determine the horizontal asymptote of the function. if none exists, state that fact.\n f(x)=\frac{4…
determine the horizontal asymptote of the function. if none exists, state that fact.\n f(x)=\frac{4 x}{x^{3}-3 x^{2}} \nselect the correct choice below and, if necessary, fill in the answer box(es) to complete your choice\na. the function has two horizontal asymptotes. the top asymptote is ( square ) and the bottom asymptote is\n(type equations.)\nb. the function has one horizontal asymptote, ( square ). (type an equation.)\nc. the function has no horizontal asymptotes
Answer
Explanation:
Step1: Analyze the degrees of numerator and denominator
The degree of the numerator (n) of (y = \frac{4x}{x^{3}-3x^{2}}) is (n = 1) (since the highest - power of (x) in the numerator is (x^1)), and the degree of the denominator (m) is (m=3) (since the highest - power of (x) in the denominator is (x^3)).
Step2: Apply the horizontal asymptote rule
The rule for horizontal asymptotes of a rational function (y=\frac{f(x)}{g(x)}) where (f(x)=a_nx^n+\cdots) and (g(x)=b_mx^m+\cdots) is: If (n\lt m), then (y = 0) is the horizontal asymptote. Since (n = 1) and (m = 3) ((1\lt3)), we find the limit as (x\to\pm\infty). We use the fact that (\lim_{x\to\pm\infty}\frac{4x}{x^{3}-3x^{2}}=\lim_{x\to\pm\infty}\frac{\frac{4x}{x^{3}}}{\frac{x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}}) (divide numerator and denominator by (x^{3})). [ \begin{align*} \lim_{x\to\pm\infty}\frac{\frac{4x}{x^{3}}}{\frac{x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}}&=\lim_{x\to\pm\infty}\frac{\frac{4}{x^{2}}}{1 - \frac{3}{x}}\ \end{align*} ] As (x\to\pm\infty), (\lim_{x\to\pm\infty}\frac{4}{x^{2}}=0) and (\lim_{x\to\pm\infty}\frac{3}{x}=0). So (\lim_{x\to\pm\infty}\frac{\frac{4}{x^{2}}}{1-\frac{3}{x}} = 0)
Answer:
B. The function has one horizontal asymptote, (y = 0)