Ratio of two polynomial functions is defined as a rational function where a polynomial is a finite length expression made up of basic mathematical operations (addition, subtraction, multiplication and non-negative exponents) which contains variables and constants. A function which evaluates a polynomial is known as polynomial function. One argument function named as g, is said to be a polynomial if it satisfies, |

*: In vertical asymptote we will have vertical lines near which the function f(x) becomes infinite.*

__Vertical asymptotes__If the denominator of a rational function has more factors of (x - a) than the numerator, then the rational function will have a vertical asymptote at x = a. Only the denominator matters in vertical asymptote.

**: Horizontal line is an asymptote only to the far left and the far right of the graph. There won't be asymptotes in the middle.**

__Horizontal asymptote__f(x) = $\frac{ax^{n}+....}{bx^{m} + .......}$

Numerator will have nth degree polynomial and denominator will have mth degree polynomial.

When n < m | x axis is the horizontal asymptote. |

When n = m | Horizontal asymptote is the line Y = $\frac{a}{b}$ |

When n > m | No horizontal asymptote |

**Oblique asymptote:**A rational function will have an oblique or slant asymptote when the degree of the numerator is greater than that of the numerator. The quotient obtained by division will yield the oblique asymptote.

A function that contains two polynomials in fraction form or written in ratios is known as rational functions. Let's see mathematical representation of rational function. In case of polynomial with one variable 'a' is said to be rational function if and only if it is written in the form:

=> f (a) = $\frac{R(a)}{S(a)}$, here both 'R' and 'S' are polynomial functions in 'a' and value of 'S(a)' is not zero. Here Domain of function (f) can be defined as set of all points of 'a' for which value of denominator is not zero. Each polynomial function can follow property of a rational function with S(a) = 1. If a function is written in form of f(a) = sin (a) then this is not a rational function.

Example of rational function is $\frac{x + 1}{ x^{2} – 1}$. Here one is divided by the other like a ratio.

**To find the derivative of a rational function, we can use the reciprocal rule of the derivative which is a combination of quotient rule and product rule.**

We do differentiation in order to reduce the equation with respect to a certain variable. Differentiating functions with real numbers except for those involve finding derivatives of rational functions or expressions including fractions are comparatively easy.

**For example,**suppose the function is given as:

f(x) = $\frac{H(x)}{G(x)}$,

Where, H(x) and G(x) are two mathematical expressions and are present in form of fraction. Now we will discuss how to find derivatives of rational functions? To resolve this we have to follow a formula given as:

$\frac{d (f(x))}{dx}$ = [(G(x) (d ($\frac{H_{(x)}}{dx}$) – H(x) (d ($\frac{G_{(x)}}{dx}$) / G

^{2}(x)],

To understand it better let’s consider an example of a function: f(x) = $\frac{(x + 1)}{ (x + 2)}$.

Here, let us say: H(x) = (x + 1) and G(x) = (x + 2).

Applying the above formula directly to differentiate the function f(x) we can write the derivative as:

$\frac{d (f(x))}{dx}$ = [((x + 2) ($\frac{d (x + 1)}{dx}$) – (x + 1) ($\frac{d (x + 2)}{dx}$)) / (x + 2)

^{2}],

$\frac{d (f(x))}{dx}$= [$\frac{(x + 2) (1) – (x + 1) (1)}{(x^2 + 4 + 4x)}$],

$\frac{d (f(x))}{dx}$= [$\frac{(x + 2 - x - 1)}{(x^2 + 4 + 4x)}$],

or $\frac{d (f(x))}{dx}$ = $\frac{1}{x^2 + 4 + 4x}$

In a proper rational function the degree of numerator will be less than the degree of denominator.

**Examples:**$\frac{x}{x^{2} - 2}$, $\frac{1}{x + 8}$, $\frac{x^{5}}{3x^{4} -2}$

In all the examples above, the degree of numerator is less than the degree of denominator.

In improper rational function the degree of numerator is greater than or equal to the denominator. Improper rational function can be simplified by polynomial long division.

