+0

# Algebra 2 Help needed.

+1
99
6
+1436

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?

Apr 11, 2021

#1
+2127
+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)

=^._.^=

Apr 11, 2021
edited by catmg  Apr 11, 2021
#2
+33727
+3

Hole at -1   asymtope at x=1

Apr 11, 2021
#3
+2127
+1

What is a hole in this function?

=^._.^=

catmg  Apr 11, 2021
#4
+33727
+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
#5
+2127
+1

Ohhh, I get it.

(x+1) = 0 when x = -1 so the hole is -1.

(x^2-1) = 0 when x = 1 so the asymptote is 1.

Thank you. :)))

=^._.^=

catmg  Apr 11, 2021
#6
+1436
+1

Thanks to both of you!

Apr 12, 2021