+0  
 
0
68
2
avatar

How many vertical asymptotes does the equation \(y=\frac{x-1}{x^2+6x-7}\) have?

 Jul 29, 2020
 #1
avatar
0

 

Whenever the denominator equals zero. 

 

x2 + 6x – 7 factors to (x+7)(x–1) so ordinarily I'd say that asymptotes are at x = –7 and x = +1

 

However, when x = +1 both the numerator and denominator equal zero..  

Maybe the (x–1) in the numerator and denominator cancel and you're left with y = 1/8.  .  

 Jul 29, 2020
edited by Guest  Jul 29, 2020
 #2
avatar+21958 
0

First, factor the denominator:  x2 + 6x - 7  =  (x + 7)(x - 1).

 

Now, the function becomes:  (x - 1) / (x2 + 6x - 7)   --->   (x - 1) / [ (x + 7)(x - 1) ]

 

The "x - 1" term in the denominator cancels the "x - 1" term in the numerator; this means that there 

will be a "hole" in the graph at the point where x = 1.

 

The function has become:  1 / (x + 7)

There will be a vertical asymptote at the point where x = -7.

 

There is one asymptote.

 Jul 29, 2020

59 Online Users

avatar
avatar
avatar