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Rational functions
Asymptotes, holes and domain — where a fraction of polynomials blows up.
A rational function is one polynomial over another. Where the denominator is zero the function has a vertical asymptote (or a hole, if the factor cancels); the degrees decide the horizontal or slant asymptote. Cancelling common factors first is what separates the two cases.
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 Precalculus 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 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 concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration 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:
\[\begin{array}{l}\text{water: }W(t)=100+10t\text{ in gallons} \\ \text{sugar: }S(t)=5+1t\text{ in pounds}\end{array}\]The concentration, \(C,\) will be the ratio of pounds of sugar to gallons of water
\[C(t)=\frac{5+t}{100+10t}\]The concentration after 12 minutes is given by evaluating \(C(t)\) at \(t=\ 12.\)
\[\begin{array}{l}C(12)=\frac{5+12}{100+10(12)} \\ =\frac{17}{220}\end{array}\]This means the concentration is 17 pounds of sugar to 220 gallons of water.
At the beginning, the concentration is
\[\begin{array}{l}C(0)=\frac{5+0}{100+10(0)} \\ =\frac{1}{20}\end{array}\]Since \(\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05,\) the concentration is greater after 12 minutes than at the beginning.
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.
\[\begin{array}{l}\begin{array}{l} \\ {x}^{2}-9=0\end{array} \\ {x}^{2}=9 \\ x=\pm 3\end{array}\]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.
\[\begin{array}{l}k(x)=\frac{5+2{x}^{2}}{2-x-{x}^{2}} \\ =\frac{5+2{x}^{2}}{(2+x)(1-x)}\end{array}\]To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:
\[\begin{array}{l}(2+x)(1-x)=0 \\ x=-2,1\end{array}\]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 Precalculus 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 leading terms of the numerator and denominator functions.
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 Precalculus 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 Precalculus 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 Precalculus 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 .
Kushanda muenzaniso: y = 1/(x^2 - 1)
Step by step
- \frac{1}{x^{2} - 1}
An expression in x. Here is what it does.
- \frac{1}{\left(x - 1\right) \left(x + 1\right)}
Factored form.
- \frac{d}{dx} = - \frac{2 x}{\left(x^{2} - 1\right)^{2}}
Derivative (slope).
Kuratidza mhinduro
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Use arrow notation to describe the end behavior and local behavior of the function graphed in .
Kuratidza mhinduro
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.
Kuratidza mhinduro
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)
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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.
Kuratidza mhinduro
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.
Kuratidza mhinduro
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 \(f(x)=\frac{1}{{(x-3)}^{2}}-4.\)
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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 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 concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?
Kuratidza mhinduro
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:
\[\begin{array}{l}\text{water: }W(t)=100+10t\text{ in gallons} \\ \text{sugar: }S(t)=5+1t\text{ in pounds}\end{array}\]The concentration, \(C,\) will be the ratio of pounds of sugar to gallons of water
\[C(t)=\frac{5+t}{100+10t}\]The concentration after 12 minutes is given by evaluating \(C(t)\) at \(t=\ 12.\)
\[\begin{array}{l}C(12)=\frac{5+12}{100+10(12)} \\ =\frac{17}{220}\end{array}\]This means the concentration is 17 pounds of sugar to 220 gallons of water.
At the beginning, the concentration is
\[\begin{array}{l}C(0)=\frac{5+0}{100+10(0)} \\ =\frac{1}{20}\end{array}\]Since \(\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05,\) the concentration is greater after 12 minutes than at the beginning.
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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.
Kuratidza mhinduro
\(\frac{12}{11}\)
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Find the domain of \(f(x)=\frac{x+3}{{x}^{2}-9}.\)
Kuratidza mhinduro
Begin by setting the denominator equal to zero and solving.
\[\begin{array}{l}\begin{array}{l} \\ {x}^{2}-9=0\end{array} \\ {x}^{2}=9 \\ x=\pm 3\end{array}\]The denominator is equal to zero when \(x=\pm 3.\) The domain of the function is all real numbers except \(x=\pm 3.\)
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Find the domain of \(f(x)=\frac{4x}{5(x-1)(x-5)}.\)
Kuratidza mhinduro
The domain is all real numbers except \(x=1\) and \(x=5.\)
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Find the vertical asymptotes of the graph of \(k(x)=\frac{5+2{x}^{2}}{2-x-{x}^{2}}.\)
Kuratidza mhinduro
First, factor the numerator and denominator.
\[\begin{array}{l}k(x)=\frac{5+2{x}^{2}}{2-x-{x}^{2}} \\ =\frac{5+2{x}^{2}}{(2+x)(1-x)}\end{array}\]To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:
\[\begin{array}{l}(2+x)(1-x)=0 \\ x=-2,1\end{array}\]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.
