4. Graphing Rational Functions
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Chapter 4
4.
Graphing Rational Functions
Rational functions are mathematical expressions that represent the ratio of two polynomials. These functions can exhibit various behaviors on a graph, such as discontinuities, which can be in the form of holes, jumps, or asymptotes. Asymptotes can be vertical, horizontal, or even oblique. The graph of a rational function can provide insights into real-world scenarios, such as modeling the annual income per capita or understanding the dynamics of road maintenance. Additionally, the graph can depict the behavior of medication in the bloodstream over time or the cost dynamics of producing items in a factory. Understanding how to graph these functions is crucial for interpreting and predicting real-world phenomena.
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Lesson Settings & Tools
| | 14 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Graphing Rational Functions
Slide of 14
A reciprocal function can be considered as the quotient where the denominator is a polynomial with a degree of 1. In other words, there is a variable in the denominator. In this lesson, rational functions in which the numerator and the denominator have a degree of at least 1 will be graphed.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Average Cost Function
LaShay's school, Jeffereson High, visits a local pencil factory. The average cost, in thousand dollars, of producing x thousand pencils is given by the following function. A(x)= 0.2x^2+50x+500/x
a Graph the average cost function.
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Discussion
Rational Functions
A rational function is a function that contains a rational expression. Any function that can be written as the quotient of two polynomial functions p(x) and q(x) is a rational function.
f(x) = p(x)/q(x), q(x) ≠ 0
For any values of x where q(x) = 0, the rational function is undefined. These values are excluded from the domain of the function. One example of a rational function is the reciprocal function. f(x) = 1/x This function has two asymptotes, the x-axis and the y-axis. The applet shows the graph of the reciprocal function and some other rational functions.
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Discussion
Discontinuity of Rational Functions
The graph of a rational function can be a smooth continuous curve, or it can have jumps, breaks, or holes. By looking at the graph of a function, functions can be categorized into two groups: continuous or discontinuous.
Concept
Discontinuous - Function
A function is said to be continuous if its graph can be drawn without lifting the pencil. Otherwise, the function is said to be discontinuous.
Concept
Point of Discontinuity
A point of discontinuity of a function is a point with the x-coordinate that makes the function value undefined. A point of discontinuity can also be considered as an excluded value of the function rule.
Removable Discontinuity
If a function can be redefined so that its point of discontinuity is removed, it is called a removable discontinuity. A removable discontinuity may occur when there is a rational expression with common factors in the numerator and denominator. Consider the following rational function. f(x) = x^2-4/x-2 For this function, x=2 is a point of discontinuity because its denominator is 0 when x=2. Next, notice that the function can be simplified using a difference of squares.
f(x) = x^2-4/x-2
▼
Simplify right-hand side
WritePow
Write as a power
f(x) = x^2-2^2/x-2
FacDiffSquares
a^2-b^2=(a+b)(a-b)
f(x) = (x+2)(x-2)/x-2
SimpQuot
Simplify quotient
f(x) = x + 2, x≠ 2
However, the function can be made continuous by redefining it at x=2. If 2 is substituted for x in the function f(x)=x+2, it is found that f(2)=4. Therefore, the following function is continuous at x=2.
g(x) = x^2-4/x-2, & if x ≠ 2 [0.7em]
4, & if x = 2
This means that x=2 is a removable discontinuity. Note that a removable discontinuity is typically a hole
in the graph.
Non-Removable Discontinuity
When a function cannot be redefined so that the point of discontinuity becomes a valid input, it is called a non-removable discontinuity. Consider, for example, y = 1x+1. Since x=-1 is not in the domain of the function, there is a point of discontinuity at x=- 1.
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Example
Modeling Annual Income Per Capita
LaShay lives in Des Moines, where the total annual amount of personal income I in millions dollars, and the population P in millions, are modeled by the following functions. I(t) & = 5000t+100 000/0.02t+1 [0.8em] P(t) & = 0.36t +10 800 In these functions, t represents the number of years that have passed since 1990.
a Write a function C(t) modeling the annual income per capita in Des Moines.
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b Identify the points of discontinuity of C(t), if any.
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Hint
a Divide the total amount of personal income by the population. Recall how to divide rational expressions.
b For which values of t is the function C(t) undefined?
Solution
a Take a look at what the given equations represent.
| Total Annual Income | Population |
|---|---|
| I(t) = 5000t+100 000/0.02t+1 | P(t) = 0.36t +10 800 |
In order to write a function for the annual income per person C(t), the total annual income I(t) must be divided by the population P(t). This will give the annual income of a person.
C(t)= I(t) ÷ P(t)
SubstituteExpressions
Substitute expressions
C(t)= (5000t+100 000/0.02t+1) ÷ (0.36t+10 800)
DivFracSL
.a/b /c.= a/b* c
C(t)= 5000t+100 000/(0.02t+1)(0.36t+10 800)
The binomials in the denominator can be multiplied.
C(t)= 5000t+100 000/(0.02t+1)(0.36t+10 800)
MultPar
Multiply parentheses
C(t)= 5000t+100 000/0.0072t^2+216t+0.36t+10 800
AddTerms
Add terms
C(t)= 5000t+100 000/0.0072t^2+216.36t+10 800
b Consider the function C(t) written in Part A.
C(t)= 5000t+100 000/(0.02t+1)(0.36t+10 800) Recall that division by zero is not defined. Therefore, the rational function is undefined when either of the parentheses equals zero.
| C(t) is undefined | |
|---|---|
| 0.02t+1=0 | 0.36t+10 800=0 |
| t= - 50 | t= - 30 000 |
This means that t=- 50 and t=- 30 000 are excluded values. Domain All real numbers except t= - 50 and t= - 30 000 If a real number a is not in the domain of a function, then the function has a point of discontinuity at x=a. Since the domain is all real numbers except t=- 50 and t=- 30 000, there are two points of discontinuity of the function. Points of Discontinuity t= - 50 and t= - 30 000
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Example
Finding Point of Discontinuity
On the way to the pencil factory, LaShay noticed that road maintenance work is being carried out to fill a hole
on the road.
The equation below models the road to the factory. R(x) = 3x^2+3x/x^3+x^2+x+1
a Find the x-coordinate of the point where the road maintenance is being completed.
b The municipality of Des Moines plans to build a new road, which is modeled by the following function.
N(x) = 2x^2-x-3/10x+10 Graph the function.
Answer
a x=- 1
b Graph:
Hint
a Start by factoring the numerator and the denominator of the function. Find the point of discontinuity of the graph.
b Check if the numerator and the denominator have a common factor.
Solution
a The terms in the numerator of the function have a common factor of 3x. Also, the denominator can be factored by grouping.
