a}_{p}{x}^{p}+{a}_{p-1}{x}^{p-1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q-1}{x}^{q-1}+...+{b}_{1}x+{b}_{0}},Q(x)\ne 0\)

Key Concepts

We can use arrow notation to describe local behavior and end behavior of the toolkit functions \(f(x)=\frac{1}{x}\) and \(f(x)=\frac{1}{{x}^{2}}.\) See .
A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote. See .
Application problems involving rates and concentrations often involve rational functions. See .
The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. See .
The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. See .
A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. See .
A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. See , , , and .
Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. See .
If a rational function has x-intercepts at \(x={x}_{1},{x}_{2},\ldots ,{x}_{n},\) vertical asymptotes at \(x={v}_{1},{v}_{2},\ldots ,{v}_{m},\) and no \({x}_{i}=\text{any }{v}_{j},\) then the function can be written in the form

\[\begin{array}{l}\begin{array}{l} \\ f(x)=a\frac{{(x-{x}_{1})}^{{p}_{1}}{(x-{x}_{2})}^{{p}_{2}}\cdots {(x-{x}_{n})}^{{p}_{n}}}{{(x-{v}_{1})}^{{q}_{1}}{(x-{v}_{2})}^{{q}_{2}}\cdots {(x-{v}_{m})}^{{q}_{n}}}\end{array}\end{array}\]

See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

\(x=3\)
\(x=2\)
\(x=-3\)
Why do we have a problem with evaluating this expression for \(x=-3\) but not for \(x=2\) ?

Determine the value for which the rational expression \(\frac{x+3}{2{x}^{2}+9x-5}\) is undefined

ਜਵਾਬ ਦਿਓ

Set the denominator equal to zero	\(2{x}^{2}+9x-5=0\)
Solve the equation	We will solve this quadratic equation by factoring \((2x-1)(x+5)=0\) \(\begin{array}{lll}2x-1=0 & & x+5=0 \\ 2x=1 & & x=-5 \\ x=\frac{1}{2} & & \end{array}\)
Answer:	Rational expression is undefined when \(x=\frac{1}{2}\) and \(x=-5\)

\(\frac{3{y}^{2}}{5x}\)
\(\frac{3x+5}{2x-3}\)
\(\frac{m-5}{{m}^{2}+m-6}\)
\(\frac{2a}{{a}^{2}-9}\)
Determine the value for which the function \(f(x)=\frac{3}{x-2}\) is undefined.
What is the connection between the value you found and the graph of this function?
Because function is not defined at \(x=2\) , we say that the Domain of this function is \(\left(-\infty ,2\right)\left(2,\infty \right)\)
The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.
Determine the domain of the function and express using interval notation.
1. ⓐ \(f(x)=\frac{x}{3x-1}\)
2. ⓑ \(f(x)=\frac{3}{{x}^{2}-5x+6}\)
\(3x-2y=6\)
\(y=5x-1\)
\(y=\frac{1}{(x-2)+3}\)
Use arrow notation to describe the end behavior and local behavior of the function graphed in .

ਜਵਾਬ ਦਿਓ

Notice that the graph is showing a vertical asymptote at \(x=2,\) which tells us that the function is undefined at \(x=2.\)
\[\text{As }x\to {2}^{-},f(x)\to -\infty ,\ \text{and as }x\to {2}^{+},\ f(x)\to \infty .\]
And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at \(y=4.\) As the inputs increase without bound, the graph levels off at 4.
\[\text{As }x\to \infty ,\ f(x)\to 4\ \text{and as }x\to -\infty ,\ f(x)\to 4.\]
Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.

ਜਵਾਬ ਦਿਓ

End behavior: as \(x\to \pm \infty ,f(x)\to 0;\) Local behavior: as \(x\to 0,f(x)\to \infty\) (there are no x- or y-intercepts)
Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.

ਜਵਾਬ ਦਿਓ

Shifting the graph left 2 and up 3 would result in the function
\[f(x)=\frac{1}{x+2}+3\]
or equivalently, by giving the terms a common denominator,
\[f(x)=\frac{3x+7}{x+2}\]
The graph of the shifted function is displayed in .
Notice that this function is undefined at \(x=-2,\) and the graph also is showing a vertical asymptote at \(x=-2.\)
\[\text{As }x\to -{2}^{-},f(x)\to -\infty ,\text{and as}x\to -{2}^{+},f(x)\to \infty .\]
As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at \(y=3.\)
\[\text{As }x\to \pm \infty ,f(x)\to 3.\]
Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.

