Which of the following is true about the given function? \(f(x) = \frac{x+1}{x^2-1} \)

a) Hole at x = 1 and x = 1

b) Vertical Asymptote at x = 1 ; Hole at x = -1

c) Vertical Asymptote at x = 1 and x = -1

d) Vertical Asymptote at x = -1 ; Hole at x = 1

I know a) and d) are incorrect. I found a hole at -1 but I also found the Vertical Asymptote to be x = 1 and x = -1. What am I doing wrong? Which one is it?

SmartMathMan Apr 11, 2021

#1**+1 **

Hello.

Idk if I'm going to be able to help much since I haven't really learned about those functions, but I'll try my best.

y = (x+1)/(x^2-1) = 1/(x-1).

That would make the vertical asympotope at x = 1, not be = -1 since x - 1 can't have x = 1. (can't divide by 0)

=^._.^=

catmg Apr 11, 2021

#2

#4**+3 **

Hole is the value of x that makes the function undefined.....this occurs when the x value makes the **denominator** = 0

the asymptope is where the** numerator** = 0

+1 is not a hole because the asymtope is there

ElectricPavlov
Apr 11, 2021