HOW TO FIND AND CLASSIFY DISCONTINUITIES OF RATIONAL FUNCTIONS

Find the points of discontinuity.  Classify each  point as a vertical asymptote or a hole.

Problem 1 :

y = x/(x² – 9)

Solution :

y = x/(x² – 9)

y = x/(x + 3)(x - 3)

We don’t see any common factor in both side in numerator and denominators. So, there is no hole.

To find vertical asymptote :

(x + 3) (x – 3) = 0

x + 3 = 0 and x – 3 = 0

x = -3 x = 3

Vertical asymptotes are at x = 3 and -3.

Problem 2 :

y = (3x² – 1)/x³

Solution :

We don’t see any common factor in both side in numerator and denominators.

To find point of discontinuity, let us equate the denominator to 0.

y = (3x² – 1)/x³

To find vertical asymptote :

x³ = 0

x = 0

Vertical asymptotes is at x = 0.

Problem 3 :

y = (6x² + 3)/(x – 1)

Solution :

y = (6x² + 3)/(x – 1)

y = 3(x² + 1)/(x - 1)

We don’t see any common factor in both side in numerator and denominators. So, there is no hole.

To find vertical asymptote :

x – 1 = 0

x = 1

Vertical asymptote is at x = 1.

Problem 4 :

y = (5x³ – 4)/(x² + 4x – 5)

Solution :

We don’t see any common factor in both side in numerator and denominators. There is no hole.

To find point of discontinuity, let us equate the denominator to 0.

y = (5x³ – 4)/(x² + 4x – 5)

y = (5x³ – 4)/(x + 5)(x - 1)

There is no common factor, so there is no hole.

To find vertical asymptote :

x + 5 = 0 and x – 1 = 0

x = -5 x = 1

Vertical asymptotes are at x = -5 and 1.

Problem 5 :

y = 7x/(x³ + 1)

Solution :

We don’t see any common factor in both side in numerator and denominators. There is no hole.

To find point of discontinuity, let us equate the denominator to 0.

y = 7x/(x³ + 1)

x³ + 1 = 0

x³ = -1

x = -1

Asymptote at x = -1.

Problem 6 :

y = (12x⁴ + 10x – 3)/3x⁴

Solution :

y = (12x⁴ + 10x – 3)/3x⁴

We don’t see any common factor in both side in numerator and denominators. There is not hole.

To find vertical asymptote :

3x⁴ = 0

x⁴ = 0

x = 0

Asymptote at x = 0.

Problem 7 :

y = (12x + 24)/(x² + 2x)

Solution :

We see common factor in both side in numerator and denominators.

y = (12x + 24)/(x² + 2x)

= 12(x + 2)/x(x + 2)

Common factor is x + 2. When x + 2 = 0, hole is at x = -2

After cancelling x + 2, we get 12/x.

At x = 0, we have vertical asymptote.

Problem 8 :

y = (x² – 1)/(x² + 3x + 2)

Solution :

y = (x² – 1)/(x² + 3x + 2)

Factoring the numerator and denominator, we get

y = (x + 1)(x – 1)/(x + 1)(x + 2)

Common factor is both numerator and denominator is x + 1.

x + 1 = 0

x = -1

So, hole is at x = -1.

Cancelling the common factor to zero, we get

y = (x - 1)/(x + 2)

x + 2 = 0

 x = -2

So, the vertical asymptote is at x = -2.

Problem 9 :

y = (x² – 1)/(x² - 2x - 3)

Solution :

y = (x² – 1)/(x² - 2x - 3)

Find the factors of numerator and denominator, we get

y = (x + 1)(x - 1) / (x - 3)(x + 1)

The common factor is x + 1.

x + 1 = 0 and x = -1

So, hole is at x = -1.

After cancelling the common factor, we get

y = (x - 1)/(x - 3)

x - 3 = 0

x = 3

S, the vertical asymptote is at x = 3.