+0  
 
+1
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6
avatar+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
avatar+1300 
+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
avatar+31586 
+3

Hole at -1   asymtope at x=1

 Apr 11, 2021
 #3
avatar+1300 
+1

What is a hole in this function?

 

=^._.^=

catmg  Apr 11, 2021
 #4
avatar+31586 
+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
avatar+1300 
+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
avatar+1436 
+1

Thanks to both of you!

 Apr 12, 2021

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