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ohaa [14]
3 years ago
15

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

Mathematics
1 answer:
DENIUS [597]3 years ago
6 0

Answer:

The points of discontinuity are: x=5 and x=1. The point of discontinuity x=5 can be removable.

Step-by-step explanation:

The points of discontinuity are those points where the function is not defined. To find such points, we should factorize the denominator of the function needs to be factorized.

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

Considering the quadratic equation of the form ax^2+bx+c =0, then using the quadratic (see attached image), where a=, b=- and c=5, we have that the roots are:

                                              x=5 and x=1.

If we simplify the fraction, by removing the term (x-5) from both the numerator and the denominator, we get: y=1/(x-1), so we removed the point of discontinuity x=5.

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7 0
3 years ago
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2 years ago
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