FIND HORIZONTAL AND VERTICAL ASYMPTOTES

Vertical Asymptote :

The vertical line x = a is called a vertical asymptote of the graph of y = f(x) if

How to find vertical asymptote?

The graph of y = f(x) will have vertical asymptotes at those values of x for which the denominator is equal to zero.

Horizontal Asymptote :

The horizontal line y = b is called a horizontal asymptote of the graph of y = f(x) if either The graph of y = f(x) will have at most one horizontal asymptote. It is found according to the following:

How to find vertical and horizontal asymptotes of rational function?

1) If

degree of numerator > degree of denominator

then the graph of y = f(x) will have no horizontal asymptote.

2) If

degree of numerator  =  degree of denominator

then the graph of y = f(x) will have a  horizontal asymptote at y  =  a_n/b_m.

3)  If

degree of numerator < degree of denominator

then the graph of y = f(x) will have a horizontal asymptote at y = 0 (i.e., the x-axis).

Find the vertical and horizontal asymptotes of the functions given below.

Example 1 :

f(x)  =  4x²/(x² + 8)

Solution :

Vertical Asymptote :

x² + 8  =  0

x²  =  -8

x  =  √-8

Since  √-8 is not a real number, the graph will have no vertical asymptotes.

Horizontal Asymptote :

The highest exponent of numerator and denominator are equal.

Vertical asymptote  y  =  4/1.

So, horizontal asymptote is y  =  4

Example 2 :

f(x)  =  (4x+5)/(4x²-9)

Solution :

Vertical Asymptote :

4x²-9  =  0

4x²  =  9

x  =  √(9/4)

x  =  ± 3/2

So, vertical asymptotes are x  =  3/2 and x  =  -3/2.

Horizontal Asymptote :

Degree of the denominator > Degree of the numerator

So, the  horizontal asymptote is y = 0.

Example 3 :

f(x)  =  (x³+2x+1)/(x²-x-12)

Solution :

Vertical Asymptote :

x²-x-12  =  0

(x-4)(x+3)  =  0

x  =  4 and -3

So, vertical asymptotes are x  =  4 and x  =  -3.

Horizontal Asymptote :

Degree of the numerator  =  3

Degree of the denominator  =  2

Degree of the numerator > Degree of the denominator

It has no horizontal asymptote.

Example 4 :

f(x)  =  (4x²-3)/(2x²-3x+1)

Solution :

Vertical Asymptote :

2x²-3x+1  =  0

(2x-1)(x-1)  =  0

x  =  1/2 and 1

So, vertical asymptotes are x  =  1/2 and x  =  1.

Horizontal Asymptote :

Degree of the numerator  =  2

Degree of the denominator  =  2

Degree of the numerator = Degree of the denominator

Horizontal asymptote  =  4/2  ==>  2

So, the horizontal asymptote is y = 2.

Example 5 :

Find the asymptote of the function 

f(x)  =  ln(2x+3)

Solution :

Vertical Asymptote :

2x+3  =  0

2x  =  -3

x  =  -3/2

So, vertical asymptote is x  =  -3/2.

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