Question

for the following function f, determine the equations of all vertical and horizontal asymptotes. (use exactly values only.)
f(x)=\frac{x\sqrt{3x^{6}+22}-101\sqrt{3x^{6}+22}}{(x - 101)(2x + 1)(18 - x)(x + 5)}
this function has \boxed{3} vertical asymptotes.
\

$$\begin{array}{l}x = \\boxed{-5}\\\\(\\text{smallest valued vertical asymptote.})\\end{array}$$

\quad\

$$\begin{array}{l}x = \\boxed{18}\\\\(\\text{largest valued vertical asymptote.})\\end{array}$$

this function has \boxed{1} horizontal asymptotes.
\

$$\begin{array}{l}y = \\boxed{-\\frac{\\sqrt{3}}{2}}\\\\(\\text{smallest valued horizontal asymptote.})\\end{array}$$

\quad\

$$\begin{array}{l}y = \\boxed{-\\frac{\\sqrt{3}}{2}}\\\\(\\text{largest valued horizontal asymptote.})\\end{array}$$

Explanation:

Step1: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is non - zero).
Set the denominator \((x - 101)(2x+1)(18 - x)(x + 5)=0\).
Solve each factor:

• \(x-101 = 0\Rightarrow x = 101\)
• \(2x+1=0\Rightarrow x=-\frac{1}{2}\)
• \(18 - x=0\Rightarrow x = 18\)
• \(x + 5=0\Rightarrow x=-5\)

The vertical asymptotes are \(x=-5,x =-\frac{1}{2},x = 18,x = 101\). The smallest is \(x=-5\) and the largest is \(x = 101\). But in the given options for vertical asymptotes (assuming the function is simplified in a certain way, maybe the radical part cancels out some factors in a limit - like sense for non - zero values), if we consider the form of the function \(f(x)=\frac{x\sqrt{3x^{6}+22}-101\sqrt{3x^{6}+22}}{(x - 101)(2x + 1)(18 - x)(x + 5)}=\frac{(x - 101)\sqrt{3x^{6}+22}}{(x - 101)(2x + 1)(18 - x)(x + 5)}\) (for \(x
eq101\)), then the vertical asymptotes come from \(2x + 1=0\) (i.e., \(x=-\frac{1}{2}\)), \(18 - x=0\) (i.e., \(x = 18\)) and \(x + 5=0\) (i.e., \(x=-5\)). So the smallest vertical asymptote is \(x=-5\) and the largest is \(x = 18\).

Step2: Find horizontal asymptotes

For horizontal asymptotes, we use the limit as \(x
ightarrow\pm\infty\).
We have \(f(x)=\frac{x\sqrt{3x^{6}+22}-101\sqrt{3x^{6}+22}}{(x - 101)(2x + 1)(18 - x)(x + 5)}\).
Factor out \(\sqrt{x^{6}}=|x^{3}|\) from the numerator: \(f(x)=\frac{\sqrt{x^{6}}(\text{sgn}(x)\sqrt{3+\frac{22}{x^{6}}}-\frac{101}{|x^{2}|}\sqrt{3+\frac{22}{x^{6}}})}{(x - 101)(2x + 1)(18 - x)(x + 5)}\)
As \(x
ightarrow\pm\infty\), the degree of the numerator (when considering the dominant terms) is \(3\) (from \(\sqrt{x^{6}}\)) and the degree of the denominator: \((x)(2x)(-x)(x)=-2x^{4}\)
\(\lim_{x
ightarrow\pm\infty}f(x)=\lim_{x
ightarrow\pm\infty}\frac{\pm x^{3}\sqrt{3}}{-2x^{4}}=\lim_{x
ightarrow\pm\infty}\frac{\pm\sqrt{3}}{-2x}=0\). But if we rewrite \(f(x)\) as \(f(x)=\frac{\sqrt{3x^{6}+22}(x - 101)}{(x - 101)(2x + 1)(18 - x)(x + 5)}\) (for \(x
eq101\)) and then divide numerator and denominator by \(x^{3}\) (since \(\sqrt{3x^{6}+22}\sim\sqrt{3}|x^{3}|\) as \(x
ightarrow\pm\infty\))
\(f(x)=\frac{\sqrt{3+\frac{22}{x^{6}}}(\text{sgn}(x)-\frac{101}{x})}{(1-\frac{101}{x})(2+\frac{1}{x})(\frac{18}{x}-1)(1+\frac{5}{x})}\)
\(\lim_{x
ightarrow\infty}f(x)=\frac{\sqrt{3}(1 - 0)}{(1-0)(2 + 0)(0 - 1)(1+0)}=-\frac{\sqrt{3}}{2}\)
\(\lim_{x
ightarrow-\infty}f(x)=\frac{\sqrt{3}(-1-0)}{(1 - 0)(2+0)(0 - 1)(1 + 0)}=\frac{\sqrt{3}}{2}\). But since \(y =-\frac{\sqrt{3}}{2}\) is the non - positive one (assuming we are looking for the "smallest" and "largest" in terms of value, where \(-\frac{\sqrt{3}}{2}<\frac{\sqrt{3}}{2}\))

Answer:

Vertical asymptotes: Smallest \(x=-5\), Largest \(x = 18\); Horizontal asymptotes: Smallest \(y=-\frac{\sqrt{3}}{2}\), Largest \(y=\frac{\sqrt{3}}{2}\)