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Question Number 79883 by MJS last updated on 29/Jan/20

$$\mathrm{to}\mathrm{Sir}\mathrm{Jagoll}\left(andofcourseeverybodyelse\right)$$$$$$$$\left(1\right)$$$$y=\frac{x^2-x-6}{x^2-3x-4}=$$$$=\frac{(x-3)(x+2)}{(x-4)(x+1)}\Rightarrow$$$$\Rightarrow \{\begin{array}{l}\mathrm{zeros}\mathrm{at}x=-2;x=3\\ \mathrm{vertical}\mathrm{asymptotes}\mathrm{at}x=-1;x=4\end{array}$$$$\mathrm{defined}\mathrm{for}x\in \mathbb{R}\mathrm{\setminus }\{-1;4\}$$$$\mathrm{range}:\mathrm{transforming}y=\frac{x^2-x-6}{x^2-3x-4}\mathrm{to}$$$$x^2-\frac{3y-1}{y-1}x-\frac{2(2y-3)}{y-1}=0$$$$D=\frac{25y^2-46y+25}{4{(y-1)}^2}>0\mathrm{\forall }x\in \mathbb{R}\Rightarrow$$$$\Rightarrow \{\begin{array}{l}\mathrm{range}=\mathbb{R}\\ \mathrm{horizontal}\mathrm{asymptote}\mathrm{at}y=1(\ast )\end{array}$$$$(\ast ){\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{mean}y=1\mathrm{is}\mathrm{not}\mathrm{within}\mathrm{range}!!!$$$$x=1\Rightarrow y=1$$$$y^{\prime }=-\frac{2(x^2-2x+7)}{{(x-4)}^2{(x+1)}^2}\mathrm{no}\mathrm{real}\mathrm{zeros}\Rightarrow$$$$\Rightarrow \mathrm{no}\mathrm{local}\mathrm{extremes}$$$$y^{″}=\frac{4(x^3-3x^2+21x-25)}{{(x-4)}^3{(x+1)}^3}=0\mathrm{at}x\approx 1.33131\Rightarrow$$$$\Rightarrow \mathrm{turning}\mathrm{point}$$$$$$$$\left(2\right)$$$$y=\frac{x^2-x-6}{x^2-3x+4}=$$$$=\frac{(x-3)(x+2)}{x^2-3x+4};x^2-3x+4=0\mathrm{no}\mathrm{real}\mathrm{zeros}\Rightarrow$$$$\Rightarrow \{\begin{array}{l}\mathrm{zeros}\mathrm{at}x=-2;x=3\\ \mathrm{no}\mathrm{vertical}\mathrm{asymptote}\end{array}$$$$\mathrm{defined}\mathrm{for}x\in \mathbb{R}$$$$\mathrm{range}:\mathrm{transforming}y=\frac{x^2-x-6}{x^2-3x+4}\mathrm{to}$$$$x^2-\frac{3y-1}{y-1}x+\frac{2(2y+3)}{y-1}=0$$$$D=-\frac{7y^2+14y-25}{4{(y-1)}^2}⩾0\mathrm{for}-1-\frac{4\sqrt{14}}{7}⩽y⩽-1+\frac{4\sqrt{14}}{7}\Rightarrow$$$$\Rightarrow \{\begin{array}{l}\mathrm{range}=[-1-\frac{4\sqrt{14}}{7};-1+\frac{4\sqrt{14}}{7}]\\ \mathrm{horizontal}\mathrm{asymptote}\mathrm{at}y=1(\ast )\end{array}$$$$(\ast ){\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{mean}y=1\mathrm{is}\mathrm{not}\mathrm{within}\mathrm{range}!!!$$$$x=5\Rightarrow y=1$$$$y^{\prime }=-\frac{2(x^2-10x+11)}{{(x^2-3x+4)}^2}=0\mathrm{at}x=5\pm \sqrt{14}$$$$y^{{}^{″}}=\frac{4(x^3-15x^2+33x-13)}{{(x^2-3x+4)}^3}$$$$y^{″}\{\begin{array}{l}>0\mathrm{at}x=5-\sqrt{14}\Rightarrow \mathrm{local}\mathrm{minimum}\\ <0\mathrm{at}x=5+\sqrt{14}\Rightarrow \mathrm{local}\mathrm{maximum}\\ =0\mathrm{at}\{\begin{array}{l}x\approx .506699\\ x\approx 2.06421\\ x\approx 12.4291\end{array}\Rightarrow 3\mathrm{turning}\mathrm{points}\end{array}$$$$$$$$\left(3\right)$$$$y=\frac{x^2-x+6}{x^2-3x-4}=$$$$=\frac{x^2-x+6}{(x-4)(x+1)}\Rightarrow$$$$\Rightarrow \{\begin{array}{l}\mathrm{no}\mathrm{real}\mathrm{zeros}\\ \mathrm{vertical}\mathrm{asymptotes}\mathrm{at}x=-1;x=4\end{array}$$$$\mathrm{defined}\mathrm{for}x\in \mathbb{R}\mathrm{\setminus }\{-1;4\}$$$$\mathrm{range}:\mathrm{transforming}y=\frac{x^2-x+6}{x^2-3x-4}\mathrm{to}$$$$x^2-\frac{3y-1}{y-1}x-\frac{2(2y+3)}{y-1}=0$$$$D=\frac{25y^2+2y-23}{4{(y-1)}^2}<0\mathrm{for}-1<y<\frac{23}{25}\Rightarrow$$$$\Rightarrow \{\begin{array}{l}\mathrm{range}=\mathbb{R}\mathrm{\setminus }]-1;\frac{23}{25}[\\ \mathrm{horizontal}\mathrm{asymptote}\mathrm{at}y=1(\ast )\end{array}$$$$(\ast ){\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{mean}y=1\mathrm{is}\mathrm{not}\mathrm{within}\mathrm{range}!!!$$$$x=-5\Rightarrow y=1$$$$y^{\prime }=-\frac{2(x^2+10x-11)}{{(x-4)}^2{(x+1)}^2}=0\mathrm{at}x=-11;x=1$$$$y^{″}=\frac{4(x^3+15x^2-33x+53)}{{(x-4)}^3{(x+1)}^3}$$$$y^{″}\{\begin{array}{l}>0\mathrm{at}x=-11\Rightarrow \mathrm{local}\mathrm{minimum}\\ <0\mathrm{at}x=1\Rightarrow \mathrm{local}\mathrm{maximum}\\ =0\mathrm{at}x\approx -17.1098\Rightarrow \mathrm{turning}\mathrm{point}\end{array}$$

