# Thread: What values of a makes f(x)= ax^2 + (a+1)/x have a horizontal asymptote when x->+/-∞?

1. ## What values of a makes f(x)= ax^2 + (a+1)/x have a horizontal asymptote when x->+/-∞?

If you could explain how to do this I would be SOOOO grateful! Thanks!!!

I have no idea where to start...

2. Originally Posted by natashabu
If you could explain how to do this I would be SOOOO grateful! Thanks!!!

I have no idea where to start...
Is it $\displaystyle f(x) = ax^2 + \frac{a+1}{x}$ ? Then you require $\displaystyle a+1 \neq 0 \, ....$

3. Originally Posted by mr fantastic
Is it $\displaystyle f(x) = ax^2 + \frac{a+1}{x}$ ? Then you require $\displaystyle a+1 \neq 0 \, ....$

Thanks I understand now (after thinking about what you wrote)

4. Originally Posted by steven_smith
F(x)=ax^2+(a+1)/x

First you need to find your critical points to do this. take the dervitive of that function and set it = to 0.

Can't remember the quotent rule right now i think it's if F(x)=a/b then f'(s) = (b')a-(a')b/(b^2)

f'(x)= 2ax .....= 0
then just solve for all values a and x that make it = 0 that will tell you where you'r critical points are. put those in the equation and see what it does.
When answering a question from a poster on Math Help Forum it is generaly considered good form to attempt to answer that question not some other question. You will also find this a usefull exam technique in its own right.

CB