## Precalculus (6th Edition) Blitzer

The function written in the fractional form as $f\left( x \right)=\frac{P\left( x \right)}{Q\left( x \right)}$ (where $P\left( x \right),Q\left( x \right)$ are the polynomial functions of x, and $Q\left( x \right)\ne 0$ ) is called the rational function. Then, the horizontal asymptotes can be found out as, • If the degree of $P\left( x \right)$ is greater than the degree of $Q\left( x \right)$ , then there is no horizontal asymptote. • If the degree of $P\left( x \right)$ is equal to the degree of $Q\left( x \right)$ , then the horizontal asymptote is given by the ratio of the coefficient of the highest degree term of x in the numerator and denominator. • If the degree of $P\left( x \right)$ is less than the degree of $Q\left( x \right)$ , then $f\left( x \right)=0$ gives the horizontal asymptote.
A function $f\left( x \right)$ is a rational function if it is written in the form $f\left( x \right)=\frac{P\left( x \right)}{Q\left( x \right)}$ …… (1) Where $P\left( x \right),Q\left( x \right)$ are the polynomial functions of x in simplest form, and $Q\left( x \right)$ cannot be a zero function. For Example, let $f\left( x \right)=\frac{5{{x}^{2}}}{3{{x}^{3}}+1}$ Compare it with equation (1), Since, the degree of $P\left( x \right)$ is less than the degree of $Q\left( x \right)$ , then $f\left( x \right)=0$ gives the horizontal asymptote. Let $f\left( x \right)=\frac{5{{x}^{3}}}{3{{x}^{3}}+1}$ Compare it with equation (1), Since, the degree of $P\left( x \right)$ is equal to the degree of $Q\left( x \right)$ , then the horizontal asymptote is given by the ratio of the coefficient of the highest degree term of x in the numerator and denominator. The horizontal asymptote is $\frac{5}{3}$. Let $f\left( x \right)=\frac{5{{x}^{4}}}{3{{x}^{3}}+1}$ Compare it with equation (1), Since, the degree of $P\left( x \right)$ is greater than the degree of $Q\left( x \right)$ , then there is no horizontal asymptote.