Thread: 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.

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.

(x+2)/(x+1)

Originally Posted by Alan306090

(x+2)/(x+1)

this might help ...

does the equation have a solution ?

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.



This means that is a horizontal asymptote and can't be part of the range.

Do you understand?
Hope I helped you