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| 14 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
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)=q(x)p(x),q(x)=0
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.
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.
Write as a power
a2−b2=(a+b)(a−b)
Simplify quotient
holein the graph.
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=x+11. Since x=-1 is not in the domain of the function, there is a point of discontinuity at x=-1.
This is a non-removable discontinuity because there is no way to redefine the function so that it becomes continuous at x=-1.Total Annual Income | Population |
---|---|
I(t)=0.02t+15000t+100000 | P(t)=0.36t+10800 |
Substitute expressions
ba/c=b⋅ca
Multiply parentheses
Add terms
C(t) is undefined | |
---|---|
0.02t+1=0 | 0.36t+10800=0 |
t=-50 | t=-30000 |
On the way to the pencil factory, LaShay noticed that road maintenance work is being carried out to fill a hole
on the road.
Factor out 3x
Factor out x2
Factor out (x+1)
Notice that it is also a removable discontinuity because x+1 is the common factor in the numerator and denominator.
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.
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)=x1 has two asymptotes — the x-axis and the y-axis.
Analyzing the diagram, the following can be observed.
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.
The degrees of polynomials in the numerator and denominator of a rational function will determine if the graph has a horizontal asymptote.
f(x)=q(x)p(x)=bxn+…axm+… | |
---|---|
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=ba. |
Vertical Asymptotes: t=-2 and t=-1
Horizontal Asymptote: y=0
Factor the numerator. What does the Factor Theorem state?
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. |
t | t3−t2−10t−8 | p(t)=t3−t2−10t−8 |
---|---|---|
1 | 13−(1)2−10(1)−8 | -18 |
-1 | -13−(-1)2−10(-1)−8 | 0 |
2 | 23−(2)2−10(2)−8 | -24 |
-2 | -23−(-2)2−10(-2)−8 | 0 |
4 | 43−6(4)2−7(4)+60 | 0 |
t3−t2−10t−8=(t+1)(t+2)(t−4)
Cross out common factors
Cancel out common factors
y=bxnaxm | Asymptote |
---|---|
m<n | y=0 |
m>n | none |
m=n | y=ba |
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)=q(x)p(x), p(x) is divided by q(x) using long division to get a quotient ax+b with a remainder r(x).4x2x3=4x
Multiply term by divisor
Subtract down
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.Vertical Asymptotes: t=-1
Oblique Asymptote: O(t)=-2t+2
Start by subtracting the rational expressions on the right-hand side of the function.
ba=b⋅2(t+1)a⋅2(t+1)
Multiply
Subtract fractions
Distribute 2t
Commutative Property of Addition
2t2-t3=-2t
Multiply term by divisor
Subtract down
2t24t2=2
Multiply term by divisor
Subtract down
Cross out common factors
Cancel out common factors
holeat x=3. Also, the graph has a vertical asymptote at x=4 because 4 is not included in the domain.
y=bxn+…axm+… | Asymptote | Asymptote Type |
---|---|---|
m<n | y=0 | Horizontal |
m>n | None | None |
m=n | y=ba | Horizontal |
m−n=1 | y= the quotient of the polynomials with no remainder | Oblique |
xx2=x
Multiply term by divisor
Subtract down
x9x=9
Multiply term by divisor
Subtract down
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.
f(x)=0
LHS⋅(x−4)=RHS⋅(x−4)
Use the Zero Product Property
(I): LHS+1=RHS+1
(II): LHS−6=RHS−6
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−4(x−1)(x+6) | f(x)=x−4(x−1)(x+6) |
---|---|---|
-11 | -11−4(-11−1)(-11+6) | -4 |
2 | 2−4(2−1)(2+6) | -4 |
6 | 6−4(6−1)(6+6) | 30 |
9 | 9−4(9−1)(9+6) | 24 |
19 | 19−4(19−1)(19+6) | 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.
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!
The factory sells 6 pencils in a rectangular box. The dimensions of the box are shown in the diagram.
