for the following function f, determine the equations of all vertical and horizontal asymptotes. (use…

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

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

Answer

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\neq101)), 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\rightarrow\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\rightarrow\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\rightarrow\pm\infty}f(x)=\lim_{x\rightarrow\pm\infty}\frac{\pm x^{3}\sqrt{3}}{-2x^{4}}=\lim_{x\rightarrow\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\neq101)) and then divide numerator and denominator by (x^{3}) (since (\sqrt{3x^{6}+22}\sim\sqrt{3}|x^{3}|) as (x\rightarrow\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\rightarrow\infty}f(x)=\frac{\sqrt{3}(1 - 0)}{(1-0)(2 + 0)(0 - 1)(1+0)}=-\frac{\sqrt{3}}{2}) (\lim_{x\rightarrow-\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})