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Elementary Functions
Part 2, Polynomials Lecture 2.6a, Rational Functions
Dr. Ken W. Smith
Sam Houston State University
2013
Smith (SHSU) Elementary Functions 2013 1 / 42
Rational Functions
A rational function f (x) is a function which is the ratio of two polynomials, that is,
f (x) = n d((xx))
where n(x) and d(x) are polynomials.
For example, f (x) = 3 x
(^2) −x− 4 x^2 − 2 x− 8 is a rational function.
In this case, both the numerator and denominator are quadratic polynomials.
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Algebra with mixed fractions
Consider the function g(x) which appeared in an earlier lecture:
g(x) :=
x + 2
2 x − 3 2 x + 1
This function, g, is a rational function. We can put g into a fraction form, as the ratio of two polynomials, by finding a common denominator.
The least common multiple of the denominators x + 2 and 2 x + 1 is simply their product, (x + 2)(2x + 1). We may write g(x) as a fraction with this denominator if we multiply the first term by 1 = 22 xx+1+1 , multiply the second term by 1 = x x+2+2 and multiply the third term by 1 = (2(2xx+1)(+1)(xx+2)+2). Then
g(x) = (
x + 2
(2x + 1) (2x + 1)
2 x − 3 2 x + 1
(x + 2) (x + 2)
(2x + 1)(x + 2) (2x + 1)(x + 2)
Combine the numerators (since there is a common denominator):
g(x) =
(2x + 1) + (2x − 3)(x + 2) + (x − 5)(2x + 1)(x + 2) (2x + 1)(x + 2)
Algebra with mixed fractions
g(x) =
(2x + 1) + (2x − 3)(x + 2) + (x − 5)(2x + 1)(x + 2) (2x + 1)(x + 2)
The numerator is a polynomial of degree 3 (it can be expanded out to 2 x^3 − 3 x^2 − 20 x − 15 ) and the denominator is a polynomial of degree 2.
The algebra of mixed fractions, including the use of a common denominator, is an important tool when working with rational functions.
Zeroes of rational functions
Given a rational function f (x) = n d((xx)) we are interested in the y- and x- intercepts.
The y-intercept occurs where x is zero and it is usually very easy to compute
f (0) = n d(0)(0).
However, the x-intercepts occur where y = 0, that is, where
0 = n d((xx)).
As a first step to solving this equation, we may multiply both sides by d(x) and so concentrate on the zeroes of the numerator, solving the equation
0 = n(x).
At this point, we have reduced the problem to finding the zeroes of a polynomial, exercises from a previous lecture! Smith (SHSU) Elementary Functions 2013 5 / 42
Algebra with mixed fractions
For example, suppose
h 1 (x) = x
(^2) − 6 x+ x^2 +x− 12. The y-intercept is (− 23 , 0) since h 1 (0) = (^) −^812 = − 23.
The x-intercepts occur where x^2 − 6 x + 8 = 0. Factoring x^2 − 6 x + 8 = (x − 4)(x − 2) tells us that x = 4 and x = 2 should be zeroes and so (4, 0) and (2, 0) are the x-intercepts. (We do need to check that they do not make the denominator zero – but they do not.)
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Poles and holes
Since rational functions have a denominator which is a polynomial, we must worry about the domain of the rational function. In particular, any real number which makes the denominator zero cannot be in the domain.
The domain of a rational function is all the real numbers except those which make the denominator equal to zero. For example
h 1 (x) = x
(^2) − 6 x+ x^2 +x− 12 =^
(x−4)(x−2) (x+4)(x−3) has domain (−∞, −4) ∪ (− 4 , 3) ∪ (3, ∞) since only x = − 4 and x = 3 make the denominator zero.
Poles and holes
The domain of a rational function is all the real numbers except those which make the denominator equal to zero. There are two types of zeroes in the denominator. One common type is a zero of the denominator which is not a zero of the numerator. In that case, the real number which makes the denominator zero is a “pole” and creates, in the graph, a vertical asymptote.
For example, using h 1 (x) = x
(^2) − 6 x+ x^2 +x− 12 from before, we see that x^2 + x − 12 = (x + 4)(x − 3) has zeroes at x = − 4 and x = 3.
Since neither x = − 4 and x = 3 are zeroes of the numerator, these values give poles of the function h 1 (x) and in the graph we will see vertical lines x = − 4 and x = 3 that are “approached” by the graph.
The lines are called asymptotes, in this case we have vertical asymptotes with equations x = − 4 and x = 3.
The sign diagram of a rational function
When we looked at graphs of polynomials, we viewed the zeroes of the polynomial as dividers or fences, separating regions of the x-axis from one another.
Within a particular region, between the zeroes, the polynomial has a fixed sign, (+) or (−), since changing sign requires crossing the x-axis.
We used this idea to create the sign diagram of a polynomial, a useful tool to guide us in the drawing of the graph of the polynomial.
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The sign diagram of a rational function
Just as we did with polynomials, we can create a sign diagram for a rational function.
In this case, we need to use both the zeroes of the rational function and the vertical asymptotes as our dividers, our “fences” between the sign changes.
To create a sign diagram of rational function, list all the x-values which give a zero or a vertical asymptote. Put them in order. Then between these x-values, test the function to see if it is positive or negative and indicate that by a plus sign or minus sign.
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The sign diagram of a rational function
For example, consider the function
h 2 (x) = x
(^2) − 6 x+ x^2 −x− 12 =^
(x−4)(x−2) (x−4)(x+3) from before. It has a zero at x = 2 and a vertical asymptote x = − 3.
The sign diagram represents the values of h 2 (x) in the regions divided by x = − 3 and x = 2. (For the purpose of a sign diagram, the hole at x = 4 is irrelevant since it does not effect the sign of the rational function.) To the left of x = − 3 , h 2 (x) is positive. Between x = − 3 and x = 1, h 2 (x) is negative. Finally, to the right of x = 1, h 2 (x) is positive.
So the sign diagram of h 2 (x) is
(+) | (−) | (+) − 3 1
Rational Functions
In the next presentation we look at the end behavior of rational functions.
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Elementary Functions
Part 2, Polynomials Lecture 2.6b, End Behavior of Rational Functions
Dr. Ken W. Smith
Sam Houston State University
2013
Smith (SHSU) Elementary Functions 2013 17 / 42
End-behavior of Rational Functions
Just as we did with polynomials, we ask questions about the “end behavior” of rational functions: what happens for x-values far away from 0, towards the “ends” of our graph?
In many cases this leads to questions about horizontal asymptotes and oblique asymptotes (sometimes called “slant asymptotes”).
Before we go very far into discussing end-behavior of rational functions, we need to agree on a basic fact.
As long as a polynomial p(x) has degree at least one (and so was not just a constant) then as x grows large p(x) also grows large in absolute value.
So, as x goes to infinity, (^) p(^1 x) goes to zero.
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End-behavior of Rational Functions
We explicitly list this as a lemma, a mathematical fact we will often use.
Lemma 1. Suppose that p(x) is a polynomial of degree at least 1. Then the rational function (^) p(^1 x) tends to zero as x gets large in absolute value.
In calculus terms, the limit as x goes to infinity of (^) p(^1 x) is zero.
Here is a slight generalization of the fact in Lemma 1:
Lemma 2. Suppose that f (x) = n d((xx)) is a rational function where the degree of n(x) is smaller than the degree of d(x).
Then the rational function n d((xx)) tends to zero as x grows large in absolute value.
In calculus terms, the limit as x goes to infinity of n d((xx)) is zero.
End-behavior of Rational Functions
Repeating the previous slide:
Lemma 2. Suppose that f (x) = n d((xx)) is a rational function where the degree of n(x) is smaller than the degree of d(x). Then the rational function n d((xx)) tends to zero as x grows large in absolute value.
This means that if f (x) = n d((xx)) is a rational function where the degree of n(x) is smaller than the degree of d(x) then as x gets large in absolute value, the graph approaches the x-axis.
The x-axis, y = 0, is a horizontal asymptote of the rational function n(x) d(x).
Horizontal asymptotes
Continuing with
h 1 (x) = x
(^2) − 6 x+ x^2 +x− 12. The same result occurs if we set x equal to negative numbers which are large in absolute value, such as x = − 1000000. More generally, as x gets large in absolute value, x^2 − 6 x+ x^2 +x− 12 begins to look like x^2 x^2 = 1. We conclude then that as x gets large in absolute value, f (x) approaches 1 and so y = 1 is a horizontal asymptote of f (x).
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Horizontal asymptotes
Below, is a graph of y = h 1 (x) = x
(^2) − 6 x+ x^2 +x− 12 , with the function drawn in blue and the various asymptotes drawn in green or red.
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Horizontal asymptotes
Example 3. Earlier we considered the rational function h 2 (x) = ((xx−−4)(4)(xx−+3)2). Like h 1 (x), this function has a horizontal asymptote y = 1 (drawn in green.)
Horizontal asymptotes
Example 4. Find the zeroes and vertical asymptotes of the rational function
g(x) =
3(x + 1)(x − 2) 4(x + 3)(x − 1) and draw the sign diagram. Then find the horizontal asymptotes.
Solution. Looking at the numerator 3(x + 1)(x − 2) of g(x) we see that the zeroes occur at x = − 1 and x = 2.
Looking at the denominator 4(x + 3)(x − 1) of g(x) we can see that the vertical asymptotes of g(x) are the lines x = − 3 and x = 1.
Horizontal asymptotes
Example 4, continued Find the zeroes and vertical asymptotes of the rational function
g(x) =
3(x + 1)(x − 2) 4(x + 3)(x − 1) and draw the sign diagram. Then find the horizontal asymptotes.
The sign diagram is
(+) | (−) | (+) | (−) | (+) − 3 − 1 1 2
There is one horizontal asymptote found by considering the end behavior of g(x). As x goes to infinity, g(x) = 3(4(xx+1)(+3)(xx−−2)1) approaches 3 x
2 4 x^2 =^
3 4 so the horizontal asymptote is the line y = 34.
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Oblique asymptotes
Consider a rational function we saw earlier: g(x) := (^) x+2^1 + 22 xx−+1^3 + x − 5 = 2 x
(^3) − 3 x (^2) − 20 x− 15 2 x^2 +5x+. If we do long division we get
x − 4 2 x^2 + 5x + 2
2 x^3 − 3 x^2 − 20 x − 15 − 2 x^3 − 5 x^2 − 2 x − 8 x^2 − 22 x − 15 8 x^2 + 20x + 8 − 2 x − 7 So g(x) = x − 4 +
− 2 x − 7 x^2 + 5x + 2 As x gets large in absolute value, g(x) ≈ x − 4 and the graph begins to look like that of y = x − 4. So the line y = x − 4 is an asymptote for g(x). In this case this asymptote is neither vertical nor horizontal; it is an slant asymptote. Slant asymptotes are also called oblique asymptotes.
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Summary
Asymptotes of a function are lines that approximate a rational function “in the large”, as we zoom out and look at global behavior of the rational function.
Vertical asymptotes occur where the denominator is zero and the numerator is not zero.
If the degree of the numerator of a rational function is less than the degree of the denominator then the rational function has horizontal asymptote y = 0.
If the degree of the numerator of a rational function is equal to the degree of the denominator then the rational function has a horizontal asymptote which can be found either by doing long division or by focusing on the leading coefficients of the numerator and denominator.
If the degree of the numerator of a rational function is one more than the degree of the denominator then the rational function has an oblique asymptote which can be found as the quotient after long division.
Summary
Remember! Asymptotes are lines! When asked for the asymptotes of a rational function, make sure to give equations of lines!
In the next presentation we work through the “Six Steps” to graphing a rational function.
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Graphing a rational function
The graph of h(x) = 10 x
(^2) − 250 x^2 +6x+8 is given below. The graph is in blue; the vertical asymptotes are in red and the horizontal asymptote is in green.
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Graphing a rational function
Example 2. Let f (x) = 3 x
(^3) +x (^2) − 12 x− 4 x^2 − 2 x− 8 Find all intercepts, zeroes and then graph this function, displaying the features found.
Solution. We will work through the six steps. 1 The denominator factors as x^2 − 2 x − 8 = (x − 4)(x + 2). So the domain is the set of real numbers where the denominator is not zero, that is, (−∞, −2) ∪ (− 2 , 4) ∪ (4, ∞). 2 Notice that when we evaluate the numerator at x = − 2 , we get zero. So (x + 2) is a factor of both the numerator and the denominator.
Recognizing that x + 2 is a factor of the numerator, we can further factor the numerator using the techniques we learned in the sections on polynomial zeroes. The numerator factors as (x + 2)(x − 2)(3x + 1). So f (x) = (x+2)((x+2)(x−2)(3x−4)x+1). The point where x = − 2 is a hole (removable singularity.) If we put the rational function into lowest terms, it becomes f (x) = (x−(2)(3x−4)x+1) , x 6 = − 2
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The sign diagram of a rational function
Continuing with Example 2 where
f (x) = 3 x
(^3) +x (^2) − 12 x− 4 x^2 − 2 x− 8
3 The y-intercept is when y = f (0) = − −^24 = 2. The x-intercepts occur when we set (x − 2)(3x + 1) equal to zero and so these occur when x = 2 and when x = − 13. 4 A hole occurs when x = − 2. Looking at the reduced form, we see that the hole has y-value (− 2 −2)(3(−2)+1) (− 2 −4) =^
(−4)(−5) − 6 =^ −^
10
There is one vertical asymptote; it is x = 4.
Graphing a rational function
5 Analyze the end behavior of the rational function. Find the horizontal or slant asymptote, if one exists. To analyze the end behavior, we do long division: 3 x − 2 x − 1
3 x^2 − 5 x − 2 − 3 x^2 + 3x − 2 x − 2 2 x − 2 − 4 and write f (x) = 3x − 2 − (^) x−^41 and so there is an oblique (slant) asymptote at y = 3x − 2.
Graphing a rational function
Continuing with Example 2 where f (x) = 3 x
(^3) +x (^2) − 12 x− 4 x^2 − 2 x− 8 6 Use a sign diagram and plot additional points, as needed, to sketch the graph of y = f (x).
The sign diagram is
The graph is drawn below in blue (with asymptotes in colors red and green.)
The graph of f (x) (in blue) with vertical asymptotes (red) and horizontal asymptote (green). The hole at (− 2 , − 10 ) is not shown.
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Rational Functions
In the next series of lectures, we move on to a new topic, exponential functions and their inverse functions (logarithms.)
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