Multiply
Distribute 4
Distribute 2
Multiply parentheses
Add and subtract terms
Substitute expressions
Write as a difference
Factor out (x+3)
holesin 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.
y=bxn+…axm+… | Asymptote |
---|---|
m<n | y=0 |
m>n | none |
m=n | y=ba |
m−n=1 | y= the quotient of the polynomials with no remainder |
E(x)=0
x=0
Calculate power
Zero Property of Multiplication
Identity Property of Addition
ba=b/(-2)a/(-2)
Now, make a table of values to graph the given function.
x | x2+x−6x2+4x−2 | E(x)=x2+x−6x2+4x−2 |
---|---|---|
-6 | (-6)2+(-6)−6(-6)2+4(-6)−2 | ≈0.417 |
-3.5 | (-3.5)2+(-3.5)−6(-3.5)2+4(-3.5)−2 | ≈-1.364 |
-2 | (-2)2+(-2)−6(-2)2+4(-2)−2 | 1.5 |
-1 | (-1)2+(-1)−6(-1)2+4(-1)−2 | ≈0.833 |
1 | (1)2+(1)−6(1)2+4(1)−2 | -0.75 |
3 | (3)2+(3)−6(3)2+4(3)−2 | ≈3.167 |
4 | (4)2+(4)−6(4)2+4(4)−2 | ≈2.143 |
5 | (5)2+(5)−6(5)2+4(5)−2 | ≈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.
Recall the Factor Theorem.
p(x) has a factor (x−a) if and only if p(a)=0. |
Substitute values
Identity Property of Multiplication
Calculate power
-a(-b)=a⋅b
Add terms
Split into factors
a⋅b=a⋅b
Calculate root
Factor out 2
Simplify quotient
Interpretation: In about 150 years, there will be no water left in the pond.
Equation: y=-15x+10
Now, push GRAPH to draw the function.
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.
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.
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.
y=bxn+…axm+… | Asymptote | Asymptote Type |
---|---|---|
m<n | y=0 | Horizontal |
m>n | None | None |
m=n | y=ba | Horizontal |
m−n=1 | y= the quotient of the polynomials with no remainder | Oblique |
15x2-x3=-15x
Multiply term by divisor
Subtract down
15x2150x2=10
Multiply term by divisor
Subtract down
y=bxn+…axm+… | Asymptote | Asymptote Type |
---|---|---|
m<n | y=0 | Horizontal |
m>n | None | None |
m=n | y=ba | Horizontal |
m−n=1 | y= the quotient of the polynomials with no remainder | Oblique |
x0.2x2=0.2x
Multiply term by divisor
Subtract down
x5x=5
Multiply term by divisor
Subtract down
A(x)=0
LHS⋅x=RHS⋅x
Use the Quadratic Formula: a=0.2,b=5,c=500
Calculate power and product
Subtract term
A table of values will be used to get a rough idea of the shape of the graph.
x | x0.2x2+5x+500 | A(x)=x0.2x2+5x+500 |
---|---|---|
-75 | -750.2(-75)2+5(-75)+500 | ≈-16.7 |
-25 | -250.2(-25)2+5(-25)+500 | -20 |
-10 | -100.2(-10)2+5(-10)+500 | -47 |
10 | 100.2(10)2+5(10)+500 | 57 |
25 | 250.2(25)2+5(25)+500 | 30 |
75 | 750.2(75)2+5(75)+500 | ≈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.
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.A 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.
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.
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.
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.
Find all points of discontinuity for each rational function. Write the answers in the form of x=a.
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
Find the vertical asymptotes of the graph of each rational function. Write the answers in the form of x=a.
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
Find the horizontal asymptote of the graph of each rational function. Write the answers in the form of y=a.
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
Find the oblique asymptote of the graph of each rational function.
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.
Identify the graph of the rational function f(x)=4x2−4x4−16.
Identify the graph of the rational function f(x)=x+5x2+8x+24.
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.
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 can be vertical, horizontal, or oblique lines.
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.
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.
The intercepts of the function are the points at which the graph intersects the axes.
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).
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).
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.
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
We will first look for vertical asymptotes and then horizontal or oblique lines.
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.
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.
The intercepts of the function are the points at which the graph intersects the axes.
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.
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).
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.
Find the asymptotes of the graph of each rational function.
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.