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

Guest Apr 26, 2015

#1**+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 !!!}

CPhill
Apr 26, 2015

#1**+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