**Examples:**$\frac{x^{5} - 4}{x^{2} + 5}$, $\frac{5x^{3} - 7}{8x^{3} - 4}$, $\frac{x^{4} - 5x} {23x + 27}$

To find the inverse of a rational function, we have to follow some steps:

If we have f(x) = G(x)/ H(x), replace f(x) by y.

Solve the equation for x in terms of y.

When we can find the value or expression for x in terms of y replace x by f$^{-1}$(y), which is the inverse of the given rational function.

**Example:**Find the inverse of the rational function f(x) = $\frac{x-7}{7x - 5}$

**Solution:**

f(x) = y = $\frac{x - 7}{7x - 5}$

y (7x - 5) = x - 7

7xy - 5y = x - 7

7xy - x = 5y - 7

x(7y - 1) = 5y - 7

x = $\frac{5y - 7}{7y -1}$

x = f$^{-1}$(y) = $\frac{5y -7}{7y -1}$

Both rational functions and rational expressions are undefined for numbers that make the denominator zero. These numbers should be excluded from being possible values for the variable. Hence, for all rational expressions or functions, the domain consists of all real numbers except those numbers that make the denominator equal to zero.

**While graphing a rational function we need to remember the following steps:**

**Step 1:**Look for the common factor of the numerator. This will give the holes of the curve.

**Step 2:**Find the vertical asymptotes by finding the zeroes of the denominator.

**Step 3:**Look for the degree of the numerator and the denominator. If degree of numerator is greater than or equal to that of the denominator, there will be horizontal asymptotes.

**Step 4:**Find the x - intercepts by equating the rational expression to 0.

**Step 5:**Find the y - intercepts by plugging in x = 0.

**Step 6:**Prepare the table with the values to the right and left of the restricted values obtained from the step 1.

**Step 7:**Graph the function by drawing the asymptotes, plot the x and y intercepts and then, plot the points from the table.

**Given below is an example:**

Graph the following function y = $\frac{3x + 5}{x - 2}$

**Solution:**

Find the vertical asymptotes by finding the zeroes of the denominator.

$\Rightarrow$ x - 2 = 0

$\Rightarrow$ x = 2

From the given equation we see that the numerator and the denominator have the same degree so the asymptotes will be horizontal. Horizontal asymptote is found by dividing the leading coefficients.

y = $\frac{3}{1}$

y = 3

Find any x or y intercepts

Put x = 0

y = $\frac{3(0) + 5}{(0) - 2}$

= $\frac{5}{-2}$

= -2.5

When y = 0,

We have 3x + 5 =0

$\Rightarrow$ x = $\frac{-5}{3}$

= - 1.67

Pick few more x values to find the corresponding y values.

x |
y = $\frac{3x + 5}{x - 2}$ |

-2 |
0.25 |

-1 | -0.67 |

1 | -8 |

5 | 6.67 |

### Solved Examples

**Question 1:**Find the derivative of $\frac{x^{2} + 2x - 5}{x^{3} - 3x + 2}$

**Solution:**

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

= $\frac{2x^{4}-6x^{2}+4x + 2x^{3}-6x + 4-[3x^{4}+6x^{3}-15x^{2}-3x^{2}-6x + 15]}{(x^{3}-3x + 2)^2}$

= $\frac{2x^{4}+2x^{3}-6x^{2}-2x + 4-[3x^{4}+6x^{3}-18x^{2}-6x + 15]}{(x^{3}-3x + 2)^2}$

= $\frac{-x^{4}-4x^{3}+12x^{2}+4x - 11}{(x^{3}-3x + 2)^2}$

**Question 2:**Find domain and vertical asymptote of f(x) = $\frac{1}{x^{2}+12x +32}$

**Solution:**

Equate the denominator to zero to find the domain of the given function

x$^{2}$ + 12x + 32 = 0

(x + 4) (x + 8) = 0

x = -4 and x = - 8

Therefore, domain is all real except - 4 and - 8.

Vertical asymptotes are x = - 4 and x = -8