Question

find all horizontal asymptotes of the following function.

f(x)=\frac{4 x^{2}+58 x+180}{6 x+27}

answer attempt 1 out of 2

one horizontal asymptote

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 \(x\):
\[ \begin{align*} \lim_{x ightarrow\pm\infty}f(x)&=\lim_{x ightarrow\pm\infty}\frac{4x^{2}+58x + 180}{6x + 27}\\ &=\lim_{x ightarrow\pm\infty}\frac{\frac{4x^{2}}{x}+\frac{58x}{x}+\frac{180}{x}}{\frac{6x}{x}+\frac{27}{x}}\\ &=\lim_{x ightarrow\pm\infty}\frac{4x + 58+\frac{180}{x}}{6+\frac{27}{x}} \end{align*} \]

Step2: Evaluate the limit

As \(x ightarrow\pm\infty\), \(\lim_{x ightarrow\pm\infty}\frac{180}{x}=0\) and \(\lim_{x ightarrow\pm\infty}\frac{27}{x}=0\).
\[ \begin{align*} \lim_{x ightarrow\pm\infty}\frac{4x + 58+\frac{180}{x}}{6+\frac{27}{x}}&=\lim_{x ightarrow\pm\infty}\frac{4x+58}{6}\\ \end{align*} \]
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 \(x ightarrow\pm\infty\), the non - linear term \(\frac{2}{3}x\) dominates. But if we consider the limit of the original function as \(x ightarrow\pm\infty\) in another way:
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x ightarrow\infty}\frac{x(4x + 58+\frac{180}{x})}{x(6+\frac{27}{x})}\\ \end{align*} \]
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 \(x ightarrow\pm\infty\), 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:
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x ightarrow\infty}\frac{4x+58+\frac{180}{x}}{6+\frac{27}{x}}=\infty\\ \lim_{x ightarrow-\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x ightarrow-\infty}\frac{4x+58+\frac{180}{x}}{6+\frac{27}{x}}=-\infty \end{align*} \]
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:
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x ightarrow\infty}\frac{x^{2}(4+\frac{58}{x}+\frac{180}{x^{2}})}{x(6+\frac{27}{x})}\\ &=\lim_{x ightarrow\infty}x\times\frac{4+\frac{58}{x}+\frac{180}{x^{2}}}{6+\frac{27}{x}}=\infty \end{align*} \]
\[ \begin{align*} \lim_{x ightarrow-\infty}\frac{4x^{2}+58x + 180}{6x + 27}&=\lim_{x ightarrow-\infty}\frac{x^{2}(4+\frac{58}{x}+\frac{180}{x^{2}})}{x(6+\frac{27}{x})}\\ &=\lim_{x ightarrow-\infty}x\times\frac{4+\frac{58}{x}+\frac{180}{x^{2}}}{6+\frac{27}{x}}=-\infty \end{align*} \]

Answer:

There is no horizontal asymptote.