Sketch the following function from the description given below.

x= -1 and y= 2 are asymptotes.
The y intercept is 3 and the x intercept is -1.5
As x -> infinity, y -> 2
As x -> negative infinity, y -> 2
As x tends to -1 from below, y -> negative infinity and as x tends to -1 from above, y -> infinity.

Help will be greatly appreciated

2. Originally Posted by needmathshelp1234
Sketch the following function from the description given below.

x= -1 and y= 2 are asymptotes.
The y intercept is 3 and the x intercept is -1.5
As x -> infinity, y -> 2
As x -> negative infinity, y -> 2
As x tends to -1 from below, y -> negative infinity and as x tends to -1 from above, y -> infinity.

...
1. Draw the asymptotes and mark those parts of the coordinate system where the graph of the function never can be.

2. Draw a smooth curve to the asymptotes in the "allowed" areas.

3. From the vertical asymptote you know that the denominator must contain the factor (x+1).
From the horizontal asymptote you know that the numerator must contain the factor (2x+3).

4. I've sketched the function $f(x) = \frac{2x+3}{x+1}$ which obviously satisfies all given conditions.

3. Originally Posted by needmathshelp1234
Sketch the following function from the description given below.

x= -1 and y= 2 are asymptotes.
The y intercept is 3 and the x intercept is -1.5
As x -> infinity, y -> 2
As x -> negative infinity, y -> 2
As x tends to -1 from below, y -> negative infinity and as x tends to -1 from above, y -> infinity.

Help will be greatly appreciated
This question was answered here. What has prompted you to ask it again?

4. The answer you posted in my first post made no sense to me. I though I would be polite and thankyou, hence my need to repost the question.

5. Originally Posted by needmathshelp1234
The answer you posted in my first post made no sense to me. I though I would be polite and thankyou, hence my need to repost the question.
Fair enough. But all you had to do was say so.

What in particular made no sense ....? Was it that you couldn't see where the equation came from? Or that you couldn't see how the equation satisfied the given requireements of the question. Or was it that you didn't know how to sketch a (rectangular) hyperbola.

You'll learn a lot more by following through on such things rather than posting again .....

Note for the future: You will not hurt my feelings (and most probably you won't hurt the feelings of 99.9% of the members of this forum) if you do as I've suggested.

6. Originally Posted by mr fantastic
Fair enough. But all you had to do was say so.

What in particular made no sense ....? Was it that you couldn't see where the equation came from? Or that you couldn't see how the equation satisfied the given requireements of the question. Or was it that you didn't know how to sketch a (rectangular) hyperbola.

You'll learn a lot more by following through on such things rather than posting again .....

Note for the future: You will not hurt my feelings (and most probably you won't hurt the feelings of 99.9% of the members of this forum) if you do as I've suggested.
Ok I will take that into consideration... thanks. I just was fully confused on the whole concept as we haven't really learnt it the way your equation was. But thanks anyway. When I draw my graph, do I put arrows on the end? Like the domain and range go to infinity?
Thanks

7. Originally Posted by needmathshelp1234
Ok I will take that into consideration... thanks. I just was fully confused on the whole concept as we haven't really learnt it the way your equation was. But thanks anyway. When I draw my graph, do I put arrows on the end? Like the domain and range go to infinity?
Thanks
When the domain and range go to infinity, yes you should. Otherwise it might look like you think there are finite endpoints ....