+1438 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?
+2401 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)
=^._.^=
+36916 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
