HOW TO FIND HOLES AND ASYMPTOTES IN RATIONAL FUNCTIONS

For a rational function, f(x) / g(x)

How to find hole ?

To find hole for a rational function, we have to factorize both numerator and denominator. 

The common factors can be considered as hole.

Equating the common factor to 0, we will get to know the hole is at which point.

How to find horizontal and vertical asymptote ?

To find horizontal asymptote, we have to consider the highest exponent of numerator and denominator.

  • If the highest exponent of the numerator is less than the highest exponent of the denominator, then x-axis or y = 0 is the horizontal asymptote.
  • If the highest exponent of the numerator and denominator are equal, then by dividing the leading coefficients, we find the equation of horizontal asymptote.
  • If the highest exponent of the numerator is greater than the highest exponent of the denominator then there will be oblique asymptote.

is s than the highest exponent of the denominator, then x-axis or y = 0 is the horizontal asymptote.

How to find vertical asymptote ?

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

Identify the holes, vertical asymptotes, x-intercepts, horizontal asymptote.

Problem 1 :

f(x)=4x-3

Solution:

Hole:

In the given rational function, clearly there is no common factor found at both numerator and denominator.

So, there is no hole for the given rational function.

Vertical asymptotes:

x - 3 = 0

x = 3

x- intercept:

Set f(x) = 0

0=4x-30=4

So, there is no x-intercept.

Horizontal asymptotes:

Comparing highest exponents,

denominator > numerator

So, horizontal asymptote is at y = 0.

Problem 2 :

f(x)=x2+7x+12-2x2-2x+12

Solution:

Hole:

In the given rational function, let us factor the numerator.

f(x)=(x+4)(x+3)(x-2)(x+3)

After having factored, the common factor found at both numerator and denominator is (x + 3).

x + 3 = 0

x = -3

So, there is a hole at x = -3.

Vertical asymptotes:

-2x2 - 2x + 12 = 0

x2 + x - 6 = 0

(x - 2) (x + 3) = 0

x = 2 or x = -3

So, vertical asymptotes is x = 2.

x- intercept:

Set f(x) = 0

0=x2+7x+12-2x2-2x+12x2+7x+12=0(x+4)(x+3)=0x=-4 or x=-3

So, x-intercept = -4.

Horizontal asymptotes:

Comparing highest exponents,

denominator = numerator

Horizontal asymptote=Coefficient of x of numeratorCoefficient of x in the denominatory=-12

So, horizontal asymptote is y = -1/2.

Problem 3 :

f(x)=1-x+4

Solution:

Hole:

In the given rational function, clearly there id no common factor found at both numerator and denominator.

So, there is no hole for the given rational function.

Vertical asymptotes:

-x + 4 = 0

-x = -4

x = 4

So, vertical asymptotes is x = 4.

x- intercept:

Set f(x) = 0

0=1-x+40=1

So, there is no x-intercept.

Horizontal asymptotes:

Comparing highest exponents,

denominator > numerator

So, horizontal asymptote is at y = 0.

Problem 4 :

f(x)=-3x+12x2-3x-4

Solution:

Hole:

In the given rational function, let us factor the numerator.

f(x)=-3(x-4)(x+1)(x-4)

After having factored, the common factor found at both numerator and denominator is (x - 4).

x - 4 = 0

x = 4

So, there is a hole at x = 4.

Vertical asymptotes:

x2 - 3x - 4 = 0

(x + 1)(x - 4) = 0

x = -1 or x = 4

So, vertical asymptotes is x = -1.

x- intercept:

Set f(x) = 0

0=-3x+12x2-3x-4-3x+12=03x=12x=4

Horizontal asymptotes:

Comparing highest exponents,

denominator > numerator

So, horizontal asymptote is at y = 0.

Problem 5 :

f(x)=-2x2+4x+16x2-5x+4

Solution:

Hole:

In the given rational function, let us factor the numerator.

f(x)=-2(x-4)(x+2)(x-1)(x-4)

After having factored, the common factor found at both numerator and denominator is (x - 4).

x - 4 = 0

x = 4

So, there is a hole at x = 4.

Vertical asymptotes:

x2 - 5x + 4 = 0

(x - 1)(x - 4) = 0

x = 1 or x = 4

So, vertical asymptotes is x = 1.

x- intercept:

Set f(x) = 0

0=-2x2+4x+16x2-5x+4-2x2+4x+16=0-2(x+2)(x-4)=0x=-2 or x=4

So, x-intercept is -2.

Horizontal asymptotes:

Comparing highest exponents,

denominator = numerator

Horizontal asymptote=Coefficient of x of numeratorCoefficient of x in the denominatory=-21y=-2

So, horizontal asymptotes is y = -2.

Problem 6 :

f(x)=x2-3x2x2+2x-12

Solution:

Hole:

In the given rational function, let us factor the numerator.

f(x)=x(x-3)2(x-2)(x+3)

So, there is no hole.

Vertical asymptotes:

2x2 + 2x - 12 = 0

2(x - 2)(x + 3) = 0

x - 2 = 0 or x + 3 = 0

x = 2 or x = -3

So, vertical asymptotes is x = 2, x = -3.

x- intercept:

Set f(x) = 0

0=x2-3x2x2+2x-12x2-3x=0x(x-3)=0x=0, x=3

So, x-intercept are x = 0, 3.

Horizontal asymptotes:

Comparing highest exponents,

denominator = numerator

Horizontal asymptote=Coefficient of x of numeratorCoefficient of x in the denominatory=12

So, horizontal asymptote is y = 1/2.

Problem 7 :

f(x)=3x+6x+3

Solution:

Hole:

In the given rational function, let us factor the numerator.

f(x)=3(x+2)x+3

So, there is no hole.

Vertical asymptotes:

x + 3 =0

x = -3

So, vertical asymptotes is x = -3.

x- intercept:

Set f(x) = 0

0=3x+6x+33x+6=03x=-6x=-2

So, x-intercept = -2.

Horizontal asymptotes:

Comparing highest exponents,

denominator = numerator

Horizontal asymptote=Coefficient of x of numeratorCoefficient of x in the denominatory=31y=3

So, horizontal asymptote is y = 3.

Problem 8 :

f(x)=x2+5x+4x2-1

Solution:

Hole:

In the given rational function, let us factor the numerator.

f(x)=(x+1)(x+4)x(x-1)

So, there is no hole.

Vertical asymptotes:

x2 - 1 = 0

x2 = 1

x = 1

So, vertical asymptotes is x = 1.

x- intercept:

Set f(x) = 0

0=x2+5x+4x2-1x2+5x+4=0(x+1)(x+4)=0x=-1 or x=-4

So, x-intercept is -4.

Horizontal asymptotes:

Comparing highest exponents,

denominator = numerator

Horizontal asymptote=Coefficient of x of numeratorCoefficient of x in the denominatory=11y=1

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