find all horizontal asymptotes of the following function.\n\n f(x)=\frac{4 x^{2}+58 x+180}{6 x+27}…

Explanation:

Step1: Divide numerator and denominator by highest power of (x) in denominator

The highest power of (x) in the denominator (6x + 27) is (x). Divide numerator (4x^{2}+58x + 180) and denominator (6x + 27) by (x): [ \begin{align*} \lim_{x\rightarrow\pm\infty}f(x)&=\lim_{x\rightarrow\pm\infty}\frac{4x^{2}+58x + 180}{6x + 27}\ &=\lim_{x\rightarrow\pm\infty}\frac{\frac{4x^{2}}{x}+\frac{58x}{x}+\frac{180}{x}}{\frac{6x}{x}+\frac{27}{x}}\ &=\lim_{x\rightarrow\pm\infty}\frac{4x + 58+\frac{180}{x}}{6+\frac{27}{x}} \end{align*} ]

Step2: Evaluate the limit

As (x\rightarrow\pm\infty), (\lim_{x\rightarrow\pm\infty}\frac{180}{x}=0) and (\lim_{x\rightarrow\pm\infty}\frac{27}{x}=0). [ \begin{align*} \lim_{x\rightarrow\pm\infty}\frac{4x + 58+\frac{180}{x}}{6+\frac{27}{x}}&=\lim_{x\rightarrow\pm\infty}\frac{4x+58}{6}\ \end{align*} ] Since the degree of the numerator ((n = 1)) is greater than the degree of the denominator ((m=0)) after simplification (in the non - constant part), we can also use polynomial long division. Divide (4x^{2}+58x + 180) by (6x + 27): [ \begin{align*} \frac{4x^{2}+58x + 180}{6x + 27}&=\frac{\frac{2}{3}x(6x + 27)+58x+180 - 18x}{6x + 27}\ &=\frac{2}{3}x+\frac{40x + 180}{6x + 27}\ &=\frac{2}{3}x+\frac{\frac{20}{3}(6x + 27)+180-180}{6x + 27}\ &=\frac{2}{3}x+\frac{20}{3} \end{align*} ] As (x\rightarrow\pm\infty), the non - linear term (\frac{2}{3}x) dominates. But if we consider the limit of the original function as (x\rightarrow\pm\infty) in another way: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x\rightarrow\infty}\frac{x(4x + 58+\frac{180}{x})}{x(6+\frac{27}{x})}\ \end{align*} ] Since the degree of the numerator ((n = 2)) is greater than the degree of the denominator ((m = 1)), we can use the fact that for a rational function (y=\frac{a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m - 1}x^{m - 1}+\cdots+b_{0}}), when (n>m), there is no horizontal asymptote. But wait, we made a mistake above. Let's start over: For a rational function (y = f(x)=\frac{a_{n}x^{n}+a_{n-1}x^{n - 1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{0}}), where (n) is the degree of the numerator and (m) is the degree of the denominator. Here (n = 2) (for (4x^{2}+58x + 180)) and (m = 1) (for (6x+27)). We use polynomial long division: [ \begin{align*} 4x^{2}+58x + 180&=(6x + 27)\times\frac{2}{3}x+(58x+180 - 18x)\ &=(6x + 27)\times\frac{2}{3}x + 40x+180\ &=(6x + 27)\times\frac{2}{3}x+(6x + 27)\times\frac{20}{3}+180 - 180\ \end{align*} ] (f(x)=\frac{4x^{2}+58x + 180}{6x + 27}=\frac{2}{3}x+\frac{20}{3}) As (x\rightarrow\pm\infty), the function (y = f(x)) behaves like (y=\frac{2}{3}x+\frac{20}{3}), so there is no horizontal asymptote. But wait, another approach: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x\rightarrow\infty}\frac{4x+58+\frac{180}{x}}{6+\frac{27}{x}}=\infty\ \lim_{x\rightarrow-\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x\rightarrow-\infty}\frac{4x+58+\frac{180}{x}}{6+\frac{27}{x}}=-\infty \end{align*} ] Wait, no! Wait, we made a mistake in the first step. The correct rule for horizontal asymptotes of a rational function (y=\frac{a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m - 1}x^{m - 1}+\cdots+b_{0}}):

• If (n<m), (y = 0) is the horizontal asymptote.
• If (n=m), (y=\frac{a_{n}}{b_{m}}) is the horizontal asymptote.
• If (n>m), there is no horizontal asymptote.

Here (n = 2) (degree of numerator (4x^{2}+58x + 180)) and (m = 1) (degree of denominator (6x + 27)). So there is no horizontal asymptote.

But wait, let's check the problem again. Maybe there was a mis - understanding. If we consider the limit as (x) approaches (\pm\infty) in another way: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x\rightarrow\infty}\frac{x^{2}(4+\frac{58}{x}+\frac{180}{x^{2}})}{x(6+\frac{27}{x})}\ &=\lim_{x\rightarrow\infty}x\times\frac{4+\frac{58}{x}+\frac{180}{x^{2}}}{6+\frac{27}{x}}=\infty \end{align*} ] [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x\rightarrow-\infty}\frac{x^{2}(4+\frac{58}{x}+\frac{180}{x^{2}})}{x(6+\frac{27}{x})}\ &=\lim_{x\rightarrow-\infty}x\times\frac{4+\frac{58}{x}+\frac{180}{x^{2}}}{6+\frac{27}{x}}=-\infty \end{align*} ]

Answer:

There is no horizontal asymptote.