3.7 Rational functions  (Page 6/16)

For these solutions, we will use $f(x)=\frac{p(x)}{q(x)}, q(x)\ne 0.$

1. $g(x)=\frac{6x^3-10x}{2x^3+5x^2}:$ The degree of $p=\text{degree of} q=3,$ so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at $y=\frac{6}{2}$ or $y=3.$
2. $h(x)=\frac{x^2-4x+1}{x+2}:$ The degree of $p=2$ and degree of $q=1.$ Since $p>q$ by 1, there is a slant asymptote found at $\frac{x^2-4x+1}{x+2}.$
   $$\begin{array}{l}\hfill \\ \begin{array}{c}-2\\ \end{array}\overline{)\begin{array}{rrr}\hfill 1& \hfill -4& \hfill 1\\ \hfill & \hfill -2& \hfill 12\end{array}}\hfill \\ \begin{array}{lll}\text{ 1}\hfill & -6\hfill & 13\hfill \end{array}\hfill \end{array}$$

   The quotient is $x–6$ and the remainder is 13. There is a slant asymptote at $y=x–6.$

3. $k(x)=\frac{x^2+4x}{x^3-8}:$ The degree of $p=2<$ degree of $q=3,$ so there is a horizontal asymptote $y=0.$

Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is $t,$ with coefficient 1. In the denominator, the leading term is $10t,$ with coefficient 10. The horizontal asymptote will be at the ratio of these values:

This function will have a horizontal asymptote at $y=\frac{1}{10}.$

This tells us that as the values of t increase, the values of $C$ will approach $\frac{1}{10}.$ In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or $\frac{1}{10}$ pounds per gallon.

First, note that this function has no common factors, so there are no potential removable discontinuities.

The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at $x=1,–2,\text{and }5,$ indicating vertical asymptotes at these values.

The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as $x\to \pm \infty , f(x)\to 0.$ This function will have a horizontal asymptote at $y=0.$ See [link] .