R(x) = 3x^2+3x/x^3+x^2+x+1
FactorOut
Factor out 3x
R(x) = 3x(x+1)/x^3+x^2+x+1
FactorOut
Factor out x^2
R(x) = 3x(x+1)/x^2(x+1)+x+1
FactorOut
Factor out (x+1)
R(x) = 3x(x+1)/(x^2+1)(x+1)
Notice that there is no x-value that can make x^2+1 equal zero. However, the other factor in the denominator makes the function undefined when x=- 1. Therefore, this value is excluded from the domain. Domain All real numbers except x= - 1 Recall that the point of discontinuity occurs at x=a when a is not in the domain of a function. Consequently, there is a point of discontinuity at x=- 1. Point of Discontinuity x= - 1 In the context of the problem, this is the x-coordinate of the point where the road is being repaired.
Notice that it is also a removable discontinuity because x+1 is the common factor in the numerator and denominator.
b Start by factoring the numerator and denominator to see if the function can be simplified.
N(x) = 2x^2-x-3/10x+10
N(x) = (2x-3)(x+1)/10(x+1)
The function is undefined when x=- 1. This means that the domain of the function is all real numbers except x=- 1. After simplification, the function will be a linear function except at x=- 1. N(x)& = (2x-3) (x+1)/10 (x+1) [0.9em] & = 2x-3/10, x ≠ - 1 Note that 2x-310 = 2x10- 310. Therefore, the graph of the given function is the graph of the function y = 2x10- 310 with a hole at x=- 1.
This graph also has a removable discontinuity at x=- 1 because x+1 is the common factor in the numerator and denominator of the rational function.
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Discussion
Asymptotes of Rational Functions
An asymptote of a graph is an imaginary line that the graph gets close to as x goes to plus or minus infinity or a particular number. For example, the graph of the rational function f(x) = 1x has two asymptotes — the x-axis and the y-axis.
Analyzing the diagram, the following can be observed.
- As x approaches infinity and as x approaches negative infinity, the value of the function approaches 0. Therefore, y = 0, or the x-axis, is a horizontal asymptote of the graph of f.
- As x approaches 0, the value of the function approaches either positive or negative infinity. Therefore, x = 0, or the y-axis, is a vertical asymptote of the graph of f.
In the coordinate plane below, the asymptotes for three different graphs are shown.
Consider a rational function where p(x) is a polynomial with a leading coefficient of a and a degree of m and where q(x) is a polynomial with a leading coefficient of b and a degree of n. f(x) = p(x)/q(x) = ax^m + .../bx^n + ... To determine the asymptotes of the function, the denominator and the degrees of the polynomials in the numerator and denominator must be analyzed.
Identifying Vertical Asymptotes
To identify the vertical asymptotes, the function should be in its simplest form. That is, if p(x) and q(x) have common factors, the function should be simplified first. The vertical asymptotes will occur at the zeros of the denominator. Check out the following examples.
Identifying Horizontal Asymptotes
The degrees of polynomials in the numerator and denominator of a rational function will determine if the graph has a horizontal asymptote.
| f(x) = p(x)/q(x) = ax^m + .../bx^n + ... | |
|---|---|
| Horizontal Asymptote | If m >n, there is no horizontal asymptote. |
| If m < n, the horizontal asymptote is y=0. | |
| If m = n, the horizontal asymptote is y = ab. |
The graph of a rational function may have one or more vertical asymptotes and at most one horizontal asymptote.
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Example
Identifying Asymptotes
LaShay takes a travel-sickness pill about 45 minutes before the school bus leaves for the field trip. The amount A in milligrams of the medication in her bloodstream is modeled by the following rational function. A(t) = 6t^2-24t/t^3-t^2-10t-8 Here, t represents the time in hours after one pill is taken. Identify the asymptotes of the function.
Answer
Vertical Asymptotes: t=- 2 and t=- 1
Horizontal Asymptote: y=0
Horizontal Asymptote: y=0
Hint
Solution
To identify the asymptotes, the numerator and denominator of the function A(t) need to be factored. Notice that the terms in the numerator have a common factor of 6t.
To factor the polynomial in the denominator, use the Factor Theorem.
|
Factor Theorem |
|
The expression x-a is a factor of a polynomial if and only if the value a is a zero of the related polynomial function. |
Now, the aim is to find all real zeros of the polynomial. Before doing so, identify the degree of the polynomial. p(t)=t^3-t^2-10t-8 It is a polynomial with a degree of 3. Subsequently, by the Fundamental Theorem of Algebra, it is known know that p(t) has exactly three roots. Additionally, by the Rational Root Theorem, integer roots must be factors of the constant term. Since the constant term of p(t) is - 8, the possible integer roots are ± 1, ± 2, ± 4, and ± 8.
| t | t^3-t^2-10t-8 | p(t)=t^3-t^2-10t-8 |
|---|---|---|
| 1 | 1^3-( 1)^2-10( 1)-8 | - 18 |
| - 1 | - 1^3-( - 1)^2-10( - 1)-8 | 0 |
| 2 | 2^3-( 2)^2-10( 2)-8 | - 24 |
| - 2 | - 2^3-( - 2)^2-10( - 2)-8 | 0 |
| 4 | 4^3-6( 4)^2-7( 4)+60 | 0 |
Three roots for p(t) were found, so there is no need for further investigation. In conclusion, - 1, - 2, and 4 are all the real roots of t^3-t^2-10t-8. Next, apply the Factor Theorem to write its factored form. t^3-t^2-10t-8 = (t+1)(t+2)(t-4) Now that the factored form of the denominator has been found, the given function can be simplified.
A(t) = 6t(t-4)/t^3-t^2-10t-8
Substitute
t^3-t^2-10t-8= (t+1)(t+2)(t-4)
A(t) = 6t(t-4)/(t+1)(t+2)(t-4)
CrossCommonFac
Cross out common factors
A(t) = 6t(t-4)/(t+1)(t+2)(t-4)
CancelCommonFac
Cancel out common factors
A(t) = 6t/(t+1)(t+2)
Since the numerator and the denominator share the factor t-4, there is a hole at t=4. Also, the denominator becomes zero when t=- 1 and t=- 2. These values make the function undefined. Therefore, the function has two vertical asymptotes. Vertical Asymptotes t = - 1 and t = - 2 To find the horizontal asymptote, compare the degrees of the polynomials in the numerator and denominator.
| y=ax^m/bx^n | Asymptote |
|---|---|
| m | y=0 |
| m>n | none |
| m=n | y=a/b |
Look at the degrees of the numerator and denominator of the function. A(t) = 6t^2-24t/t^3-t^2-10t-8 The degree of the denominator is greater than the degree of the numerator, m < n. Therefore, the horizontal asymptote is y=0. Horizontal Asymptotes y =0
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Discussion
Oblique Asymptote
An asymptote can be neither vertical nor horizontal. It can be a slanted line. In such cases, it is called an oblique asymptote or slant asymptote.
