HOW TO FIND EQUATIONS OF VERTICAL ASYMPTOTES

Vertical Asymptotes :

The Vertical Asymptotes of a rational function are found using the zeros of the denominator.

Describe the vertical asymptotes and holes for the graph of each rational function.

Problem 1 :

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

Solution :

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

We see x – 2 as a common factor in both numerator and denominator.

After cancel x – 2. We get,

1/(x + 2)

To find vertical asymptote, we equate the denominator to 0.

x + 2 = 0

x = -2

We see the hole at x = 2 in the graph.

x – 2 = 0 we will get hole at x = 2.

So, the vertical asymptote at x = -2; hole at x = 2.

Problem 2 :

y = x/x(x - 1)

Solution :

y = x/x(x - 1)

We see x is a common factor in both numerator and denominator.

After cancel x .We get,

1/x – 1

To find vertical asymptote, we equate the denominator to 0.

x – 1 = 0

x = 1

We see the hole at x = 0 in the graph.

We will get hole at x = 0.

So, the vertical asymptote at x = 1; hole at x = 0.

Problem 3 :

y = (5 - x)/(x² - 1)

Solution :

Given, y = (5 - x)/(x² - 1)

To find vertical asymptote, we equate the denominator to 0.

x² – 1 = 0

(x + 1) (x – 1) = 0

x + 1 = 0 and x – 1 = 0

x = -1 x = 1

So, the vertical asymptotes at x = 1 and x = -1.

Problem 4 :

y = (x² - 2)/(x + 2)

Solution :

Given, y = (x² - 2)/(x + 2)

To find vertical asymptote, we equate the denominator to 0.

x + 2 = 0

x = -2

So, the vertical asymptote at x = -2.

Problem 5 :

y = (x² - 4)/(x² + 4)

Solution :

y = (x² – 4)/(x² + 4)

We see no common factor in both numerator and denominator.

To find vertical asymptote, we equate the denominator to 0.

x² + 4 = 0

x² = -4

So, no vertical asymptotes or holes.

Problem 6 :

y = (x + 3)/(x² - 9)

Solution :

y = (x + 3)/(x² - 9)

y = (x + 3)/(x + 3) (x – 3)

We see x+3 as a common factor in both numerator and denominator.

After cancel x + 3. We get,

1/(x – 3)

To find vertical asymptote, we equate the denominator to 0.

x – 3 = 0

x = 3

We see the hole at x = -3 in the graph.

x + 3 = 0 we will get hole at x = -3.

So, the vertical asymptote at x = 3; hole at x = -3.

Problem 7 :

y = (x² - 25)/(x – 4)

Solution :

y = (x² - 25)/(x – 4)

To find vertical asymptote, we equate the denominator to 0.

x – 4 = 0

x = 4

So, the vertical asymptote at x = 4.

Problem 8 :

y = (x - 2) (2x + 3)/(5x + 4) (x – 3)

Solution :

y = (x - 2) (2x + 3)/(5x + 4) (x – 3)

To find vertical asymptote, we equate the denominator to 0.

5x + 4 = 0

5x = -4

x = -4/5

x – 3 = 0

x = 3

So, the vertical asymptotes at x = -4/5 and x = 3.

Problem 9 :

y = (15x² - 7x - 2)/(x² - 4)

Solution :

y = (15x² - 7x - 2)/(x² - 4)

To find vertical asymptote, we equate the denominator to 0.

x² – 4 = 0

(x + 2) (x – 2) = 0

x + 2 = 0 and x – 2 = 0

x = -2 x = 2

So, the vertical asymptotes at x = 2 and x = -2.