Whenever we see the word rational it comes to our mind that there is a numerator along with a non zero denominator. Same is in case of rational functions.

When we have two functions such that they are in the form y = $\frac{f (x)}{g (x)}$ with restriction that $g (x)$ is not equal to zero, then ‘y’ is called a rational function.

We can reduce our given expression in lowest terms by factorizing the given polynomials and then taking out common factors from both to remove them and getting the lowest possible rational function equivalent to the original function in rational form.

When we have two functions such that they are in the form y = $\frac{f (x)}{g (x)}$ with restriction that $g (x)$ is not equal to zero, then ‘y’ is called a rational function.

We can reduce our given expression in lowest terms by factorizing the given polynomials and then taking out common factors from both to remove them and getting the lowest possible rational function equivalent to the original function in rational form.

We know that y = $\frac{f (x)}{g (x)}$ is a rational function if and only if:

The rational function in which the degree of the function in the numerator is greater than or equal to the degree of the function in the denominator is known as an improper rational function.

In case of rational functions we have a concept of asymptotes which
are of three types: vertical, horizontal, slant. A given function in
rational form can have multiple vertical asymptotes, at most one
horizontal asymptote and at most one slant or oblique asymptote.

In
order to find vertical asymptotes we substitute the denominator equal
to zero. For finding the horizontal asymptote the degree of the
numerator function should be less than or equal to the degree of the
denominator function. When the degrees are same horizontal asymptote is
given by dividing the leading coefficients of both the functions. When
the degree of numerator is less than denominator then the horizontal
asymptote is given by y = 0.

To find slant asymptote the degree of numerator should be one more than that of denominator.

To find slant asymptote the degree of numerator should be one more than that of denominator.

If so, then to
divide the polynomials and the quotient so obtained is the value of
slant asymptote. In cases when the degree of numerator is greater by 2
or more than that of denominator then there is no horizontal or slant
asymptote existing.

$\frac{(x + 2)}{(x^2 – 4)}$

= $\frac{(x + 2)}{[(x – 2) (x + 2)]}$

= $\frac{1}{(x – 2)}$

Hence the vertical asymptote is given by denominator equal to zero. That is,

x – 2 = 0

$\rightarrow$ x = 2 is the vertical asymptote.

Horizontal asymptote is given by y = 0 as degree of numerator function is less than degree of denominator function.

And no slant asymptote exists in the case of given rational function.

**Example 2:** Find the roots of f(x) = $\frac{(6x-18)(x+2)}{x+1}$

Solution: To find the roots of the given rational function we simply substitute numerator equal to zero. If possible, first reduced the function to its lowest terms.

f(x) = $\frac{(6x-18)(x+2)}{x+1}$

To find roots of given rational function follow steps:

__Step 1:__ Set numerator of the function equal to zero.

(6x-18)(x+2) = 0

__Step 2:__ Isolate the value of x.

6x - 18 = 0 or x = 3

x + 2 = 0 or x = -2

x = -2 and 3 are roots of given rational function.

**Example 3:** Reduce $\frac{x^3-5x^2+6x}{x-3}$

**Solution:**

$\frac{x^3-5x^2+6x}{x-3}$ = $\frac{x(x^2-5x+6}{x-3}$

= $\frac{x(x -2)(x-3)}{x-3}$

= x(x - 2) (Cancel common terms)

or $\frac{x^3-5x^2+6x}{x-3}$ = x(x - 2)

Solution: To find the roots of the given rational function we simply substitute numerator equal to zero. If possible, first reduced the function to its lowest terms.

f(x) = $\frac{(6x-18)(x+2)}{x+1}$

To find roots of given rational function follow steps:

(6x-18)(x+2) = 0

6x - 18 = 0 or x = 3

x + 2 = 0 or x = -2

x = -2 and 3 are roots of given rational function.

$\frac{x^3-5x^2+6x}{x-3}$ = $\frac{x(x^2-5x+6}{x-3}$

= $\frac{x(x -2)(x-3)}{x-3}$

= x(x - 2) (Cancel common terms)

or $\frac{x^3-5x^2+6x}{x-3}$ = x(x - 2)