An oblique asymptote of a rational function occurs when the degree of the numerator is 1 more than that of the denominator. Therefore, a function with an oblique asymptote can never have a horizontal asymptote. To find the equation of the oblique asymptote of a rational function f(x)= p(x)q(x), p(x) is divided by q(x) using long division to get a quotient ax+b with a remainder r(x). f(x) = p(x)/q(x) ⇓ f(x) = ax+b+r(x) The equation of the oblique asymptote is of the form y = ax + b, where a is a non-zero real number.
Example
To illustrate how the technique is used, the asymptotes of the following function will be found. f(x) = x^3+2x/4x^2-1 The difference between the degrees of the numerator and denominator is 1. This means that its graph does not have a horizontal asymptote but an oblique asymptote. To find the oblique asymptote, polynomial long division will be used.
l r 4x^2 - 1 & |l x^3 + 2x
▼
Divide
x^3/4x^2= x/4
r x4 r 4x^2-1 & |l x^3+2x
Multiply term by divisor
r x4 rl 4x^2 - 1 & |l x^3+2x & x^3- x4
Subtract down
r x4 r 4x^2 - 1 & |l 9x4
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Example
Modeling Interest in The Trip
The students spent about 3 hours at the factory. At the end of this period, LaShay's teacher made a survey about the trip and drew a graph relating the students' interest in the trip and the time elapsed at the factory. The greater the S(t) value, the greater the interest of the students.
The graph above is the part of the graph of the following function in the first quadrant. S(t) = t/t+1-t^3/2(t+1)^2 Identify the asymptotes of the function.
Answer
Vertical Asymptotes: t=- 1
Oblique Asymptote: O(t)=- t/2+2
Oblique Asymptote: O(t)=- t/2+2
Hint
Start by subtracting the rational expressions on the right-hand side of the function.
Solution
First, the rational expressions on the right-hand side will be subtracted. The least common denominator of the expressions is 2(t+1)^2. Therefore, the first expression needs to be multiplied by 2(t+1)2(t+1) so that both have the same denominator.
S(t) = t/t+1-t^3/2(t+1)^2
▼
Simplify right-hand side
ExpandFrac
a/b=a * 2(t+1)/b * 2(t+1)
S(t) = t * 2(t+1)/(t+1)* 2(t+1)-t^3/2(t+1)^2
Multiply
Multiply
S(t) = 2t(t+1)/2(t+1)^2-t^3/2(t+1)^2
SubFrac
Subtract fractions
S(t) = 2t(t+1)-t^3/2(t+1)^2
Distr
Distribute 2t
S(t) = 2t^2+2t-t^3/2(t+1)^2
CommutativePropAdd
Commutative Property of Addition
S(t) = - t^3+2t^2+2t/2(t+1)^2
▼
Rewrite the denominator
S(t) = - t^3+2t^2+2t/2t^2+4t+2
The numerator and the denominator do not have any common factors. Therefore, the function does not have a point of discontinuity, or hole, in its graph. However, it has a vertical asymptote at t=- 1 because the function is undefined at that value. Vertical Asymptote t = -1 Now, to determine if the given rational function has a horizontal or oblique asymptote, the degrees of the polynomials in the numerator and the denominator will be compared. S(t) = - t^3+2t^2+2t/2(t+1)^2 ⇕ S(t) = - t^3+2t^2+2t/2t^2+4t+2 The degree of the numerator is greater than the degree of the denominator, so the graph has no horizontal asymptotes. However, since the degree of the numerator is 1 more than that of the denominator, the graph has an oblique asymptote. Its equation is found by dividing the numerator by the denominator. Use long division to do this.
l r 2t^2+4t+2 & |l - t^3+2t^2+2t
▼
Divide
- t^3/2t^2= - t/2
r - t2 r 2t^2+4t+2 & |l - t^3+2t^2+2t
Multiply term by divisor
r - t2 rl 2t^2+4t+2 & |l - t^3+2t^2+2t & - t^3 -2t^2-t
Subtract down
r - t2 r 2t^2+4t+2 & |l 4t^2+3t
▼
Divide
4t^2/2t^2= 2
r - t2 + 2 r 2t^2+4t+2 & |l 4t^2+3t
Multiply term by divisor
r- t2 + 2 rl 2t^2+4t+2 & |l 4t^2+3t & 4t^2+8t+4
Subtract down
r - t2 + 2 r 2t^2+4t+2 & |l - 5t-4
The given function can be rewritten as follows. S(t) = - t/2 + 2 + - 5t-4/2t^2+4t+2 Therefore, the equation of the oblique asymptote is O(t)=- t2+2.
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Discussion
Graphing Rational Functions
Rational functions may have more than one vertical asymptote, which can make graphing them a bit more difficult than graphing reciprocal functions. However, the asymptotes are useful when graphing a rational function because they can describe the end behavior. Consider graphing the following function. f(x) =(x^2+5x-6)(x-3)/x^2-7x+12 To graph the rational function, these four steps can be followed.
1
Draw the Asymptotes
expand_more First, the vertical asymptote(s) of the function will be found. If the numerator and the denominator of a rational function have no common factors, then the vertical asymptote is the x-value that makes the denominator zero. Start by factoring the denominator to see candidates for asymptotes.
f(x) =(x^2+5x-6)(x-3)/x^2-7x+12
f(x) =(x^2+5x-6)(x-3)/(x-4)(x-3)
The rational function is undefined where x-4=0 and x-3=0. This means that x= 4 and x=3 are not included in the domain. Notice that x-3 is a common factor between the numerator and the denominator. Next, check if there are any other common factors and cancel them.
f(x) =(x^2+5x-6)(x-3)/(x-4)(x-3)
f(x) =(x-1)(x+6)(x-3)/(x-4)(x-3)
▼
Simplify
CrossCommonFac
Cross out common factors
f(x) =(x-1)(x+6)(x-3)/(x-4)(x-3)
CancelCommonFac
Cancel out common factors
f(x) =(x-1)(x+6)/x-4, x ≠ 3
Since the factor x-3 was canceled, the graph has a hole
at x=3. Also, the graph has a vertical asymptote at x=4 because 4 is not included in the domain. Hole: & x =3 Vertical Asymptote: & x = 4
The degrees of the polynomials in the numerator and denominator determine whether the function has a horizontal or oblique asymptote. In the following table, m and n are the degrees of the polynomials in the numerator and denominator, and a and b are their respective leading coefficients.
| y=ax^m+.../bx^n+... | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
Now consider the degrees of the numerator and denominator for the given function. The simplified form of the function can also be considered at this point because both have the same graph. f(x) =(x-1)(x+6)/x-4, x≠ 3 ⇓ f(x) =x^2+5x-6/x^1-4, x ≠ 3 The degree of the numerator minus the degree of the denominator equals 1. Therefore, the function has an oblique asymptote. The equation of the oblique asymptote is found by dividing the numerator by the denominator without considering the remainder. To divide the polynomials, long division will be used.
l r x-4 & |l x^2+5x-6
▼
Divide
x^2/x= x
r x r x-4 & |l x^2+5x-6
Multiply term by divisor
r x rl x-4 & |l x^2+5x-6 & x^2-4x
Subtract down
r x r x-4 & |l 9x-6
▼
Divide
9x/x= 9
r x+9 r x-4 & |l 9x-6
Multiply term by divisor
rx + 9 rl x-4 & |l 9x - 6 & 9x-36
Subtract down
r x+9 r x-4 & |l 30
The equation of the oblique asymptote is y=x+9. Finally, draw the asymptotes on a coordinate plane.
