# Finding the equation of a graph with asymptotes (help please)

• jenni2010
In summary, the conversation discusses a rational function with vertical and horizontal asymptotes and three given points on the graph. The task is to determine the equation for the function from five given options, with the process of elimination used to narrow it down to option D. However, further calculations show that none of the options are suitable, with option F having a typo and option E being omitted. The conversation also reveals a lack of knowledge on limits and difficulty understanding the course material.
jenni2010

## Homework Statement

This is the question:

Shown in the figure below is the graph of a rational function with vertical asymptotes x=2, x=6, and horizontal asymptote y= -2 . (All x-intercepts of the graph of f are also shown, and a point on the graph is indicated.) The equation for f(x) has one of the five forms shown below. Choose the appropriate form for f(x) , and then write the equation.

I can't get the graph on but point are at (-4,0); (0,2); (3,0).

## Homework Equations

A) f(x)= a / x-b

B) f(x)= a(x-b) / x-c

C) f(x)= a / (x-b)(x-c)

D) f(x)= a(x-b) / (x-c)(x-d)

E) f(x)= a(x-b)(x-c) / (x-d)(x-e)

## The Attempt at a Solution

What I first did was try and get rid of equations that couldn't possible work. So I knew I had vertical asymptotes at x=2,6. I had two asymptotes so I knew that options A and B couldn't work. I then got rid of option C because dividing a by x to get my horizontal asymptote would not give me y= -2 it would give me y=0. I then got rid of option E by factoring the top part through. Since having ax^2 as my leading coificent would give me a diagonal asymptote I got rid of it. So I got D as my answer but then I have to find the equation. So I got (x-2)(x-6) for the bottom because those would give me vertical asymptotes at x=2,6 But I don't know how to find the top. I got -2(x-12) divided by (x-2)(x-6) because sticking 0 in for x would give me 2. So I had the point (0,2) but I can't get any of the others to match up. I'm not sure how to figure it out.

Last edited:
You are right, that the other points don't match up. To put it in a mathematical way: the set of equations for a, b, c... you get from plugging in the points is overdetermined and does not have a solution. Therefore D is wrong.

I think your argument for getting rid of F is not entirely correct. You are saying, that your leading behaviour is ~ x^2. However, you also have something ~ x^2 in the denominator.
Did you learn to take limits yet? Can you properly calculate
$$\lim_{x \to \infty} \frac{a (x - b)(x - c) }{ (x - d)(x - c) }$$

Also note that you will need all the points to determine the constants. To take your (wrong) example of option D:
if you plug in x = 0 then you get
f(0) = - a b / 12 = 3.
Although you get a relation a = -36 / b, this alone will not allow you to determine a and b, you will need at least one other equation.

Does option F really have "(x- c)" in both numerator and denominator? And what happened to option E?

HallsofIvy said:
Does option F really have "(x- c)" in both numerator and denominator?
Ah right, I assumed that was a typo and it should be (x - e) in the denominator.

And what happened to option E?
lol that I didn't even notice.

No I really don't know how to do limits. We've touched on it a little bit last year. But I am in an online pre-calculus course and the program doesn't really explain things in terms I can understand. The program has never shown me how to do this so I have no idea how to do it. I'm completely confused and lost. I'm slow at learning math so most of the time I need it in the simplest terms possible.

## 1. How do I determine the equation of a graph with asymptotes?

To find the equation of a graph with asymptotes, you will need to identify the type of asymptote (vertical, horizontal, or slant) and the location of the asymptote(s). Then, you can use the general form of the equation for that type of asymptote to write the equation of the graph.

## 2. What is the general form of the equation for a vertical asymptote?

The general form of the equation for a vertical asymptote is x = a, where a is the x-coordinate of the vertical asymptote. This means that the graph will approach but never touch the line x = a as x gets larger or smaller.

## 3. How do I find the equation of a graph with a horizontal asymptote?

If the graph has a horizontal asymptote, you will need to determine the y-value that the graph approaches as x gets larger or smaller. This will be the equation of the horizontal asymptote. In some cases, you may also need to adjust the equation of the graph to match the behavior near the asymptote.

## 4. Can a graph have more than one asymptote?

Yes, a graph can have multiple asymptotes. It is common for a graph to have both vertical and horizontal asymptotes. It is also possible for a graph to have multiple vertical or horizontal asymptotes, as well as a combination of both.

## 5. Are there any special cases when finding the equation of a graph with asymptotes?

Yes, there are a few special cases that may arise when finding the equation of a graph with asymptotes. These include slant asymptotes, which occur when the degree of the numerator is one more than the degree of the denominator, and removable discontinuities, where the graph has a hole where the asymptote would normally be. In these cases, additional steps may be required to accurately write the equation of the graph.

• Precalculus Mathematics Homework Help
Replies
3
Views
782
• Precalculus Mathematics Homework Help
Replies
21
Views
1K
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
6
Views
845
• Precalculus Mathematics Homework Help
Replies
2
Views
944
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
12
Views
616
• Precalculus Mathematics Homework Help
Replies
13
Views
443
• Precalculus Mathematics Homework Help
Replies
5
Views
1K
• Precalculus Mathematics Homework Help
Replies
21
Views
3K