read the entry directions carefully! the purpose of this question is to do the preliminary work to make a…

read the entry directions carefully! the purpose of this question is to do the preliminary work to make a well - labeled sketch of the graph of the function $f(x)=\frac{2x^{2}+22x + 65}{x^{2}+11x - 102}$. first, find the critical values (there should be three): two of the critical values should be associated with asymptotes of this function. the remaining critical value should be a local maximum. what is the local maximum for this graph: $x = $ $y = $ in addition to the vertical asymptotes, there is another asymptote. give the equation for this asymptote: hint: what is $lim_{x\rightarrowpminfty}f(x)$? finally, give the range for the function $f$. range = in your written work for this problem, please make a well - labled sketch of the function $y = f(x)$.

read the entry directions carefully! the purpose of this question is to do the preliminary work to make a well - labeled sketch of the graph of the function $f(x)=\frac{2x^{2}+22x + 65}{x^{2}+11x - 102}$. first, find the critical values (there should be three): two of the critical values should be associated with asymptotes of this function. the remaining critical value should be a local maximum. what is the local maximum for this graph: $x = $ $y = $ in addition to the vertical asymptotes, there is another asymptote. give the equation for this asymptote: hint: what is $lim_{x\rightarrowpminfty}f(x)$? finally, give the range for the function $f$. range = in your written work for this problem, please make a well - labled sketch of the function $y = f(x)$.

Answer

Explanation:

Step1: Find vertical asymptotes

Set the denominator $x^{2}+11x - 102=(x + 17)(x-6)=0$. So the vertical - asymptotes occur at $x=-17$ and $x = 6$. These are two of the critical values.

Step2: Find the horizontal asymptote

Since the degrees of the numerator and denominator are the same (both degree 2), we find the limit as $x\to\pm\infty$. $\lim_{x\to\pm\infty}\frac{2x^{2}+22x + 65}{x^{2}+11x - 102}=\lim_{x\to\pm\infty}\frac{2+\frac{22}{x}+\frac{65}{x^{2}}}{1+\frac{11}{x}-\frac{102}{x^{2}}}=2$. The equation of the horizontal asymptote is $y = 2$.

Step3: Find the derivative of the function

Use the quotient rule. If $f(x)=\frac{u(x)}{v(x)}$ where $u(x)=2x^{2}+22x + 65$ and $v(x)=x^{2}+11x - 102$, then $u^\prime(x)=4x + 22$ and $v^\prime(x)=2x + 11$. The quotient rule states that $f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v^{2}(x)}$. [ \begin{align*} f^\prime(x)&=\frac{(4x + 22)(x^{2}+11x - 102)-(2x^{2}+22x + 65)(2x + 11)}{(x^{2}+11x - 102)^{2}}\ &=\frac{(4x^{3}+44x^{2}-408x+22x^{2}+242x-2244)-(4x^{3}+22x^{2}+44x^{2}+242x+130x + 715)}{(x^{2}+11x - 102)^{2}}\ &=\frac{4x^{3}+66x^{2}-166x-2244-(4x^{3}+66x^{2}+372x + 715)}{(x^{2}+11x - 102)^{2}}\ &=\frac{4x^{3}+66x^{2}-166x-2244 - 4x^{3}-66x^{2}-372x - 715}{(x^{2}+11x - 102)^{2}}\ &=\frac{-538x-2959}{(x^{2}+11x - 102)^{2}} \end{align*} ] Set $f^\prime(x)=0$, then $-538x-2959 = 0$, so $x=-\frac{2959}{538}\approx - 5.5$.

Step4: Find the local - maximum value

Substitute $x =-\frac{2959}{538}$ into $f(x)$: [ \begin{align*} f\left(-\frac{2959}{538}\right)&=\frac{2\left(-\frac{2959}{538}\right)^{2}+22\left(-\frac{2959}{538}\right)+65}{\left(-\frac{2959}{538}\right)^{2}+11\left(-\frac{2959}{538}\right)-102}\ \end{align*} ] [ \begin{align*} f\left(-\frac{2959}{538}\right)&=\frac{2\times\frac{8755681}{289444}- \frac{65098}{538}+65}{\frac{8755681}{289444}-\frac{32549}{538}-102}\ &=\frac{\frac{17511362}{289444}-\frac{3475394}{289444}+\frac{18813860}{289444}}{\frac{8755681}{289444}-\frac{17376446}{289444}-\frac{29523288}{289444}}\ &=\frac{\frac{17511362-3475394 + 18813860}{289444}}{\frac{8755681-17376446-29523288}{289444}}\ &=\frac{32850828}{-38144053}\approx - 0.86 \end{align*} ]

Step5: Find the range

The function has a horizontal asymptote at $y = 2$ and a local - maximum. The range is $(-\infty,-0.86]\cup(2,\infty)$.

Answer:

  • Critical values (vertical asymptotes): $x=-17,x = 6$
  • Local maximum: $x\approx - 5.5,y\approx - 0.86$
  • Non - vertical asymptote: $y = 2$
  • Range: $(-\infty,-0.86]\cup(2,\infty)$