ਜਵਾਬ ਦਿਓ

The function and the asymptotes are shifted 3 units right and 4 units down. As \(x\to 3,f(x)\to \infty ,\) and as \(x\to \pm \infty ,f(x)\to -4.\)
The function is \(f(x)=\frac{1}{{(x-3)}^{2}}-4.\)
After running out of pre-packaged supplies, a nurse in a refugee camp is preparing an intravenous sugar solution for patients in the camp hospital. A large mixing tank currently contains 100 gallons of distilled water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of distilled water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the ratio of sugar to water, in pounds per gallon in the tank after 12 minutes. Is that a greater ratio of sugar to water, in pounds per gallon than at the beginning?

ਜਵਾਬ ਦਿਓ

Let \(t\) be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:
\[\begin{array}{lll}\text{water: }W(t) & = & 100+10t\ \text{in gallons} \\ \text{sugar: }S(t) & = & 5+1t\ \text{in pounds}\end{array}\]
The ratio of sugar to water, in pounds per gallon, \(C\) , will be the ratio of pounds of sugar to gallons of water
\[C(t)=\frac{5+t}{100+10t}\]
The ratio of sugar to water, in pounds per gallon after 12 minutes is given by evaluating \(C(t)\) at \(t=\ 12.\)
\[\begin{array}{lll}C(12) & = & \frac{5+12}{100+10(12)} \\ & = & \frac{17}{220}\end{array}\]
This means the ratio of sugar to water, in pounds per gallon is 17 pounds of sugar to 220 gallons of water.
At the beginning, the ratio of sugar to water, in pounds per gallon is
\[\begin{array}{lll}C(0) & = & \frac{5+0}{100+10(0)} \\ & = & \frac{1}{20}\end{array}\]
Since \(\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05,\) the ratio of sugar to water, in pounds per gallon is greater after 12 minutes than at the beginning.
There are 1,200 first-year and 1,500 second-year students at a rally at noon. After 12 p.m., 20 first-year students arrive at the rally every five minutes while 15 second-year students leave the rally. Find the ratio of first-year to second-year students at 1 p.m.

ਜਵਾਬ ਦਿਓ

\(\frac{12}{11}\)
Find the domain of \(f(x)=\frac{x+3}{{x}^{2}-9}.\)

ਜਵਾਬ ਦਿਓ

Begin by setting the denominator equal to zero and solving.
\[\begin{array}{lll}{x}^{2}-9 & = & 0 \\ {x}^{2} & = & 9 \\ x & = & \pm 3\end{array}\]
The denominator is equal to zero when \(x=\pm 3.\) The domain of the function is all real numbers except \(x=\pm 3.\)
Find the domain of \(f(x)=\frac{4x}{5(x-1)(x-5)}.\)

ਜਵਾਬ ਦਿਓ

The domain is all real numbers except \(x=1\) and \(x=5.\)
Find the vertical asymptotes of the graph of \(k(x)=\frac{5+2{x}^{2}}{2-x-{x}^{2}}.\)

ਜਵਾਬ ਦਿਓ

First, factor the numerator and denominator.
\[\begin{array}{lll}k(x) & = & \frac{5+2{x}^{2}}{2-x-{x}^{2}} \\ & = & \frac{5+2{x}^{2}}{(2+x)(1-x)}\end{array}\]
To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:
\[\begin{array}{lll}(2+x)(1-x) & = & 0 \\ x & = & -2,1\end{array}\]
Neither \(x=-2\) nor \(x=1\) are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in confirms the location of the two vertical asymptotes.
Find the vertical asymptotes and removable discontinuities of the graph of \(k(x)=\frac{x-2}{{x}^{2}-4}.\)

ਜਵਾਬ ਦਿਓ

Factor the numerator and the denominator.
\[k(x)=\frac{x-2}{(x-2)(x+2)}\]
Notice that there is a common factor in the numerator and the denominator, \(x-2.\) The zero for this factor is \(x=2.\) This is the location of the removable discontinuity.
Notice that there is a factor in the denominator that is not in the numerator, \(x+2.\) The zero for this factor is \(x=-2.\) The vertical asymptote is \(x=-2.\) See .
The graph of this function will have the vertical asymptote at \(x=-2,\) but at \(x=2\) the graph will have a hole.
Find the vertical asymptotes and removable discontinuities of the graph of \(f(x)=\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}.\)