The given function has one vertical asymptote. Therefore, its graph will consist of two parts, one to the left and the other to the right of the vertical asymptote. These parts may lie above or below the oblique asymptote.
2
Find the Intercepts
expand_more The x-intercepts are the points where the graph intersects the x-axis. The value of the y-coordinate is zero at these points. Substitute 0 for f(x) in the given function and solve for x. The simplified function can be used to find the zeros.
f(x) =(x-1)(x+6)/x-4
Substitute
f(x)= 0
0=(x-1)(x+6)/x-4
▼
Solve for x
MultEqn
LHS * (x-4)=RHS* (x-4)
0=(x-1)(x+6)
ZeroProdProp
Use the Zero Product Property
lcx-1=0 & (I) x+6=0 & (II)
AddEqn
(I): LHS+1=RHS+1
lx=1 x+6 = 0
SubEqn
(II): LHS-6=RHS-6
lx=1 x=- 6
The graph has two x-intercepts. The y-intercept is the point where the graph intersects the y-axis. At this point, the value of the x-coordinate is zero.
There is a y-intercept at (0,1.5). Show the intercepts on the same coordinate plane.
3
Make a Table of Values
expand_more Some additional points are needed to get a good sense of the shape of the graph.
Make sure to only use values included in the domain of the function.
| x | (x-1)(x+6)/x-4 | f(x)=(x-1)(x+6)/x-4 |
|---|---|---|
| - 11 | ( - 11-1)( - 11+6)/- 11-4 | - 4 |
| 2 | ( 2-1)( 2+6)/2-4 | - 4 |
| 6 | ( 6-1)( 6+6)/6-4 | 30 |
| 9 | ( 9-1)( 9+6)/9-4 | 24 |
| 19 | ( 19-1)( 19+6)/19-4 | 30 |
It is almost done. Plot the points and imagine how the shape of the graph should look!
As can be seen, one part of the graph will lie to the left of the vertical asymptote and below the oblique asymptote. The other part of the graph will be to the right of the vertical asymptote and above the oblique asymptote.
4
Draw the Graph
expand_more The graph can now be drawn by connecting the points with a smooth curve. It must approach the asymptotes. Do not forget to plot the hole at x=3!
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Example
Graphing the Function for the Surface Area to Volume Ratio of a Box
The factory sells 6 pencils in a rectangular box. The dimensions of the box are shown in the diagram.
a Write a function of the form E(x)= S(x)V(x), where S(x) and V(x) are the surface area and volume of the box, respectively.
b Graph the function E(x).
Answer
a E(x) = x^2+4x-2/x^2+x-6
b Graph:
Hint
a The surface area of a prism is the sum of the areas of its surfaces. The volume of a prism is the product of its dimensions.
b Start by determining the vertical and horizontal asymptotes. Then, make a table of values.
Solution
a First, the surface area and volume of the given box will be found.
The surface area of a rectangular box is the total area of all the surfaces of the shape. Since the dimensions are given in terms of x, the surface area is a function of x.
S(x)= 2( 2)( x-2) + 2( 2x+6)( x-2)+2( 2)( 2x+6)
▼
Simplify right-hand side
Multiply
Multiply
S(x)= 4(x-2) + 2(2x+6)(x-2)+4(2x+6)
Distr
Distribute 4
S(x)= 4x-8 + 2(2x+6)(x-2)+8x+24
Distr
Distribute 2
S(x)= 4x-8 + (4x+12)(x-2)+8x+24
MultPar
Multiply parentheses
S(x)= 4x-8 +4x^2-8x+12x-24+8x+24
AddSubTerms
Add and subtract terms
S(x)= 4x^2+16x-8
Next, the volume of the pencil box will be calculated. Recall that the volume of a rectangular prism is the product of its dimensions.
V(x) = ( 2x+6)( x-2)( 2)
V(x) = 4x^2+4x-24
Finally, the function E(x), the ratio of S(x) to V(x), can be written.
E(x) = S(x)/V(x)
SubstituteExpressions
Substitute expressions
E(x) = 4x^2+16x-8/4x^2+4x-24
▼
Simplify right-hand side
E(x) = x^2+4x-2/x^2+x-6
b To graph the given rational function, its asymptotes and intercepts will be found. A table of values will then be used to draw the graph.
Asymptotes
Consider the function written in the previous part. E(x)=x^2+4x-2/x^2+x-6 First, factor the denominator of this function to identify its domain.
E(x)=x^2+4x-2/x^2+x-6
WriteDiff
Write as a difference
E(x)=x^2+4x-2/x^2+3x-2x-6
E(x)=x^2+4x-2/x(x+3)-2(x+3)
FactorOut
Factor out (x+3)
E(x)=x^2+4x-2/(x-2)(x+3)
The rational function is undefined where x-2=0 and x+3=0 because division by zero is not defined. This means that x=2 and x=- 3 are not included in the domain.
Domain All real numbers except x=- 3 and x= 2
Asymptotes can be vertical, horizontal, or slanted lines. Once again, consider the function.
E(x) = x^2+4x-2/(x-2)(x+3)
It seems that the denominator and the numerator do not have a common factor. To find the factors of the numerator, its zeros can be found using the Quadratic Formula.
E(x) = (x+2+sqrt(6))(x+2-sqrt(6))/(x-2)(x+3)
As shown, they do not have common factors. Therefore, there are no holes
in the graph. Recall that if the real number a is not included in the domain, there is a vertical asymptote at x=a. In this case, there are two vertical asymptotes, one at x=- 3 and the other at x=2. Vertical Asymptotes x=- 3 and x =2
To find whether the graph has a horizontal or oblique asymptote, review how its type is determined using the degrees of the polynomials in the numerator and the denominator.
| y=ax^m+.../bx^n+... | Asymptote |
|---|---|
| m | y=0 |
| m>n | none |
| m=n | y=a/b |
| m-n=1 | y= the quotient of the polynomials with no remainder |
Look at the degrees of the numerator and denominator of the function. E(x)=1x^2+4x-2/1x^2+x-6 The degree of the denominator is equal to the degree of the numerator. Therefore, the line y= 1 1=1 is the horizontal asymptote of the function.
Intercepts
To find the x-intercepts, 0 is substituted for E(x) into the function. It will be better to use the form in which the numerator and the denominator are factored.
