Asked by Ciaria
What are the points of discontinuity? Are they all removable? Please show your work.

y=(x-5) / x^2 - 6x +5

y=(x-5) / x^2 - 6x +5
Answers
Answered by
Jai
y = (x-5) / (x^2 - 6x + 5)
Simplifying,
y = (x-5) / (x - 5)(x - 1)
y = 1 / (x-1)
The root of the denominator is discontinuity:
x = 1 : an asymptotic discontinuity
The other roots (before simplification) is also a discontinuity:
x = 5 : a removable discontinuity
If you plot this, the graph is asymptotic to x = 1, that is, as the graph approaches to x = 1, the value of y approaches infinity. Also, the graph has a hole as point x = 5, a removable/point discontinuity.
hope this helps~ `u`
Simplifying,
y = (x-5) / (x - 5)(x - 1)
y = 1 / (x-1)
The root of the denominator is discontinuity:
x = 1 : an asymptotic discontinuity
The other roots (before simplification) is also a discontinuity:
x = 5 : a removable discontinuity
If you plot this, the graph is asymptotic to x = 1, that is, as the graph approaches to x = 1, the value of y approaches infinity. Also, the graph has a hole as point x = 5, a removable/point discontinuity.
hope this helps~ `u`
Answered by
Ciaria
Thank you !!!
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