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Find the vertical asymptotes and removable discontinuities of the graph of \(k(x)=\frac{x-2}{{x}^{2}-4}.\)
Kuratidza mhinduro
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.
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Find the vertical asymptotes and removable discontinuities of the graph of \(f(x)=\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}.\)
Kuratidza mhinduro
Removable discontinuity at \(x=5.\) Vertical asymptotes: \(x=0,\ x=1.\)
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For the functions below, identify the horizontal or slant asymptote.
- ⓐ \(g(x)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}\)
- ⓑ \(h(x)=\frac{{x}^{2}-4x+1}{x+2}\)
- ⓒ \(k(x)=\frac{{x}^{2}+4x}{{x}^{3}-8}\)
Kuratidza mhinduro
For these solutions, we will use \(f(x)=\frac{p(x)}{q(x)},\ q(x)\ne 0.\)
- ⓐ \(g(x)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}:\) 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 \(y=3.\)
- ⓑ \(h(x)=\frac{{x}^{2}-4x+1}{x+2}:\) The degree of \(p=2\) and degree of \(q=1.\) Since \(p>q\) by 1, there is a slant asymptote found at \(\frac{{x}^{2}-4x+1}{x+2}.\)
\[\begin{array}{lllll} \\ \begin{array}{l}-2 \\ \end{array}\begin{array}{lll}1 & -4 & 1 \\ & -2 & 12\end{array} \\ \begin{array}{lll}\text{ 1} & -6 & 13\end{array}\end{array}\]
The quotient is \(x-6\) and the remainder is 13. There is a slant asymptote at \(y=x-6.\)
- ⓒ \(k(x)=\frac{{x}^{2}+4x}{{x}^{3}-8}:\) The degree of \(p=2\ <\) degree of \(q=3,\) so there is a horizontal asymptote \(y=0.\)
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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.
Kuratidza mhinduro
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.
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Find the horizontal and vertical asymptotes of the function
\[f(x)=\frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}\]Kuratidza mhinduro
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 .
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Find the vertical and horizontal asymptotes of the function:
\(f(x)=\frac{(2x-1)(2x+1)}{(x-2)(x+3)}\)
Kuratidza mhinduro
Vertical asymptotes at \(x=2\) and \(x=-3;\) horizontal asymptote at \(y=4.\)
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Find the intercepts of \(f(x)=\frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}.\)
Kuratidza mhinduro
We can find the y-intercept by evaluating the function at zero
\[\begin{array}{l}f(0)=\frac{(0-2)(0+3)}{(0-1)(0+2)(0-5)} \\ =\frac{-6}{10} \\ =-\frac{3}{5} \\ \ =-0.6\end{array}\]The x-intercepts will occur when the function is equal to zero:
\[\begin{array}{ll} & \\ 0=\frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)} & \text{This is zero when the numerator is zero}. \\ 0=(x-2)(x+3) & \\ x=2,\ -3 & \end{array}\]The y-intercept is \((0,-0.6),\) the x-intercepts are \((2,0)\) and \((-3,0).\) See .
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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.
Kuratidza mhinduro
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 \((0,\frac{-35}{9}).\)
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Sketch a graph of \(f(x)=\frac{(x+2)(x-3)}{{(x+1)}^{2}(x-2)}.\)
Kuratidza mhinduro
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:
\[\begin{array}{l}f(0)=\frac{(0+2)(0-3)}{{(0+1)}^{2}(0-2)} \\ =3\end{array}\]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.
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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.
Kuratidza mhinduro
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.
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Write an equation for the rational function shown in .
