sketching x/x+2

[aurao2003]

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

I am trying to understand how to sketch the above named graph.


2. Relevant equations



3. The attempt at a solution
I have obtained the vertical asymptote as x = -2 but not sure of the horizontal asymptote. Examining the graph as x tends to infinty, it seems the horizontal asymptote is 1. But I am stuck at this stage. Any help or suggestions? I am not sure of the shape of the graph also. Thanks.

[TylerH]

1: What's the rule for determining the horizontal asymptote when the highest power of x on the bottom is the same as that on the top? This will help you learn WHY there is a horizontal asymptote at y=1. You MUST say it is at y=1, just saying 1 would be counted wrong on a test or quiz.

2: Make dotted or slashed lines for the horizontal and vertical asymptote.

3: Determine whether it goes to plus or minus infinity. Do for both sides of vertical asymptote.

4: Pick a point, on both sides, that will be easy to go through, such as x = -3 and x = -1, since it will be a nice integer point.

5: Do your best, remember to go through those points, as they help you stay on track and remember which side goes which way. Also, remember that the farther away from the vertical asymptote, the closer you get to the horizontal asymptote.

This is just to give you an idea of how to solve this type of problem, It's by no means all inclusive.

[aurao2003]

1: What's the rule for determining the horizontal asymptote when the highest power of x on the bottom is the same as that on the top? This will help you learn WHY there is a horizontal asymptote at y=1. You MUST say it is at y=1, just saying 1 would be counted wrong on a test or quiz.

2: Make dotted or slashed lines for the horizontal and vertical asymptote.

3: Determine whether it goes to plus or minus infinity. Do for both sides of vertical asymptote.

4: Pick a point, on both sides, that will be easy to go through, such as x = -3 and x = -1, since it will be a nice integer point.

5: Do your best, remember to go through those points, as they help you stay on track and remember which side goes which way. Also, remember that the farther away from the vertical asymptote, the closer you get to the horizontal asymptote.

This is just to give you an idea of how to solve this type of problem, It's by no means all inclusive.
Well, thanks. I am studying independently. So, the only test is the exam or by myself. I am not sure about the rule. Kindly tell me. Thanks.

[TylerH]

Here's a table of them: http://en.wikipedia.org/wiki/Asymptote#Asymptotes_for_rational_functions You're looking for the "= 0" rule, because the degree of the numerator of your function is 1 and the degree of the denominator is 1 and 1 - 1 = 0.

[aurao2003]

Here's a table of them: http://en.wikipedia.org/wiki/Asymptote#Asymptotes_for_rational_functions You're looking for the "= 0" rule, because the degree of the numerator of your function is 1 and the degree of the denominator is 1 and 1 - 1 = 0.

Hmm! This is freshly baked bread on a Sunday morning! Nice and thanks a million.

[Mentallic]

Right, no wonder I was scratching my head thinking why there is so much discussion on this question. It's actually x/(x+2), not x/x+2=1+2 :biggrin:

[aurao2003]

Right, no wonder I was scratching my head thinking why there is so much discussion on this question. It's actually x/(x+2), not x/x+2=1+2 :biggrin:
One of those days???:tongue2:

[Mentallic]

One of those days???:tongue2:

Pretty much :cry:

By the way, to show there is an asymptote at y=1, you can try this -

y=\frac{x}{x+2}

=\frac{x+2-2}{x+2}

=\frac{x+2}{x+2}-\frac{2}{x+2}

=1-\frac{2}{x+2}

Now all you need to do is take the limit as x \to \infty and show this leaves y=1. So then there is an asymptote there.

edit: or even more simply, multiply the fraction by 1=\frac{\left(\frac{1}{x}\right)}{\left(\frac{1}{x }\right)} and then again take the limit as x\to \infty :wink:

[aurao2003]

Pretty much :cry:

By the way, to show there is an asymptote at y=1, you can try this -

y=\frac{x}{x+2}

=\frac{x+2-2}{x+2}

=\frac{x+2}{x+2}-\frac{2}{x+2}

=1-\frac{2}{x+2}

Now all you need to do is take the limit as x \to \infty and show this leaves y=1. So then there is an asymptote there.

edit: or even more simply, multiply the fraction by 1=\frac{\left(\frac{1}{x}\right)}{\left(\frac{1}{x }\right)} and then again take the limit as x\to \infty :wink:

Cool! Thanks.