ਜਵਾਬ ਦਿਓ

Removable discontinuity at \(x=5.\) Vertical asymptotes: \(x=0,\ x=1.\)
For the functions listed, identify the horizontal or slant asymptote.
1. ⓐ \(g(x)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}\)
2. ⓑ \(h(x)=\frac{{x}^{2}-4x+1}{x+2}\)
3. ⓒ \(k(x)=\frac{{x}^{2}+4x}{{x}^{3}-8}\)
ਜਵਾਬ ਦਿਓ
For these solutions, we will use \(f(x)=\frac{p(x)}{q(x)},q(x)\ne 0.\)
1. ⓐ \(g(x)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}:\) The degree of \(p=\text{degree of}q=3,\) so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at \(y=\frac{6}{2}\) or \(y=3.\)
2. ⓑ \(h(x)=\frac{{x}^{2}-4x+1}{x+2}:\) The degree of \(p=2\) and degree of \(q=1.\) Since \(p>q\) by 1, there is a slant asymptote found at \(\frac{{x}^{2}-4x+1}{x+2}.\)
  The quotient is \(x-6\) and the remainder is 13. There is a slant asymptote at \(y=x-6.\)
3. ⓒ \(k(x)=\frac{{x}^{2}+4x}{{x}^{3}-8}:\) The degree of \(p=2\ <\) degree of \(q=3,\) so there is a horizontal asymptote \(y=0.\)
In the sugar concentration problem earlier, we created the equation \(C(t)=\frac{5+t}{100+10t}.\)
Find the horizontal asymptote and interpret it in context of the problem.

ਜਵਾਬ ਦਿਓ

Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is \(t,\) with coefficient 1. In the denominator, the leading term is \(10t,\) with coefficient 10. The horizontal asymptote will be at the ratio of these values:
\[t\to \infty ,C(t)\to \frac{1}{10}\]
This function will have a horizontal asymptote at \(y=\frac{1}{10}.\)
This tells us that as the values of t increase, the values of \(C\) will approach \(\frac{1}{10}.\) In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or \(\frac{1}{10}\) pounds per gallon.
Find the horizontal and vertical asymptotes of the function
\[f(x)=\frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}\]

ਜਵਾਬ ਦਿਓ

First, note that this function has no common factors, so there are no potential removable discontinuities.
The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at \(x=1,-2,\text{and }5,\) indicating vertical asymptotes at these values.
The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as \(x\to \pm \infty ,f(x)\to 0.\) This function will have a horizontal asymptote at \(y=0.\) See .
Find the vertical and horizontal asymptotes of the function:
\(f(x)=\frac{(2x-1)(2x+1)}{(x-2)(x+3)}\)

ਜਵਾਬ ਦਿਓ

Vertical asymptotes at \(x=2\) and \(x=-3;\) horizontal asymptote at \(y=4.\)
Find the intercepts of \(f(x)=\frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}.\)

ਜਵਾਬ ਦਿਓ

We can find the y-intercept by evaluating the function at zero
\[\begin{array}{lll}f(0) & = & \frac{(0-2)(0+3)}{(0-1)(0+2)(0-5)} \\ & = & \frac{-6}{10} \\ & = & -\frac{3}{5} \\ & = & -0.6\end{array}\]
The x-intercepts will occur when the function is equal to zero:
\[\begin{array}{llll}0 & = & \frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)} & \ \text{This is zero when the numerator is zero}. \\ 0 & = & (x-2)(x+3) & \\ x & = & 2,-3 & \end{array}\]
The y-intercept is \((0,-0.6),\) the x-intercepts are \((2,0)\) and \((-3,0).\) See .
Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x- and y-intercepts and the horizontal and vertical asymptotes.

