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Question Number 113952 by bemath last updated on 16/Sep/20

$$findallasymptotesoffunction$$$$f\left(x\right)={\left(\frac{x^2+4x-5}{x^2+x+3}\right)}^{x+5}$$

Commented by MJS_new last updated on 16/Sep/20

$$\mathrm{forgot}\mathrm{the}\mathrm{squares}???$$

Commented by bemath last updated on 16/Sep/20

$$yessir.sorry$$

Commented by MJS_new last updated on 16/Sep/20

$$\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}$$$$f\left(x\right)={\left(\frac{x^2+4x-5}{x^2+x+3}\right)}^{x+5}$$$$\mathrm{I}\mathrm{get}\underset{x\to \pm \mathrm{\infty }}{\lim }f\left(x\right)={\mathrm{e}}^3$$$$\Rightarrow \mathrm{horizontal}\mathrm{asymptote}y={\mathrm{e}}^3$$$$\mathrm{for}\mathrm{the}\mathrm{rest}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}\mathrm{as}\mathrm{we}\mathrm{still}\mathrm{get}\mathrm{some}$$$$\mathrm{real}\mathrm{solutions}\mathrm{for}\frac{x^2+4x-5}{x^2+x+3}<0$$

Commented by MJS_new last updated on 16/Sep/20

$$...\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{vertical}\mathrm{asymptote}x=-5$$

Answered by 1549442205PVT last updated on 16/Sep/20

$$\mathrm{we}\mathrm{find}\mathrm{horizontal}\mathrm{asymptote}:$$$$\mathrm{y}=\underset{\mathrm{x}\to \mathrm{\infty }}{\mathrm{limf}\left(\mathrm{x}\right)}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{(1+\frac{3\mathrm{x}-8}{{\mathrm{x}}^2+\mathrm{x}+3})}^{\mathrm{x}+5}$$$$=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }(1+\frac{1}{\frac{{\mathrm{x}}^2+\mathrm{x}+3}{3\mathrm{x}-8}})\frac{{\mathrm{x}}^2+\mathrm{x}+3}{3\mathrm{x}-8}.\left(\frac{(3\mathrm{x}-8)(\mathrm{x}+5)}{{\mathrm{x}}^2+\mathrm{x}+3}\right)$$$$={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{{3\mathrm{x}}^2+7\mathrm{x}-40}{{\mathrm{x}}^2+\mathrm{x}+3}}={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }(3+\frac{4\mathrm{x}-49}{{\mathrm{x}}^2+\mathrm{x}+3})}$$$$={\mathrm{e}}^3\mathrm{since}\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }(3+\frac{4\mathrm{x}-49}{{\mathrm{x}}^2+\mathrm{x}+3})=3$$$$\mathrm{Therefore}\mathrm{graph}\mathrm{of}\mathrm{function}$$$$f\left(x\right)={\left(\frac{x^2+4x-5}{x^2+x+3}\right)}^{x+5}\mathrm{has}\mathrm{the}$$$$\mathrm{horizontal}\mathrm{asymptote}\mathrm{y}={\mathrm{e}}^3$$

Answered by bobhans last updated on 16/Sep/20

$$\left(1\right)horizontalasymptotes$$$$y=\underset{x\to \pm \mathrm{\infty }}{\lim }f\left(x\right)=\underset{x\to \pm \mathrm{\infty }}{\lim }{\left(\frac{x^2+4x-5}{x^2+x+3}\right)}^{x+5}$$$$=e^{\underset{x\to \pm \mathrm{\infty }}{\lim }(\frac{x^2+4x-5}{x^2+x+3}-1).(x+5)}$$$$=e^{\underset{x\to \pm \mathrm{\infty }}{\lim }\left(\frac{(3x-8)(x+5)}{x^2+x+3}\right)}=e^3$$$$\left(2\right)verticalasymptotes$$$$\underset{x\to a}{\lim }f\left(x\right)=\pm \mathrm{\infty }.weapllyformula$$$$Iff\left(x\right)={\left(\frac{h\left(x\right)}{g\left(x\right)}\right)}^{u\left(x\right)};verticalasymptotes$$$$wegetfromh\left(x\right)=0,inthiscase$$$$h\left(x\right)=x^2+4x-5=(x+5)(x-1)=0$$$$sotheverticalasymptotesisx=-5\wedge x=1$$

Commented by MJS_new last updated on 16/Sep/20

$$\mathrm{but}f\left(1\right)=0\mathrm{is}\mathrm{defined},\mathrm{so}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{asymptote}$$

Commented by bobhans last updated on 16/Sep/20

$$howaboutf(-5)sir?$$$$oof(-5)=0^{0}sir.yesx=-5is$$$$verticalasymptotes$$

Commented by bobhans last updated on 16/Sep/20

Commented by bobhans last updated on 16/Sep/20

$$iapplygeogebra.thisgraph$$

Commented by MJS_new last updated on 16/Sep/20

$$f(-5)={\left(\frac{0}{17}\right)}^0=0^0\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{defined}$$$$\mathrm{many}\mathrm{graphic}\mathrm{calculators}\mathrm{use}{0}^0=1\mathrm{and}\mathrm{in}$$$$\mathrm{this}\mathrm{case}\underset{x\to -5}{\lim }f\left(x\right)=1\Rightarrow \mathrm{different}\mathrm{situation}$$$$\mathrm{from}f\left(1\right)=0$$$$\mathrm{we}\mathrm{would}\mathrm{also}\mathrm{not}\mathrm{say}y=x^2\mathrm{has}\mathrm{an}\mathrm{asymptote}$$$$y=0,\mathrm{would}\mathrm{we}?$$

Commented by bemath last updated on 16/Sep/20

$$sohorizontalasymptotesisy=e^3$$$$andverticalasymptoteisx=-5.$$$$thatrightsir?$$

Commented by MJS_new last updated on 16/Sep/20

$$\mathrm{yes}\mathrm{in}\mathrm{my}\mathrm{opinion}.$$

Commented by bemath last updated on 16/Sep/20

$$thankyousir.iagree$$

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