# Asymptotes of (X-5)/(X^2+X-6)

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1. Sep 16, 2015

### theo12349

Good afternoon

1. The problem statement, all variables and given/known data

Draw graph of the following equation

2. Relevant equations
$$\frac{X-5}{X^2+X-6} = y$$

3. The attempt at a solution
my problem is searching for the vertical asymptote

from what I know the way to find the vertical asymptote is by limiting the equation near to ∞. and based on my calculation, the vertical asymptote should be 0

but when you put X=5, the graph will show that it crosses (5,0)

Last edited: Sep 16, 2015
2. Sep 16, 2015

### Dick

I have no idea what your calculation is or how it tells you that the vertical asymptote should be 1. Vertical asymptotes are where the function grows indefinitely large at nearby values. I think you need to review the definition of 'vertical asymptote'.

3. Sep 16, 2015

### theo12349

I do know that vertical asymptotes are where the function grows indefinitely large at nearby values. and it will reache the value of the asymptotes when X=∞

from what I read at Schaum's outline book, you can find the vertical asymptote by using $\lim_{n\rightarrow +\infty} f(X)$

I use this formula and get 0 as the answer

I did put the wrong value, sorry for my mistake. I am recalling it from my memory

anyway, when I insert X=5 it reaches the value 0

Last edited: Sep 16, 2015
4. Sep 17, 2015

### Staff: Mentor

No -- that's a horizontal asymptote.
That would be the x-intercept.

5. Sep 17, 2015

### Ray Vickson

I doubt very much that the Schaum's outline book tells you to take $\lim_{n\rightarrow +\infty} f(X)$, because that expression makes no sense at all, and Schaum's outlines are never so careless as that. There is a difference between horizontal and vertical asymptotes; do you know how to distinguish between them?

6. Sep 17, 2015

### epenguin

On the substance, you can easily factorise the denominator and that should hopefully immediately reveal to you a lot about the 'vertical asymptotes'.

7. Sep 17, 2015

### RUber

For starters, look at $y = \frac1x$. This has a vertical asymptote at x = 0, since it is undefined at this value and for very small x, 1/x is very large.
Your confusion earlier, as Mark pointed out was that you were using the definition for a horizontal asymptote instead of the one for vertical.
The horizontal asymptote of 1/x is y=0. Which the function never reaches but nears as x gets very large.
Note that vertical asymptotes are almost always defined by values of x where the function is undefined...you may have covered these in exercises about the domain of a function.

8. Sep 17, 2015

### theo12349

Mod note: The OP started a new thread, but I have merged the new one to the existing thread.

some of you might already seen this question in my previous post, but since there is a lot of mistakes that I made. I decided to make a new thread. I'm terribly sorry for the confusion before.

1. The problem statement, all variables and given/known data

Draw graph of the following equation

2. Relevant equations
$\frac {X−5} {X^2+X−6}=y$

3. The attempt at a solution
my problem is searching for the Horizontal asymptote

from what I know (froum schaum's Outlines College Mathemathics 3rd Edition) the way to find the Horizontal asymptote is by limiting the equation near to ∞. and based on my calculation, the horizontal asymptote should be 0

but when you put X=5, the graph will show that it crosses (5,0)

[sorry for my english]

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Last edited by a moderator: Sep 17, 2015
9. Sep 17, 2015

### Staff: Mentor

Yes, that's correct. It is also useful to find the limit of this function as x approaches -∞. Some functions have different horizontal asymptotes as x → ∞ and as x → -∞.
The horizontal asymptote indicates behavior for very large values of the independent variable x, or for very negative values. The graph can still cross the x-axis at reasonably small values of x, such as x = 5.

10. Sep 17, 2015

### RUber

How does your graph look? There should be two vertical asymptotes and you should be able to determine where the horizontal asymptotes are for x going to positive and negative infinity. I mean, is the plot above or below the horizontal asymptotes in the limit?
You should be able to describe the plot generally by its behavior near these limiting and critical values.