# 3.4 Rational function  (Page 4/6)

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2. Find horizontal asymptote :

$f\left(x\right)=\frac{-x\left(x+1\right)}{{x}^{4}+16}\phantom{\rule{1em}{0ex}}$

The order of highest power term is 2 in numerator and 4 in denominator. Thus, n<m. Hence, x-axis is horizontal asymptote.

$y=0$

3. Find horizontal asymptote :

$f\left(x\right)=\frac{x\left({x}^{2}-1\right)}{x+2}$

The order of highest power term is 3 in numerator and 1 in denominator. Here, n>m. Hence, there is no horizontal asymptote.

## Slant asymptotes

Slant asymptote is a line that the graph approaches. This line is neither vertical nor horizontal. A rational function has a slant or oblique asymptote when order of numerator (n) is greater than order of denominator (m).

The equation of slant asymptote is obtained by dividing numerator polynomial by denominator polynomial. The quotient of division is equation of asymptote. Clearly, asymptote is a straight line. As such, quotient should be a linear expression. The requirement that asymptote is a straight line implies that the order of numerator polynomial is higher than order of denominator polynomial by 1 i.e. n=m+1.

In the nutshell, slant asymptote exists when n=m+1. The slant asymptote is obtained by dividing numerator and denominator. We neglect remainder. The equation of the slant asymptote is given by quotient equated to “y”.

Find slant asymptote :

$f\left(x\right)=\frac{{x}^{2}}{x+1}$

Division here yields quotient as “x-1”. Hence, equation of slant asymptote is :

$y=x-1$

## X-intercepts

The x-intercepts are also known as zeroes of function or real roots of corresponding equation when function is equated to zero. Since function is many-one, there can be more than one x-intercept. On graphs, x-intercepts are points on x-axis, where graph intersects it. Thus, x-intercepts are x-values where function value becomes zero.

$f\left(x\right)=0$

In the case of rational function, x-intercepts exist when numerator turns zero. In other words, x-intercepts are x-values for which numerator of the function turns zero.

$g\left(x\right)=\frac{f\left(x\right)}{h\left(x\right)}=0$ $g\left(x\right)=0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)=0$

We determine x-intercepts by solving equation formed by equating function to zero. It is hepful to know that real polynomial of odd degree has a real root and, therefore, at least one x-intercept. In the case of rational function, the function is not defined for values of x when denominator turns zero. It means that there will be no x-intercept corresponding to linear factor which is common to denominator. Consider the function given here :

$h\left(x\right)=\frac{\left(x-1\right)\left(x+2\right)}{{\left(x-1\right)}^{2}\left(x+1\right)}$

Equating numerator to zero, we have :

$⇒\left(x-1\right)\left(x+2\right)=0$ $⇒x=1,-2$

But (x-1) is also linear factor in denominator. It means that point x=1 is a singularity. Hence, x-intercept is only x=-2 as function is not defined at x=1.

Find x-intercepts of reciprocal function :

$f\left(x\right)=\frac{1}{x}$

Here numerator is 1 and can not be zero. Thus, reciprocal function does not have x-intercepts.

Find x-intercepts of function given by :

$\frac{{x}^{2}-3x+2}{{x}^{2}-2x-3}$

Factorizing, we have :

$\frac{{x}^{2}-3x+2}{{x}^{2}-2x-3}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+1\right)\left(x-3\right)}$

Equating numerator to zero, we have :

$⇒\left(x-1\right)\left(x-2\right)=0$ $⇒x=1\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}2$

Thus, x-intercepts are 1 and 2.

## Y-intercepts

This is function or y value when x is zero. Functions are many-one relation. Thus, there can be only one y-intercept. The y-intercept is calculated as :

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Period of sin^6 3x+ cos^6 3x