1. ## Asymptotes and Range

Function = (X+2)\(X+1)

When X = -1 for a function, it says "ERROR" on the calculator.

So if this is an asymtote, doesn't that mean the range would be all real numbers as the values between -2 and -1 would reach infinity?

Can someone explain why this reasoning is wrong? The real answer is supposed to be all real numbers except 1.

2. Originally Posted by Alan306090
When X = -1 for a function, it says "ERROR" on the calculator.

So if this is an asymtote, doesn't that mean the range would be all real numbers as the values between -2 and -1 would reach infinity?

Can someone explain why this reasoning is wrong? The real answer is supposed to be all real numbers except 1.
I want to help you, but it's hard to without knowing the function you are talking about. If you post the function, I'll try my best to help you.

3. (x+2)/(x+1)

4. Originally Posted by Alan306090
(x+2)/(x+1)
this might help ...

does the equation $\displaystyle \frac{x+2}{x+1} = 1$ have a solution ?

5. Originally Posted by Alan306090
(x+2)/(x+1)
Yes, your reasoning is correct, since there is a vertical asymptote, there will be no upper- or lower limit to the range. However, you need to check for horizontal asymptotes.

$\displaystyle \lim_{x\rightarrow \pm \infty} \frac{x+2}{x+1} = \frac {1}{1} = 1$

This means that $\displaystyle y=1$ is a horizontal asymptote and can't be part of the range.

Do you understand?
Hope I helped you