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what are the points of discontinuity? are they all removable?  y= (x-5) / x^2 - 6x + 5

 Apr 26, 2015

Best Answer 

 #1
avatar+129690 
+10

Note......

(x - 5) / [ x^2 - 6x + 5] =

(x - 5) / [ (x - 1) (x -5) ] =

1 / (x - 1)

There will be a "hole" at x = 5......but the discontinuity at x = 1  isn't "removable".....there's a vertical asymptote at that point......

Here's the graph.......

https://www.desmos.com/calculator/m598ibdrql

 

{Believe me....there is a "hole" at x = 5  !!!}

 

  

 Apr 26, 2015
 #1
avatar+129690 
+10
Best Answer

Note......

(x - 5) / [ x^2 - 6x + 5] =

(x - 5) / [ (x - 1) (x -5) ] =

1 / (x - 1)

There will be a "hole" at x = 5......but the discontinuity at x = 1  isn't "removable".....there's a vertical asymptote at that point......

Here's the graph.......

https://www.desmos.com/calculator/m598ibdrql

 

{Believe me....there is a "hole" at x = 5  !!!}

 

  

CPhill Apr 26, 2015
 #2
avatar+118622 
0

I have never heard of the term "removable" in relation to discontinuities  

 Apr 26, 2015

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