E(x) = (x+2+sqrt(6))(x+2-sqrt(6))/(x-2)(x+3)
Substitute
E(x)= 0
0 = (x+2+sqrt(6))(x+2-sqrt(6))/(x-2)(x+3)
▼
Solve for x
MultEqn
LHS * (x-2)(x+3)=RHS* (x-2)(x+3)
0 = (x+2+sqrt(6))(x+2-sqrt(6))
ZeroProdProp
Use the Zero Product Property
lcx+2+sqrt(6)=0 & (I) x+2-sqrt(6)=0 & (II)
SubEqn
(I): LHS-2-sqrt(6)=RHS-2-sqrt(6)
lx=- 2- sqrt(6) x+2-sqrt(6)=0
SubEqn
(II): LHS-2+sqrt(6)=RHS-2+sqrt(6)
lx=- 2 - sqrt(6) x = - 2+sqrt(6)
UseCalc
Use a calculator
lx=- 4.449489 ... x = 0.449489 ...
RoundDec
Round to 1 decimal place(s)
lx ≈ - 4.4 x ≈ 0.4
To find the y-intercept, substitute 0 for x in the function and solve for E(0). This time the form of the function found in Part A can be used.
E(x)=x^2+4x-2/x^2+x-6
Substitute
x= 0
E( 0)=0^2+4( 0)-2/0^2+ 0-6
▼
Solve for E(0)
CalcPow
Calculate power
E(0)=0+4* 0-2/0+0-6
ZeroPropMult
Zero Property of Multiplication
E(0)=0+0-2/0+0-6
IdPropAdd
Identity Property of Addition
y=- 2/- 6
ReduceFrac
a/b=.a /(- 2)./.b /(- 2).
y=1/3
There is a y-intercept at (0, 13 ). Finally, all the intercepts can be plotted on the coordinate plane.
Table of Values
Now, make a table of values to graph the given function.
| x | x^2+4x-2/x^2+x-6 | E(x)=x^2+4x-2/x^2+x-6 |
|---|---|---|
| - 6 | ( - 6)^2+4( - 6)-2/( - 6)^2+( - 6)-6 | ≈ 0.417 |
| - 3.5 | ( - 3.5)^2+4( - 3.5)-2/( - 3.5)^2+( - 3.5)-6 | ≈ - 1.364 |
| - 2 | ( - 2)^2+4( - 2)-2/( - 2)^2+( - 2)-6 | 1.5 |
| - 1 | ( - 1)^2+4( - 1)-2/( - 1)^2+( - 1)-6 | ≈ 0.833 |
| 1 | ( 1)^2+4( 1)-2/( 1)^2+( 1)-6 | - 0.75 |
| 3 | ( 3)^2+4( 3)-2/( 3)^2+( 3)-6 | ≈ 3.167 |
| 4 | ( 4)^2+4( 4)-2/( 4)^2+( 4)-6 | ≈ 2.143 |
| 5 | ( 5)^2+4( 5)-2/( 5)^2+( 5)-6 | ≈ 1.792 |
Finally, the graph of the function can be drawn by plotting the found points and connecting them with a smooth curve. Recall that a rational function can cross the horizontal asymptote but cannot cross the vertical asymptotes. In this case, the horizontal asymptote is crossed.
Extra
Factoring The Numerator of E(x)Recall the Factor Theorem.
|
p(x) has a factor (x−a) if and only if p(a)=0. |
In other words, the polynomial function has zeros at the a-values. Since the numerator of E(x) is a quadratic expression its zeros can be found by using the Quadratic Formula. E(x) = x^2+4x-2/x^2+x-6 The values of a, b, and c of the numerator are 1, 4, and - 2, respectively. Substitute these values into the formula.
x = - b ± sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x = - 4 ± sqrt(4^2-4(1)(- 2))/2(1)
▼
Evaluate right-hand side
IdPropMult
Identity Property of Multiplication
x = - 4 ± sqrt(4^2-4(- 2))/2
CalcPow
Calculate power
x = - 4 ± sqrt(16-4(- 2))/2
MultNegNegOnePar
- a(- b)=a* b
x = - 4 ± sqrt(16+8)/2
AddTerms
Add terms
x = - 4 ± sqrt(24)/2
SplitIntoFactors
Split into factors
x = - 4 ± sqrt(4 * 6)/2
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
x = - 4 ± sqrt(4) * sqrt(6)/2
CalcRoot
Calculate root
x = - 4 ± 2 * sqrt(6)/2
FactorOut
Factor out 2
x = 2( - 2 ± sqrt(6))/2
SimpQuot
Simplify quotient
x = - 2 ± sqrt(6)
The zeros of the numerator are x = - 2 -sqrt(6) and x= - 2 + sqrt(6). By the Factor Theorem, the following is the factored form of the numerator. x^2+4x-2 ⇕ (x+2+sqrt(6))(x+2-sqrt(6))
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Example
Modeling the Amount of Water Pumped Into the Factory
Factory managers plan to pump water into the factory from a pond near the factory. The managers modeled the amount of water pumped in thousands of barrels per year x years after pumping starts to be as follows. W(x) =- x^3+140x^2+1500x+5250/15x^2+150x
a Use a graphing calculator to find the x-intercepts rounded to the nearest integer. Interpret the x-intercepts.
b Determine if the graph has horizontal or oblique asymptotes. If the graph has an oblique asymptote, find its equation.
Answer
a x-intercept: (150,0)
Interpretation: In about 150 years, there will be no water left in the pond.
b The graph has an oblique asymptote because the difference between the degrees of the numerator and the denominator is 1.
Equation: y = - x/15+10
Hint
a Use the zero option in the calculator. It may be necessary to re-size the viewing window in order to see the entire graph.
b Compare the degrees of the numerator and the denominator.
Solution
File:Slide A2 U4 C4 Example6 1.svg
Now, push GRAPH to draw the function.
File:Slide A2 U4 C4 Example6 2.svg
It appears that only one part of the graph is shown, but it is not possible to know that until the size of the viewing window is changed.
The graph does not seem to have any zeros. However, note that as x increases in the first quadrant, the graph gets closer to the x-axis and could cross it. To check this, zoom in on this part by changing the window settings. Push WINDOW and change the settings as shown below. Then push GRAPH once more to draw the equation with these new settings.
Now it can be seen that the graph intersects the x-axis and there is a zero. To find it, use the zero option in the calculator. It can be found by pressing 2ND and then CALC.
File:Slide A2 U4 C4 Example6 7.svg
After selecting the zero
option, choose left and right boundaries for the zero. Finally, the calculator asks for a guess where the zero might be. After that, it will calculate the exact point.
