Question

4 4. Given the function f(x)- + 3)2 (a) Find any vertical and horizontal asymptotes of the function f(x) (b) Sketch the graph of f(x) 10 10 (c) Find a such that m f(x) 00. a (d) lim /(r) irm

Answer 1

Solution : ( a )

$\mathrm{Vertical\:asymptotes\:of\:}\frac{2}{\left(x+3\right)^2}-4:$

$\mathrm{Go\:over\:every\:undefined\:point\:}x=a\mathrm{\:and\:check\:if\:at\:least\:one\:of\:the\:following\:statements\:is\:true:}$

$\lim _{x\to a^-}f\left(x\right)=\pm \infty$

$\lim _{x\to a^+}f\left(x\right)=\pm \infty$

$\mathrm{Find\:undefined\:\left(singularity\right)\:points}:$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{2}{\left(x+3\right)^2}-4\mathrm{\:and\:compare\:to\:zero}$

$\mathrm{Solve\:}\:\left(x+3\right)^2=0:$

$\left(x+3\right)^2=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

$x+3=0$

$\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}$

$x+3-3=0-3$

$\mathrm{Simplify}$

$x=-3$

$\mathrm{The\:final\:solution\:to\:the\:quadratic\:equation\:is:}$

$x=-3$

$\mathrm{The\:following\:points\:are\:undefined}$

$x=-3$

$\mathrm{The\:vertical\:asymptotes\:are\quad\:}x=-3$

$\mathrm{and}$

$\mathrm{Horizontal\:Asymptotes\:of\:}\frac{2}{\left(x+3\right)^2}-4:$

$\mathrm{Check\:if\:at\:}x\to \pm \infty \mathrm{\:the\:function\:}y=\frac{2}{\left(x+3\right)^2}-4$

$\mathrm{\:behaves\:as\:a\:line,\:}y=mx+b$

$\mathrm{Find\:an\:asymptote\:for\:}x\to -\infty \::$

$\mathrm{Compute}\:\lim _{x\to -\infty \:}\frac{f\left(x\right)}{x}\:\mathrm{to\:find\:m:}$

$\lim _{x\to -\infty \:}\frac{f\left(x\right)}{x}=\lim _{x\to -\infty \:}\frac{\frac{2}{\left(x+3\right)^2}-4}{x}\quad :$

$\lim _{x\to \:-\infty \:}\left(\frac{\frac{2}{\left(x+3\right)^2}-4}{x}\right)=0$

$\mathrm{Now,}$

$\mathrm{Compute}\:\lim _{x\to -\infty \:}f\left(x\right)-mx\:\mathrm{to\:find\:b:}$

$\lim _{x\to -\infty \:}f\left(x\right)-mx=\lim _{x\to -\infty \:}\frac{2}{\left(x+3\right)^2}-4-0x\quad :$

$\lim _{x\to \:-\infty \:}\left(\frac{2}{\left(x+3\right)^2}-4-0*x\right)=-4$

$y=-4$

$\mathrm{So,\quad\:}\mathrm{\:an\:asymptote\:for\:}x\to -\infty\mathrm{\: \:is\:}\quad y=-4$

$\mathrm{and}$

$\mathrm{Find\:an\:asymptote\:for\:}x\to \infty \::\quad$

$\mathrm{Compute}\:\lim _{x\to \infty \:}\frac{f\left(x\right)}{x}\:\mathrm{to\:find\:m:}$

$\lim _{x\to \infty \:}\frac{f\left(x\right)}{x}=\lim _{x\to \infty \:}\frac{\frac{2}{\left(x+3\right)^2}-4}{x}\quad :$

$\lim _{x\to \infty \:}\left(\frac{\frac{2}{\left(x+3\right)^2}-4}{x}\right)=0$

$\mathrm{Now,}$

$\mathrm{Compute}\:\lim _{x\to \infty \:}f\left(x\right)-mx\:\mathrm{to\:find\:b:}$

$\lim _{x\to \infty \:}f\left(x\right)-mx=\lim _{x\to \infty \:}\frac{2}{\left(x+3\right)^2}-4-0x\quad :$

$\lim _{x\to \infty \:}\left(\frac{2}{\left(x+3\right)^2}-4-0*x\right)=-4$

$y=-4$

$\mathrm{So,\quad\:}\mathrm{\:an\:asymptote\:for\:}x\to \infty\mathrm{\: \:is\:}\quad y=-4$

$\mathrm{The\:horizontal\:asymptote\:is\quad\:}y=-4$

$\mathrm{Hence,}$

$\mathrm{Vertical\:asymptotes\:of\:the\:function\:of\:}f(x):\quad\:x=-3,$

$\mathrm{Horizontal\:asymptotes\:of\:the\:function\:of\:}f(x):\quad\:y=-4$

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Solution : ( d )

$\lim _{x\to \infty \:}f(x)=\lim _{x\to \infty \:}\left(\frac{2}{\left(x+3\right)^2}-4\right)$

Steps:

lim (-アー4

$\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)$

With the exception of indeterminate form

$=\lim _{x\to \infty \:}\left(\frac{2}{\left(x+3\right)^2}\right)-\lim _{x\to \infty \:}\left(4\right)$

$=2* \lim \:_{x\to \infty \:}\left(\frac{1}{\left(x+3\right)^2}\right)-\lim _{x\to \infty \:}\left(4\right)$

3 2 T+0

1 A

$=2* \frac{1}{\infty \:}-4$

$=0-4$

$=-4$

$\mathrm{Hence,}$

$\lim _{x\to \infty \:}f(x)=-4$

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Solution : ( e )

$\lim _{x\to -\infty \:}f(x)=\lim _{x\to -\infty \:}\left(\frac{2}{\left(x+3\right)^2}-4\right)$

Steps:

$\lim _{x\to -\infty \:}\left(\frac{2}{\left(x+3\right)^2}-4\right)$

$=\lim _{x\to \:-\infty \:}\left(\frac{2}{x^2+6x+9}-4\right)$

$\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)$

With the exception of indeterminate form

$=\lim _{x\to \:-\infty \:}\left(\frac{2}{x^2+6x+9}\right)-\lim _{x\to \:-\infty \:}\left(4\right)$

$=2*\lim \:_{x\to \:-\infty \:}\left(\frac{1}{x^2+6x+9}\right)-\lim _{x\to \:-\infty \:}\left(4\right)$

$=2*\frac{\lim _{x\to \:-\infty \:}\left(1\right)}{\lim _{x\to \:-\infty \:}\left(x^2+6x+9\right)}-\lim _{x\to \:-\infty \:}\left(4\right)$

$=2*\frac{1}{\infty \:}-4$

$=2*0-4$

$=0-4$

$=-4$

$\mathrm{Hence,}$

$\lim _{x\to -\infty \:}f(x)=-4$