ਜਵਾਬ ਦਿਓ

For the transformed reciprocal squared function, we find the rational form. \(f(x)=\frac{1}{{(x-3)}^{2}}-4=\frac{1-4{(x-3)}^{2}}{{(x-3)}^{2}}=\frac{1-4({x}^{2}-6x+9)}{(x-3)(x-3)}=\frac{-4{x}^{2}+24x-35}{{x}^{2}-6x+9}\)
Because the numerator is the same degree as the denominator we know that as \(x\to \pm \infty ,f(x)\to -4;\text{so}y=-4\) is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is \(x=3,\) because as \(x\to 3,f(x)\to \infty .\) We then set the numerator equal to 0 and find the x-intercepts are at \((2.5,0)\) and \((3.5,0).\) Finally, we evaluate the function at 0 and find the y-intercept to be at \((0,\frac{-35}{9}).\)
Sketch a graph of \(f(x)=\frac{(x+2)(x-3)}{{(x+1)}^{2}(x-2)}.\)

ਜਵਾਬ ਦਿਓ

We can start by noting that the function is already factored, saving us a step.
Next, we will find the intercepts. Evaluating the function at zero gives the y-intercept:
\[\begin{array}{lll}f(0) & = & \frac{(0+2)(0-3)}{{(0+1)}^{2}(0-2)} \\ & = & 3\end{array}\]
To find the x-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find x-intercepts at \(x=-2\) and \(x=3.\) At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.
We have a y-intercept at \((0,3)\) and x-intercepts at \((-2,0)\) and \((3,0).\)
To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when \(x+1=0\) and when \(x-2=0,\) giving us vertical asymptotes at \(x=-1\) and \(x=2.\)
There are no common factors in the numerator and denominator. This means there are no removable discontinuities.
Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at \(y=0.\)
To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x-intercepts between the vertical asymptotes, and the y-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in .
The factor associated with the vertical asymptote at \(x=-1\) was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.
For the vertical asymptote at \(x=2,\) the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See . After passing through the x-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.
Given the function \(f(x)=\frac{{(x+2)}^{2}(x-2)}{2{(x-1)}^{2}(x-3)},\) use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.

ਜਵਾਬ ਦਿਓ

Horizontal asymptote at \(y=\frac{1}{2}.\) Vertical asymptotes at \(x=1\text{and}x=3.\) y-intercept at \((0,\frac{4}{3}.)\)
x-intercepts at \((2,0)\ \text{and }(-2,0).\) \((-2,0)\) is a zero with multiplicity 2, and the graph bounces off the x-axis at this point. \((2,0)\) is a single zero and the graph crosses the axis at this point.
Write an equation for the rational function shown in .

ਜਵਾਬ ਦਿਓ

The graph appears to have x-intercepts at \(x=-2\) and \(x=3.\) At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at \(x=-1\) seems to exhibit the basic behavior similar to \(\frac{1}{x},\) with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at \(x=2\) is exhibiting a behavior similar to \(\frac{1}{{x}^{2}},\) with the graph heading toward negative infinity on both sides of the asymptote. See .
We can use this information to write a function of the form
\[f(x)=a\frac{(x+2)(x-3)}{(x+1){(x-2)}^{2}}\]
To find the stretch factor, we can use another clear point on the graph, such as the y-intercept \((0,-2).\)
\[\begin{array}{lll}-2 & = & a\frac{(0+2)(0-3)}{(0+1){(0-2)}^{2}} \\ -2 & = & a\frac{-6}{4} \\ a & = & \frac{-8}{-6}=\frac{4}{3}\end{array}\]
This gives us a final function of \(f(x)=\frac{4(x+2)(x-3)}{3(x+1){(x-2)}^{2}}.\)
What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?

ਜਵਾਬ ਦਿਓ

The rational function will be represented by a quotient of polynomial functions.
What is the fundamental difference in the graphs of polynomial functions and rational functions?
If the graph of a rational function has a removable discontinuity, what must be true of the functional rule?

ਜਵਾਬ ਦਿਓ

The numerator and denominator must have a common factor.
Can a graph of a rational function have no vertical asymptote? If so, how?
Can a graph of a rational function have no x-intercepts? If so, how?

ਜਵਾਬ ਦਿਓ

Yes. The numerator of the formula of the functions would have only complex roots and/or factors common to both the numerator and denominator.
\(f(x)=\frac{x-1}{x+2}\)

Symbols used here

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Rational Functions

Use arrow notation.
Solve applied problems involving rational functions.
Find the domains of rational functions.
Identify vertical asymptotes.
Identify horizontal asymptotes.
Graph rational functions.
Determine the values for which a rational expression is undefined (IA 7.1.1)
Find x- and y-intercepts (IA 3.1.4)

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

ਆਪਣਾ ਹੀ ਕੋਸ਼ਿਸ਼ ਕਰੋ

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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