File:Slide A2 U4 C4 Example6 8.svg
The function intersects the x-axis at about (150,0). Since x represents the numbers of years that have passed since pumping starts, the x-intercept means that in about 150 years, there will be no water left in the pond.
b Recall the table used to determine what type of asymptote a rational function can have.
| y=ax^m+.../bx^n+... | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
Now, compare the degrees of the numerator and denominator of the given function. W(x) =- x^3+140x^2+1500x+5250/15x^2+150x The degree of the numerator minus the degree of the denominator equals 1. Therefore, the function has an oblique asymptote. The equation of this asymptote is found by dividing the numerator by the denominator and disregarding any remainder. To divide the polynomials, use long division.
l r 15x^2+150x & |l - x^3+140x^2+1500x+5250
▼
Divide
- x^3/15x^2= - x/15
r - x15 r 15x^2+150x & |l - x^3+140x^2+1500x+5250
Multiply term by divisor
r - x15 rl 15x^2+150x & |l - x^3+140x^2+1500x+5250 & - x^3 - 10x^2
Subtract down
r - x15 r 15x^2+150x & |l 150x^2+1500x+5250
▼
Divide
150x^2/15x^2= 10
r - x15 + 10 r 15x^2+150x & |l 150x^2+1500x+5250
Multiply term by divisor
r- x10 + 10 rl 15x^2+150x & |l 150x^2+1500x+5250 & 150x^2+1500x
Subtract down
r - x15 + 10 r 15x^2+150x & |l 5250
The given function can be rewritten as follows. W(x) = - x/15 + 10 + 5250/15x^2+150x Therefore, the equation of the oblique asymptote is y=- x15+10.
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Closure
Graphing the Average Cost Function
Rational functions can have more than one vertical asymptote but at most one horizontal or oblique asymptote. Also, if the numerator and denominator have a common factor (x-a), then there is a hole at x=a. With these tips in mind, the graph of the function that LaShay wrote at the beginning of the lesson can now be graphed. A(x)= 0.2x^2+5x+500/x The function gives the average cost A(x) in thousands dollars for producing x thousand pencils.
a Graph the average cost function.
Answer
a Graph:
b Domain: (- ∞, 0) ⋃ (0,∞)
Range: (- ∞, - 15] ⋃ [25, ∞)
Range: (- ∞, - 15] ⋃ [25, ∞)
Hint
a Start by identifying asymptotes. Then, make a table of values using values that are around the vertical asymptote.
b Examine the graph and see if the graph is symmetric about the point of intersection of the asymptotes. Determine all the y-values for which the function is graphed.
Solution
a The steps to draw a rational function can be listed as follows.
- Draw the asymptotes.
- Find any intercepts.
- Make a table of values.
- Draw the graph.
Drawing the Asymptotes
The vertical asymptotes of a rational function are the x-values where the denominator of the function is zero. A(x) = 0.2x^2+5x+500/x The denominator is 0 when x=0. This means that this value is not included in the domain. Domain All real numbers except x =0 Since the numerator and denominator do not share a common factor, the graph does not contain any holes. Additionally, the graph has a vertical asymptote at the x-value excluded from the domain. Vertical Asymptote x =0 Next, recall the table for determining the other asymptotes.
| y=ax^m+.../bx^n+... | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
Examine the degrees of the numerator and denominator. A(x) = 0.2x^2+5x+500/x^1 The degree of the numerator minus the degree of the denominator equals 1. Therefore, the function has an oblique asymptote. To find its equation, divide the numerator by the denominator using long division.
l r x & |l 0.2x^2+5x+500
▼
Divide
0.2x^2/x= 0.2x
r 0.2x r x & |l 0.2x^2+5x+500
Multiply term by divisor
r 0.2x rl x & |l 0.2x^2+5x+500 & 0.2x^2
Subtract down
r 0.2x r x & |l 5x+500
▼
Divide
5x/x= 5
r 0.2x + 5 r x & |l 5x+500
Multiply term by divisor
r0.2x + 5 rl x & |l 5x+500 & 5x
Subtract down
r 0.2x+5 r x & |l 500
The equation of the oblique asymptote is y=0.2x+5. Oblique Asymptote y = 0.2x+5 The asymptotes can be drawn on a coordinate plane.
Finding the Intercepts
The x-intercept and y-intercept are the points where a graph crosses the x- and y-axes, respectively. Subsequently, the x-intercepts occurs when A(x)=0.
A(x) =0.2x^2+5x+500/x
Substitute
A(x)= 0
0=0.2x^2+5x+500/x
MultEqn
LHS * x=RHS* x
0= 0.2x^2+5x+500
▼
Solve using the quadratic formula
UseQuadForm
Use the Quadratic Formula: a = 0.2, b= 5, c= 500
x=- 5±sqrt(5^2 -4( 0.2)( 500))/2( 0.2)
CalcPowProd
Calculate power and product
x=- 5±sqrt(25 -400)/0.4
SubTerm
Subtract term
x=- 5±sqrt(- 375)/0.4
Notice that there is a negative number under the radical symbol. Therefore, the function has no real solutions. In other words, its graph does not cross the x-axis. Furthermore, the graph has no y-intercept as the domain does not contain 0.
Making a Table of Values
A table of values will be used to get a rough idea of the shape of the graph.
| x | 0.2x^2+5x+500/x | A(x)=0.2x^2+5x+500/x |
|---|---|---|
| - 75 | 0.2( - 75)^2+5( - 75)+500/- 75 | ≈ - 16.7 |
| - 25 | 0.2( - 25)^2+5( - 25)+500/- 25 | - 20 |
| - 10 | 0.2( - 10)^2+5( - 10)+500/- 10 | - 47 |
| 10 | 0.2( 10)^2+5( 10)+500/10 | 57 |
| 25 | 0.2( 25)^2+5( 25)+500/25 | 30 |
| 75 | 0.2( 75)^2+5( 75)+500/75 | ≈ 26.7 |
Plot the points and imagine how the shape of the graph should look!
Finally, draw the graph by connecting the points. It must approach, but not cross, the asymptotes.
b The domain of the function was identified when graphing the function.
Domain: (- ∞,0) ⋃ (0,∞) To identify the range of the function, take a look at its graph.
As shown, the range does not contain all real numbers. It appears that the minimum point of the part of the graph in the first quadrant is (50,25). Notice also that the graph is symmetric about the point of intersection of the asymptotes, (0,5). Using this fact, the maximum point of the other part can be identified.
It is the point (- 50,- 15). Therefore, the range is all the y-values less than or equal to - 15 and greater than or equal to 25. Range: (∞,- 15] ⋃ [25,∞)
Checking Our Answer
Use a Graphing CalculatorA graphing calculator can be used to check the answers. The graph of the function will be drawn first. Press the Y= button and type the equation in the first row.
![File:Slide A2 U4 C4 Closure 1.svg]()
![File:Slide A2 U4 C4 Closure 4.svg]()
![File:Slide A2 U4 C4 Closure 5.svg]()
File:Slide A2 U4 C4 Closure 1.svg
By pushing GRAPH, the calculator will draw the equation. For this function to be visible on the screen, re-size the standard window by pushing the WINDOW button. Change the settings to a more appropriate size and then push GRAPH.
