1. ## Asymptote

I have this equation here that needs graphing but it has two restriction. Does that mean it will have two asymptotes? Any help is appreciated, you dont need to draw the graph just tell me how to do it!

The equation is $\displaystyle y=(x+5)/(x+2)$
which eaquals $\displaystyle y=(-7/x+2) +1$ with restriction of x cannot be -4 and -2

3. ## Re: Asymptote

thanks but
can you provide a rough sketch please >.<

4. ## Re: Asymptote

Your function to graph is $\displaystyle y = 1 - \frac{7}{x+2}, x \notin \{-4, -2 \}$.

What I see is:
1/x (I know what that its graph looks like),
then shifted horizontally left 2 units (now it's 1/(x+2)),
then stretched veritcally by a factor of 7 (now it's 7/(x+2)),
then reflected about the x-axis (now it's - 7/(x+2)),
then shifted up vertically 1 unit (now it's 1 - 7/(x+2)).

Oh yeah, and then I have to pull x=-4 out of the domain.

I hope that helps you graph it.

5. ## Re: Asymptote

Hello, sakonpure6!

I have this equation here that needs graphing but it has two restrictions.
Does that mean it will have two asymptotes?

The equation is: .$\displaystyle y\:=\:\frac{x+5}{x+2}$ . . . . which equals: .$\displaystyle y\:=\:\frac{-7}{x+2} +1$ . This is wrong!

. . with restriction of x cannot be -4 and -2. . $\displaystyle {\color{blue}\text{Who said }x \ne -4\,?}$

We have: .$\displaystyle y \;\;=\;\;\frac{x+5}{x+2} \;\;=\;\;\frac{x+2 + 3}{x+2} \;\;=\;\;\frac{x+2}{x+2} + \frac{3}{x+2}\;\;=\;\;1 + \frac{3}{x+2}$

We are expected to know the graph of $\displaystyle y \:=\:\frac{3}{x}$

This graph is translated 2 units to the left and 1 unit up.

6. ## Re: Asymptote

i under stand now, my teacher explained it to us in class!! thank you all for your help!