consider the following curve.\n$f(x)=\\frac{x - 5}{x^{2}}$\ndetermine the domain of the curve. (enter your…

consider the following curve.\n$f(x)=\\frac{x - 5}{x^{2}}$\ndetermine the domain of the curve. (enter your answer using interval notation.)\nfind the intercepts. (enter your answers as comma - separated lists. if an answer does not exist, enter dne.)\ny - intercept\nx - intercept\ndetermine the symmetry of the graph. (select all that apply.)\n□ symmetric with respect to the x - axis\n□ symmetric with respect to the y - axis\n□ symmetric with respect to the origin\n□ no apparent symmetry\nfind $y(x)$.\n$y(x)=$\nfind $y(x)$.\n$y(x)=$\nfind the vertical and horizontal asymptotes. (enter your answers as comma - separated lists. if an answer does not exist, enter dne.)\n$y=$

consider the following curve.\n$f(x)=\\frac{x - 5}{x^{2}}$\ndetermine the domain of the curve. (enter your answer using interval notation.)\nfind the intercepts. (enter your answers as comma - separated lists. if an answer does not exist, enter dne.)\ny - intercept\nx - intercept\ndetermine the symmetry of the graph. (select all that apply.)\n□ symmetric with respect to the x - axis\n□ symmetric with respect to the y - axis\n□ symmetric with respect to the origin\n□ no apparent symmetry\nfind $y(x)$.\n$y(x)=$\nfind $y(x)$.\n$y(x)=$\nfind the vertical and horizontal asymptotes. (enter your answers as comma - separated lists. if an answer does not exist, enter dne.)\n$y=$

Answer

Explanation:

Step1: Find the domain

The function ( f(x)=\frac{x - 5}{x^{2}} ) is undefined when the denominator ( x^{2}=0 ), i.e., ( x = 0 ). The domain is ( (-\infty,0)\cup(0,\infty) ).

Step2: Find the ( y )-intercept

Set ( x = 0 ). But ( x=0 ) is not in the domain. So ( y )-intercept is DNE.

Step3: Find the ( x )-intercept

Set ( y = 0 ), then ( \frac{x - 5}{x^{2}}=0 ). Since ( x^{2}\neq0 ), we solve ( x-5=0 ), so ( x = 5 ).

Step4: Check symmetry

  • For ( x )-axis symmetry: Replace ( y ) with ( -y ), ( -y=\frac{x - 5}{x^{2}}), ( y=-\frac{x - 5}{x^{2}}\neq\frac{x - 5}{x^{2}}) (except when ( x = 5 )).
  • For ( y )-axis symmetry: Replace ( x ) with ( -x ), ( y=\frac{-x - 5}{x^{2}}\neq\frac{x - 5}{x^{2}}) (except when ( x = 0 )).
  • For origin symmetry: Replace ( x ) with ( -x ) and ( y ) with ( -y ), ( -y=\frac{-x - 5}{x^{2}}), ( y=\frac{x + 5}{x^{2}}\neq\frac{x - 5}{x^{2}}). So the graph has no apparent symmetry.

Step5: Find ( y^{\prime}(x) )

Use the quotient rule ( (\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}} ), where ( u=x - 5), ( u^\prime=1 ), ( v=x^{2}), ( v^\prime = 2x ). ( y^{\prime}(x)=\frac{1\times x^{2}-(x - 5)\times2x}{x^{4}}=\frac{x^{2}-2x^{2}+10x}{x^{4}}=\frac{-x^{2}+10x}{x^{4}}=\frac{-x + 10}{x^{3}})

Step6: Find ( y^{\prime\prime}(x) )

Use the quotient rule again. Let ( u=-x + 10), ( u^\prime=-1 ), ( v=x^{3}), ( v^\prime=3x^{2}). ( y^{\prime\prime}(x)=\frac{-1\times x^{3}-(-x + 10)\times3x^{2}}{x^{6}}=\frac{-x^{3}+3x^{3}-30x^{2}}{x^{6}}=\frac{2x^{3}-30x^{2}}{x^{6}}=\frac{2x - 30}{x^{4}})

Step7: Find the asymptotes

  • Vertical asymptote: ( x = 0 ) (since ( \lim_{x\rightarrow0}\frac{x - 5}{x^{2}}=\pm\infty ))
  • Horizontal asymptote: ( \lim_{x\rightarrow\pm\infty}\frac{x - 5}{x^{2}}=\lim_{x\rightarrow\pm\infty}\frac{1}{x}-\frac{5}{x^{2}}=0 )

Answer:

Domain: ( (-\infty,0)\cup(0,\infty) ) ( y )-intercept: DNE ( x )-intercept: ( 5 ) Symmetry: no apparent symmetry ( y^{\prime}(x)=\frac{-x + 10}{x^{3}} ) ( y^{\prime\prime}(x)=\frac{2x - 30}{x^{4}} ) Vertical asymptote: ( x = 0 ) Horizontal asymptote: ( y = 0 )