Next, the local maximum and local minimum will be found. To do so, push 2nd, then TRACE, and choose the maximum
option.
File:Slide A2 U4 C4 Closure 4.svg
When using the maximum
feature, choose the left and right bounds. The calculator will then provide a best guess as to where the maximum might be.
File:Slide A2 U4 C4 Closure 5.svg
The maximum point of the graph in the third quadrant is (- 50,- 15). To find the minimum, repeat the same process but choose the minimum
option.
The minimum point of the part in the first quadrant is (50,25). As a result, the range is (∞,- 15] ⋃ [25,∞), and the domain is all real numbers except 0.
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Graphing Rational Functions
Exercise 1.1
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In order to find the points of discontinuity of the function, we will first find its domain. Consider the given function. f(x) = x-3/x^2+7x+10 To find the domain, we will start by factoring the denominator.
We have fully factored the denominator. f(x)=x-3/(x+ 2)(x+ 5) Recall that division by zero is not defined. Therefore, the rational function is undefined where x+ 2=0 and where x+ 5=0.
| f(x) is undefined | |
|---|---|
| x+ 2=0 ⇕ x= - 2 | x+ 5=0 ⇕ x= - 5 |
This means that neither x= -2 nor x= -5 are included in the domain. Domain All real numbers except x= -2 and x= -5 If a real number a is not in the domain of a function, then the function has a point of discontinuity at x=a. Since the domain is all real numbers except x= -2 and x= -5, we can write the points of discontinuity of our function. Points of Discontinuity x= -2 and x= -5
We will find the points of discontinuity for the given function by following a similar fashion. We will first find the domain of the function.
g(x) = x+4/x^3-8x^2-48x
Let's start by factoring the denominator. Notice that x is a common factor for all the terms in the denominator. We will start by factoring it out.
Having fully factored the denominator, we can find the values that make the rational function undefined.
| g(x) is undefined | ||
|---|---|---|
| x = 0 | x+4=0 | x-12=0 |
| x= 0 | x= - 4 | x = 12 |
This means that x = 0, x= -4 , and x= 12 are not included in the domain. Domain All real numbers except x = 0, x= -4, and x= 12 If a real number a is not in the domain of a function, then the function has a point of discontinuity at x=a. Since the domain is all real numbers except x= 0, x= -4, and x= 12, we can write the points of discontinuity of our function. Points of Discontinuity x = 0, x= -4, and x= 12
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We will begin by factoring the denominator to find the values that make the denominator to be 0.
Recall that division by zero is not defined. Therefore, the rational function is undefined where x+5=0 and x-5=0.
| w(x) is undefined | ||
|---|---|---|
| x+5=0 | x-5=0 | |
| x=- 5 | x=5 | |
The function is not defined when x=-5 and x=5. Recall that these points are also points of discontinuity of the function. Points of Discontinuity x = - 5 and x = 5 This means we have either a hole or a vertical asymptote. Notice that the numerator and denominator do not have any common factor so the rational expression cannot be simplified. The function has non-removable discontinuity. Therefore the lines x=-5 and x=5 are vertical asymptotes. Vertical Asymptotes x = - 5 and x = 5
Following a similar procedure, we start by factoring the denominator.
Since division by zero is not defined, the rational function is undefined where x= 0, x-3=0, and x+1=0.
| h(x) is undefined | ||
|---|---|---|
| x=0 | x-3=0 | x+1=0 |
| x=0 | x=3 | x=- 1 |
At these values of x, we have either a hole
or a vertical asymptote. Let's now check if the numerator and denominator have any common factor.
After simplifying, the factor x+1 is still in the denominator. Therefore the function has removable and non-removable discontinuity.
| Points Discontinuity of h(x) | |
|---|---|
| Removable Discontinuity | Non-Removable Discontinuity |
| x=0 and x=3 | x =- 1 |
This indicates that the line x=- 1 is a vertical asymptote. Vertical Asymptote x = - 1
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The horizontal asymptotes of a rational function can be determined by comparing the terms with the highest degree from the numerator and denominator.
| y=ax^m/bx^n | Asymptote |
|---|---|
| m | y=0 |
| m>n | none |
| m=n | y=a/b |
Let's now consider the given function. y = 7x^2+49/3x^4-27x^2 Since the degree of the denominator, 4, is greater than the degree of the numerator, 2, the line y=0 is a horizontal asymptote. Horizontal Asymptote y = 0
By using the same rules as in the previous part, we can determine the asymptotes of the given function.
| y=ax^m/bx^n | Asymptote |
|---|---|
| m | y=0 |
| m>n | none |
| m=n | y=a/b |
In the given function, we can see that the degrees of the numerator and the denominator are equal. y = 10x^2-13x-3/2x^2-x-3 Because the degrees of the numerator and the denominator are equal, the line y = 10 2 is a horizontal asymptote of the rational function. Horizontal Asymptote y = 102=5
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We will begin by looking at the rules that help determine if a rational function has a horizontal or oblique asymptote. These rules compare the terms with the highest degree of the numerator and denominator of the function.
| y=ax^m/bx^n | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | none | none |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials excluding the remainder | Oblique |
Using this information, we can verify if the given function does indeed have an oblique asymptote. f(x)=x^3/x^2-4 We can see that the numerator and the denominator have no common factors. Also, the degree of the numerator minus the degree of the denominator equals 1. Therefore, the function has an oblique asymptote. To find its equation, we first need to use polynomial long division.
Using the quotient and the remainder, we can rewrite the given function. f(x) = x^3/x^2-4 ⇔ f(x) = x + - 4x/x^2-4 Since the equation of an oblique asymptote is the quotient of the polynomials excluding any remainder, the oblique asymptote for the given function is y= x.
When we look at the degrees of the numerator and denominator of our function, we can see that their difference is 1.
g(x)=x^2-6x+18/x^1-3
Notice also that the numerator and the denominator have no common factors. Therefore, the function has an oblique asymptote. To find its equation, we first need to use polynomial long division.
The rational function can be rewritten in terms of its quotient and remainder. g(x) = x^2-6x+18/x-3 ⇕ g(x) = x-3 + 9/x-3 Because the oblique asymptote is given by the quotient of the polynomials excluding any remainder, the equation for the oblique asymptote is y= x-3.
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To graph the given rational function, we need to find its domain, asymptotes, and intercepts. Then we will evaluate the function rule to create a table of values. Finally, we will plot and connect those points.
Domain
Consider the given function. f(x)=x^4-16/4x^2-4 This first thing we need to do is factor the denominator of this function.
Recall that division by zero is not defined. Therefore, the rational function is undefined where x+1=0 and x-1=0. c|c x+1=0 & x-1=0 ⇕ & ⇕ x=- 1 & x=1 This means that x=- 1 and x=1 are not in the domain of the function. Domain All real numbers except x=- 1 and x=1
Asymptotes
Asymptotes can be vertical, horizontal, or oblique lines.