Kuratidza mhinduro
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 \((0,-2).\)
\[\begin{array}{l}-2=a\frac{(0+2)(0-3)}{(0+1){(0-2)}^{2}} \\ -2=a\frac{-6}{4} \\ a=\frac{-8}{-6}=\frac{4}{3}\end{array}\]This gives us a final function of \(f(x)=\frac{4(x+2)(x-3)}{3(x+1){(x-2)}^{2}}.\)
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What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?
Kuratidza mhinduro
The rational function will be represented by a quotient of polynomial functions.
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What is the fundamental difference in the graphs of polynomial functions and rational functions?
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If the graph of a rational function has a removable discontinuity, what must be true of the functional rule?
Kuratidza mhinduro
The numerator and denominator must have a common factor.
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Can a graph of a rational function have no vertical asymptote? If so, how?
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Can a graph of a rational function have no x-intercepts? If so, how?
Kuratidza mhinduro
Yes. The numerator of the formula of the functions would have only complex roots and/or factors common to both the numerator and denominator.
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\(f(x)=\frac{x-1}{x+2}\)
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\(f(x)=\frac{x+1}{{x}^{2}-1}\)
Kuratidza mhinduro
\(\text{All reals }x\ne -1,\ 1\)
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\(f(x)=\frac{{x}^{2}+4}{{x}^{2}-2x-8}\)
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\(f(x)=\frac{{x}^{2}+4x-3}{{x}^{4}-5{x}^{2}+4}\)
Kuratidza mhinduro
\(\text{All reals }x\ne -1,\ -2,\ 1,\ 2\)
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\(f(x)=\frac{4}{x-1}\)
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\(f(x)=\frac{2}{5x+2}\)
Kuratidza mhinduro
V.A. at \(x=-\frac{2}{5};\) H.A. at \(y=0;\) Domain is all reals \(x\ne -\frac{2}{5}\)
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\(f(x)=\frac{x}{{x}^{2}-9}\)
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\(f(x)=\frac{x}{{x}^{2}+5x-36}\)
Kuratidza mhinduro
V.A. at \(x=4,\ -9;\) H.A. at \(y=0;\) Domain is all reals \(x\ne 4,\ -9\)
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\(f(x)=\frac{3+x}{{x}^{3}-27}\)
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\(f(x)=\frac{3x-4}{{x}^{3}-16x}\)
Kuratidza mhinduro
V.A. at \(x=0,\ 4,\ -4;\) H.A. at \(y=0;\) Domain is all reals \(x\ne 0,4,\ -4\)
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\(f(x)=\frac{{x}^{2}-1}{{x}^{3}+9{x}^{2}+14x}\)
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\(f(x)=\frac{x+5}{{x}^{2}-25}\)
Kuratidza mhinduro
V.A. at \(x=5;\) H.A. at \(y=0;\) Domain is all reals \(x\ne 5,-5\)
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\(f(x)=\frac{x-4}{x-6}\)
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\(f(x)=\frac{4-2x}{3x-1}\)
Kuratidza mhinduro
V.A. at \(x=\frac{1}{3};\) H.A. at \(y=-\frac{2}{3};\) Domain is all reals \(x\ne \frac{1}{3}.\)
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\(f(x)=\frac{x+5}{{x}^{2}+4}\)
Symbols used here
Not a number: "grows without bound" in limits and intervals.
Chance of A; chance of A given that B happened.
Instantaneous rate of change; slope of the graph.
Equal to the precision shown, not exactly.
Both signs at once: x = 3 ± 2 means 5 and 1.
Inequalities that allow equality; < and > exclude it.
The two sides are different.
Least upper bound, greatest lower bound.
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
Number of k-element subsets of n things: n!/(k!(n−k)!).
Ratio of a circle's circumference to its diameter, 3.14159…
2.71828…, the base whose exponential is its own derivative.
i² = −1.
The usual name for an angle.
The exponent b must be raised to for x; ln uses base e.
A quantity with magnitude and direction; a column of 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.
- On the left branch of the graph, the curve approaches the
- As the graph approaches
Questions people ask
What is a function, really?
A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.
Why do we need complex numbers?
Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.
Tarisa yako
Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
More in Precalculus
Complex numbersPolynomial functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates