4.1: Rational Functions
In the previous sections, we have built polynomials based on the positive whole number power functions. In this section, we explore functions based on power functions with negative integer powers, called rational functions.
You plan to drive 100 miles. Find a formula for the time the trip will take as a function of the speed you drive.
Solution
You may recall that multiplying speed by time will give you distance. If we let \(t\) represent the drive time in hours, and \(v\) represent the velocity (speed or rate) at which we drive, then \(vt\) = distance. Since our distance is fixed at 100 miles, \(vt=100\). Solving this relationship for the time gives us the function we desired:
\[t(v)=\dfrac{100}{v} =100v^{-1}\nonumber \]
While this type of relationship can be written using the negative exponent, it is more common to see it written as a fraction.
This particular example is one of an inversely proportional relationship – where one quantity is a constant divided by the other quantity, like \(y=\dfrac{5}{x}\).
Notice that this is a transformation of the reciprocal toolkit function, \(f(x)=\dfrac{1}{x}\)
Several natural phenomena, such as gravitational force and volume of sound, behave in a manner inversely proportional to the square of another quantity. For example, the volume, \(V\), of a sound heard at a distance \(d\) from the source would be related by \(V=\dfrac{k}{d^{2} }\) for some constant value \(k\).
These functions are transformations of the reciprocal squared toolkit function \(f(x)=\dfrac{1}{x^{2} }\).
We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. These graphs have several important features.
\(f(x) = \dfrac{1}{x}\) \(f(x) = \dfrac{1}{x^2}\)
Let’s begin by looking at the reciprocal function, \(f(x)=\dfrac{1}{x}\). As you well know, dividing by zero is not allowed and therefore zero is not in the domain, and so the function is undefined at an input of zero.
Short run behavior:
As the input values approach zero from the left side (taking on very small, negative values), the function values become very large in the negative direction (in other words, they approach negative infinity).
We write: as\(x\to 0^{-}\), \(f(x)\to -\infty\).
As we approach zero from the right side (small, positive input values), the function values become very large in the positive direction (approaching infinity).
We write: as\(x\to 0^{+}\),\(f(x)\to \infty\).
This behavior creates a vertical asymptote . An asymptote is a line that the graph approaches. In this case the graph is approaching the vertical line \(x = 0\) as the input becomes close to zero.
Long run behavior:
As the values of \(x\) approach infinity, the function values approach 0.
As the values of \(x\) approach negative infinity, the function values approach 0.
Symbolically: as\(x\to \pm \infty\),\(f(x)\to 0\)
Based on this long run behavior and the graph we can see that the function approaches 0 but never actually reaches 0, it just “levels off” as the inputs become large. This behavior creates a horizontal asymptote . In this case the graph is approaching the horizontal line \(f(x)=0\) as the input becomes very large in the negative and positive directions.
A vertical asymptote of a graph is a vertical line \(x = a\) where the graph tends towards positive or negative infinity as the inputs approach a . As\(x \to a\),\(f(x) \to \pm \infty\).
A horizontal asymptote of a graph is a horizontal line \(y=b\) where the graph approaches the line as the inputs get large. As\(x \to \pm \infty\),\(f(x) \to b\).
Use symbolic notation to describe the long run behavior and short run behavior for the reciprocal squared function.
- Answer
-
Long run behavior, as \(x\to\pm\infty ,\quad f(x)\to 0\)
Short run behavior, as \(x\to 0,\quad f(x)\to\infty \)(there are no horizontal or vertical 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.
Solution
Transforming the graph left 2 and up 3 would result in the function
\(f(x)=\dfrac{1}{x+2} +3\), or equivalently, by giving the terms a common denominator,
\(f(x)=\dfrac{3x+7}{x+2}\).
Shifting the toolkit function would give us this graph. Notice that this equation is undefined at \(x = -2\), and the graph also is showing a vertical asymptote at \(x = -2\).
As\(x \to -2^{-}\),\(f(x) \to -\infty\), and
as \(x \to -2^{+}\),\(f(x) \to \infty\)
As the inputs grow large, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at \(y=3\).
As \(x \to \pm \infty\), \(f(x) \to 3\).
Notice that horizontal and vertical asymptotes get shifted left 2 and up 3 along with the function.
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.
- Answer
-
The function and the asymptotes are shifted 3 units right and 4 units down.
As \(x \to 3\), \(f(x) \to \infty\) and as \(x \to \pm \infty\), \(f(x) \to -4\)
In the previous example, we shifted a toolkit function in a way that resulted in a function of the form \(f(x)=\dfrac{3x+7}{x+2}\). This is an example of a more general rational function.
A rational function is a function that can be written as the ratio of two polynomials, \(P(x)\) and \(Q(x)\).
\[f(x)=\dfrac{P(x)}{Q(x)} =\dfrac{a_{0} +a_{1} x+a_{2} x^{2} +\cdots +a_{p} x^{p} }{b_{0} +b_{1} x+b_{2} x^{2} +\cdots +b_{q} x^{q} }\]
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 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 \(t\) minutes.
Solution
Notice that 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:
\[\text{water}=100+10t\nonumber\]
\[\text{sugar}=5+1t\nonumber\]
The concentration, \(C\), will be the ratio of pounds of sugar to gallons of water
\[C(t)=\dfrac{5+t}{100+10t}\nonumber \]
Finding Asymptotes and Intercepts
Given a rational function, as part of investigating the short run behavior we are interested in finding any vertical and horizontal asymptotes, as well as finding any vertical or horizontal intercepts, as we have done in the past.
To find vertical asymptotes, we notice that the vertical asymptotes in our examples occur when the denominator of the function is undefined. With one exception, a vertical asymptote will occur whenever the denominator is undefined.
Find the vertical asymptotes of the function \(k(x)=\dfrac{5+2x^{2} }{2-x-x^{2} }\)
Solution
To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:
\[\begin{array}{l} {2-x-x^{2} =0} \\ {(2+x)(1-x)=0} \\ {x=-2,\; 1} \end{array}\nonumber\]
This indicates two vertical asymptotes, which a look at a graph confirms.
The exception to this rule can occur when both the numerator and denominator of a rational function are zero at the same input.
Find the vertical asymptotes of the function \(k(x)=\dfrac{x-2}{x^{2} -4}\).
Solution
To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:
\[\begin{array}{l} {x^{2} -4=0} \\ {x^{2} =4} \\ {x=-2,\; 2} \end{array}\nonumber \]
However, the numerator of this function is also equal to zero when \(x = 2\). Because of this, the function will still be undefined at 2, since \(\dfrac{0}{0}\) is undefined, but the graph will not have a vertical asymptote at \(x = 2\).
The graph of this function will have the vertical asymptote at \(x = -2\), but at \(x = 2\) the graph will have a hole: a single point where the graph is not defined, indicated by an open circle.
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.
A hole occurs in the graph of a rational function if an input causes both numerator and denominator to be zero. In this case, factor the numerator and denominator and simplify; if the simplified expression still has a zero in the denominator at the original input the original function has a vertical asymptote at the input, otherwise it has a hole.
To find horizontal asymptotes, we are interested in the behavior of the function as the input grows large, so we consider long run behavior of the numerator and denominator separately. Recall that a polynomial’s long run behavior will mirror that of the leading term. Likewise, a rational function’s long run behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions.
There are three distinct outcomes when this analysis is done:
Case 1 : The degree of the denominator > degree of the numerator
\[f(x)=\dfrac{3x+2}{x^{2} +4x-5}\nonumber \]
In this case, the long run behavior is \(f(x)\approx \dfrac{3x}{x^{2} } =\dfrac{3}{x}\). This tells us that as the inputs grow large, this function will behave similarly to the function \(g(x)=\dfrac{3}{x}\). As the inputs grow large, the outputs will approach zero, resulting in a horizontal asymptote at \(y=0\).
As \(x\to \pm \infty\), \(f(x)\to 0\)
Case 2: The degree of the denominator < degree of the numerator
\[f(x)=\dfrac{3x^{2} +2}{x-5}\nonumber \]
In this case, the long run behavior is\(f(x) \approx \dfrac{3x^{2} }{x} =3x\). This tells us that as the inputs grow large, 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.
As\(x\to \pm \infty\),\(f(x)\to \pm \infty\), respectively.
Case 3: The degree of the denominator = degree of the numerator
\[f(x)=\dfrac{3x^{2} +2}{x^{2} +4x-5}\nonumber \]
In this case, the long run behavior is \(f(x) \approx \dfrac{3x^{2} }{x^{2} } =3\). This tells us that as the inputs grow large, this function will behave like the function \(g(x)=3\), which is a horizontal line. As \(x \to \pm \infty\), \(f(x) \to 3\), resulting in a horizontal asymptote at \(y=3\).
The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.
Degree of denominator > degree of numerator: Horizontal asymptote at \(y=0\)
Degree of denominator < degree of numerator: No horizontal asymptote
Degree of denominator = degree of numerator: Horizontal asymptote at ratio of leading coefficients.
In the sugar concentration problem from earlier, we created the equation
\[C(t)=\dfrac{5+t}{100+10t}\nonumber \]
Find the horizontal asymptote and interpret it in context of the scenario.
Solution
Both the numerator and denominator are linear (degree 1), so since 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 10\(t\), with coefficient 10. The horizontal asymptote will be at the ratio of these values: As \(t \to \infty\), \(C(t) \to \dfrac{1}{10}\). This function will have a horizontal asymptote at \(y=\dfrac{1}{10}\).
This tells us that as the input gets large, the output values will approach 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 1/10 pounds per gallon.
Find the horizontal and vertical asymptotes of the function
\[f(x)=\dfrac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}\nonumber \]
Solution
First, note this function has no inputs that make both the numerator and denominator zero, so there are no potential holes. 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, 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 that 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 \(y=0\).
Find the vertical and horizontal asymptotes of the function \(f(x)=\dfrac{(2x-1)(2x+1)}{(x-2)(x+3)}\)
- Answer
-
Vertical asymptotes at \(x = 2\) and \(x = -3\); horizontal asymptote at \(y = 4\)
Intercepts
As with all functions, a rational function will have a vertical intercept when the input is zero, if the function is defined at zero. It is possible for a rational function to not have a vertical intercept if the function is undefined at zero.
Likewise, a rational function will have horizontal intercepts at the inputs that cause the output to be zero (unless that input corresponds to a hole). It is possible there are no horizontal intercepts. Since a fraction is only equal to zero when the numerator is zero, horizontal intercepts will occur when the numerator of the rational function is equal to zero.
Find the intercepts of \(f(x)=\dfrac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}\)
Solution
We can find the vertical intercept by evaluating the function at zero
\[f(0)=\dfrac{(0-2)(0+3)}{(0-1)(0+2)(0-5)} =\dfrac{-6}{10} =-\dfrac{3}{5}\nonumber \]
The horizontal intercepts will occur when the function is equal to zero:
\[0=\dfrac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}\nonumber \] This is zero when the numerator is zero
\[\begin{array}{l} {0=(x-2)(x+3)} \\ {x=2,\; -3} \end{array}\nonumber \]
Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function and find the horizontal and vertical intercepts and the horizontal and vertical asymptotes.
- Answer
-
For the transformed reciprocal squared function, we find the rational form.
\[f(x) = \dfrac{1}{(x - 3)^2} - 4 = \dfrac{1 - 4(x - 3)^2}{(x - 3)^2} = \dfrac{1 - 4(x^2 - 6x + 9)}{(x - 3)(x - 3)} = \dfrac{-4x^2 + 24x - 35}{x^2 - 6x + 9}\nonumber \]
Since the numerator is the same degree as the denominator we know that as \(x \to \pm \infty\), \(f(x) \to -4\), \(y = -4\) is the horizontal asymptote. Next, we set the denominator equal to zero to find the vertical asymptote at \(x = 3\), because as \(x \to 3\), \(f(x) \to \infty\). We set the numerator equal to 0 and find the horizontal intercepts are at (2.5, 0) and (3.5, 0), then we evaluate at 0 and the vertical intercept is at \((0, \dfrac{-35}{9})\)
From the previous example, you probably noticed that the numerator of a rational function reveals the horizontal intercepts of the graph, while the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Happily, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.
When factors of the denominator have integer powers greater than one, the behavior at the corresponding vertical asymptote will mirror one of the two toolkit reciprocal functions.
We get this behavior 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.
We get this behavior 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.
F or example, the graph of \(f(x)=\dfrac{(x+1)^{2} (x-3)}{(x+3)^{2} (x-2)}\) is shown here.
At the horizontal intercept \(x = -1\) corresponding to the \((x+1)^{2}\)factor of the numerator, the graph bounces at the intercept, consistent with the quadratic nature of the factor.
At the horizontal intercept \(x = 3\) corresponding to the \((x-3)\)factor of the numerator, the graph passes through the axis as we’d 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 \(\dfrac{1}{x^{2} }\) toolkit.
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, consistent with the behavior of the \(\dfrac{1}{x}\) toolkit.
Sketch a graph of \(f(x)=\dfrac{(x+2)(x-3)}{(x+1)^{2} (x-2)}\).
Solution
We can start our sketch by finding intercepts and asymptotes. Evaluating the function at zero gives the vertical intercept:
\[f(0)=\dfrac{(0+2)(0-3)}{(0+1)^{2} (0-2)} =3\nonumber \]
Looking at when the numerator of the function is zero, we can determine the graph will have horizontal intercepts at \(x = -2\) and \(x = 3\). At each, the behavior will be linear, with the graph passing through the intercept.
Looking at when the denominator of the function is zero, we can determine the graph will have vertical asymptotes at \(x = -1\) and \(x = 2\).
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 horizontal intercepts between the vertical asymptotes, and the vertical intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph.
Since the factor associated with the vertical asymptote at \(x = -1\) was squared, we know the graph will have the same behavior on both sides of the asymptote. Since the graph heads towards positive infinity as the inputs approach the asymptote on the right, the graph will head towards 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.
After passing through the horizontal intercepts, the graph will then level off towards an output of zero, as indicated by the horizontal asymptote.
Given the function \(f(x)=\dfrac{(x+2)^{2} (x-2)}{2(x-1)^{2} (x-3)}\), use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.
- Answer
-
Horizontal asymptote at \(y = 1/2\).
Vertical asymptotes are at \(x = 1\), and \(x = 3\).
Vertical intercept at \((0, 4/3)\),
Horizontal intercepts \((2, 0)\) and \((-2, 0) (-2, 0)\) is a double zero and the graph bounces off the axis at this point.
\((2, 0)\) is a single zero and crosses the axis at this point.
Since a rational function written in factored form will have a horizontal intercept where each factor of the numerator is equal to zero, we can form a numerator that will pass through a set of horizontal intercepts by introducing a corresponding set of factors. Likewise, since 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.
If a rational function has horizontal intercepts at \(x=x_{1} ,x_{2} ,\ldots ,x_{n}\), and vertical asymptotes at \(x=v_{1} ,v_{2} ,\ldots ,v_{m}\) then the function can be written in the form
\[f(x)=a\dfrac{(x-x_{1} )^{p_{1} } (x-x_{2} )^{p_{2} } \cdots (x-x_{n} )^{p_{n} } }{(x-v_{1} )^{q_{1} } (x-v_{2} )^{q_{2} } \cdots (x-v_{m} )^{q_{n} } }\]
where the powers p\({}_{i}\) or q\({}_{i}\) on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor a can be determined given a value of the function other than the horizontal intercept, or by the horizontal asymptote if it is nonzero.
Write an equation for the rational function graphed here.
Solution
The graph appears to have horizontal 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 \(\dfrac{1}{x}\), 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 \(\dfrac{1}{x^{2} }\), with the graph heading toward negative infinity on both sides of the asymptote.
Utilizing this information indicates an function of the form
\[f(x)=a\dfrac{(x+2)(x-3)}{(x+1)(x-2)^{2} }\nonumber \]
To find the stretch factor, we can use another clear point on the graph, such as the vertical intercept (0,-2):
\[\begin{array}{l} {-2=a\dfrac{(0+2)(0-3)}{(0+1)(0-2)^{2} } } \\ {-2=a\dfrac{-6}{4} } \\ {a=\dfrac{-8}{-6} =\dfrac{4}{3} } \end{array}\nonumber \]
This gives us a final function of \(f(x)=\dfrac{4(x+2)(x-3)}{3(x+1)(x-2)^{2} }\)
Oblique Asymptotes
Earlier we saw graphs of rational functions that had no horizontal asymptote, which occurs when the degree of the numerator is larger than the degree of the denominator. We can, however, describe in more detail the long-run behavior of a rational function.
Describe the long-run behavior of \(f(x)=\dfrac{3x^{2} +2}{x-5}\)
Solution
Earlier we explored this function when discussing horizontal asymptotes. We found the long-run behavior is\(f(x) \approx \dfrac{3x^{2} }{x} =3x\), meaning that \(x\to \pm \infty\),\(f(x)\to \pm \infty\), respectively, and there is no horizontal asymptote.
If we were to do polynomial long division, we could get a better understanding of the behavior as \(x\to \pm \infty\).
This means \(f(x)=\dfrac{3x^{2} +2}{x-5}\) can be rewritten as \(f(x)=3x+15+\dfrac{77}{x-5}\).
As \(x \to \pm \infty\), the term \(\dfrac{77}{x-5}\) will become very small and approach zero, becoming insignificant. The remaining \(3x+15\)then describes the long-run behavior of the function: as \(x\to \pm \infty\), \(f(x) \to 3x+15\).
We call this equation \(y=3x+15\) the oblique asymptote of the function.
In the graph, you can see how the function is approaching the line on the far left and far right.
To explore the long-run behavior of a rational function,
1) Perform polynomial long division (or synthetic division)
2) The quotient will describe the asymptotic behavior of the function
When this result is a line, we call it an oblique asymptote , or slant asymptote .
Find the oblique asymptote of \(f(x)=\dfrac{-x^{2} +2x+1}{x+1}\)
Solution
Performing polynomial long division:
This allows us to rewrite the function as
\[f(x)=-x+3-\dfrac{2}{x+1}\nonumber\].
The quotient, \(y=-x+3\), is the oblique asymptote of \(f(x)\). Just like functions we saw earlier approached their horizontal asymptote in the long run, this function will approach this oblique asymptote in the long run.
Find the oblique asymptote of \(f(x)=\dfrac{1+7x-2x^{2} }{x-2}\)
- Answer
-
Using long division:
The oblique asymptote is \(y=-2x+3\)
While we primarily concern ourselves with oblique asymptotes, this same approach can describe other asymptotic behavior.
Describe the long-run shape of \(f(x)=\dfrac{-x^{3} -x^{2} +4x+2}{x+1}\)
Solution
We could rewrite this using long division as
\[f(x)=-x^{2} +4+\dfrac{2}{x+1}\nonumber \]
Just looking at the quotient gives us the asymptote, \(y=-x^{2} +4\).
This suggests that in the long run, the function will behave like a downwards opening parabola. The function will also have a vertical asymptote at \(x=-1\).
Important Topics of this Section
Inversely proportional; Reciprocal toolkit function
Inversely proportional to the square; Reciprocal squared toolkit function
Horizontal Asymptotes
Vertical Asymptotes
Rational Functions
Finding intercepts, asymptotes, and holes.
Given equation sketch the graph
Identifying a function from its graph
Oblique Asymptotes