Vertical Asymptotes
To see if the function has vertical asymptotes, we will investigate if there are common factors.
f(x)=x^4-16/4x^2-4
Because we cannot cancel out common factors, there are no holes.
Also, if a real number a is not included in the domain of the function, there is a vertical asymptote at x=a. This means that we have vertical asymptotes at x=- 1 and x=1.
Horizontal and Oblique Asymptotes
To find the horizontal and oblique asymptotes, we can use following set of rules. In this case, m and n will represent the degree of the numerator and the denominator, respectively.
| y=ax^m/bx^n | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
Let's look at the degrees of the numerator and denominator for our function. f(x)=x^4-16/4x^2-4 We can see that the degree of the numerator is higher than the degree of the denominator. Therefore, there are no horizontal or oblique asymptotes.
Intercepts
The intercepts of the function are the points at which the graph intersects the axes.
x-intercepts
The x-intercepts are the points where the graph intersects the x-axis. At these points the value of the y-coordinate is zero. Let's substitute 0 for f(x) in the given function and solve for x.
There are x-intercepts at (2,0) and (- 2,0).
y-intercept
The y-intercept is the point where the graph intersects the y-axis. At this point, the value of the x-coordinate is zero. Let's substitute 0 for x in the given function and solve for f(x).
There is a y-intercept at (0,2).
Graph
Let's make a table of values to graph the given function. Make sure to only use values included in the domain of the function.
| x | x^4-16/4x^2-4 | f(x)=x^4-16/4x^2-4 |
|---|---|---|
| - 4 | ( - 4)^4-16/4( - 4)^2-4 | 4 |
| - 3 | ( - 3)^4-16/4( - 3)^2-4 | ≈ 2 |
| - 1.5 | ( - 1.5)^4-16/4( - 1.5)^2-4 | ≈ - 2.2 |
| - 0.5 | ( - 0.5)^4-16/4( - 0.5)^2-4 | ≈ 5.3 |
| 0.5 | ( 0.5)^4-16/4( 0.5)^2-4 | ≈ 5.3 |
| 1.5 | ( 1.5)^4-16/4( 1.5)^2-4 | ≈ - 2.2 |
| 3 | ( 3)^4-16/4( 3)^2-4 | ≈ 2 |
| 4 | ( 4)^4-16/4( 4)^2-4 | 4 |
Finally, let's plot and connect the points. Do not forget to draw the asymptotes and to plot the intercepts and holes, if any.
This graph corresponds to option A.
By following a similar procedure, we will begin by determining the domain, asymptotes, and intercepts of the given function.
Domain
Consider the given function. y=x^2 + 8x+24/x+5 Recall that division by zero is not defined. Therefore, the rational function is undefined where x+5=0. This means that x= - 5 is not included in the domain of the function. Domain All real numbers except x= - 5
Asymptotes
We will first look for vertical asymptotes and then horizontal or oblique lines.
Vertical Asymptotes
Once again, let's consider the given function.
y=x^2 + 8x+24/x+5
Note that we cannot cancel out common factors. Therefore, there are no holes.
Also, if the real number a is not included in the domain, there is a vertical asymptote at x=a. In this case, we have a vertical asymptote at x=- 5.
Horizontal and Oblique Asymptotes
To find the horizontal and oblique asymptotes we can use a set of rules. To properly use these rules, in the following m and n must be the highest degree of the numerator and denominator.
| y=ax^m/bx^n | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
Now, consider the function one more time. Let's look at the degrees of the numerator and denominator for our function. y=x^2 + 8x+24/x^1+5 We can see that the numerator and the denominator have no common factors. Also, the degree of the numerator minus the degree of the denominator equals 1. Therefore, the function has an oblique asymptote. The equation of this asymptote is found by dividing the numerator by the denominator, with no remainder.
The equation of the oblique asymptote is y= x+3.
Intercepts
The intercepts of the function are the points at which the graph intersects the axes.
x-intercepts
The x-intercepts are the points where the graph intersects the x-axis. At these points, the value of the y-coordinate is zero. Let's substitute 0 for f(x) in the given function and solve for x.
Notice that there is a negative number under the radical symbol. Therefore, the function has no real solutions. In other words, there are no x-intercepts.
y-intercept
The y-intercept is the point where the graph intersects the y-axis. At this point, the value of the x-coordinate is zero. Let's substitute 0 for x in the given function and solve for f(x).
There is a y-intercept at (0,4.8).
Graph
Let's make a table of values to graph the given function. Make sure to only use values included in the domain of the function.
| x | x^2+8x+24/x+5 | f(x)=x^2+8x+24/x+5 |
|---|---|---|
| - 20 | ( - 20)^2+8( - 20)+24/- 20 + 5 | - 17.6 |
| - 15 | ( - 15)^2+8( - 15)+24/- 15 + 5 | - 12.9 |
| - 10 | ( - 10)^2+8( - 10)+24/- 10 + 5 | - 8.8 |
| - 6 | ( - 6)^2+8( - 6)+24/- 6 + 5 | - 12 |
| - 4 | ( - 4)^2+8( - 4)+24/- 4 + 5 | 8 |
| 5 | ( 5)^2+8( 5)+24/5 + 5 | 8.9 |
| 10 | ( 10)^2+8( 10)+24/10 + 5 | 13.6 |
| 15 | ( 15)^2+8( 15)+24/15 + 5 | 18.45 |
Finally, let's plot and connect the points. Do not forget to draw the asymptotes, the intercepts, and holes, if any.
Note that this corresponds to graph C.
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We want to find the asymptotes of the graph of the rational function. Let's analyze the graph. As x goes to -7 and 2, the function value approaches either positive or negative infinity.
This means that x=-7 and x=2 are vertical asymptotes of the graph of the rational function. Vertical Asymptote x = - 7 and x =2 We can also see that as x goes to positive or negative infinity, it approaches the line y=2.
This line is the horizontal asymptote of the graph of the function. Horizontal Asymptote y = 2
We can find the asymptotes of the graph of the rational function by following a similar fashion. Note that as x goes to 5, the function value approaches either positive or negative infinity.
This means that x=5 is the vertical asymptote of the graph of the rational function. Vertical Asymptote x =5 We can also see that as x goes to positive or negative infinity, the graph behaves like a line.
Therefore, this graph has an oblique asymptote. To find its equation, let's take a closer look at the ends of the graph. The points (14,9) and (18,11), for example, seem to lie on the line we draw. Additionally, it intercepts the y-axis at the point (0,2).
The slope and the y -intercept of the line are m=0.5 and b=2, respectively. Using this information, we can write its equation by using the slope-intercept form. y = mx + b ⇔ y = 0.5x + 2 This equation is the equation of the oblique asymptote.
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Graphing Rational Functions
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