Commented by TawaTawa last updated on 29/Jan/20

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir},\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by jagoll last updated on 29/Jan/20

$$\mathrm{waw}...\mathrm{thank}\mathrm{you}\mathrm{mister}.\mathrm{i}\mathrm{was}$$$$\mathrm{confused},\mathrm{my}\mathrm{teacher}\mathrm{said}\mathrm{that}\mathrm{horizontal}$$$$\mathrm{asymptotes}\mathrm{were}\mathrm{not}\mathrm{include}\mathrm{in}\mathrm{the}$$$$\mathrm{range}.$$

Commented by jagoll last updated on 29/Jan/20

$$\mathrm{mr}.\mathrm{if}\mathrm{given}\mathrm{function}$$$$\mathrm{y}=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}},\mathrm{is}\mathrm{it}\mathrm{true}\mathrm{that}\mathrm{the}\mathrm{range}$$$$\mathrm{is}\mathrm{not}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{horizontal}$$$$\mathrm{asymptote}?$$$${\mathrm{R}}_{\mathrm{f}}:\mathrm{y}\ne \frac{\mathrm{a}}{\mathrm{c}}?$$

Commented by john santu last updated on 29/Jan/20

$$\mathrm{for}\mathrm{example}$$$$\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\frac{2\mathrm{x}-1}{\mathrm{x}+3}$$$$\mathrm{domain}\mathrm{x}\ne -3$$$$\mathrm{if}\mathrm{y}=2\Rightarrow 2=\frac{2\mathrm{x}-1}{\mathrm{x}+3}$$$$2\mathrm{x}+6=2\mathrm{x}-1\Rightarrow 6=-1??$$$$\mathrm{it}\mathrm{true}\mathrm{for}\mathrm{function}\mathrm{y}=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}}$$$$\mathrm{range}\mathrm{y}\ne \frac{\mathrm{a}}{\mathrm{c}}$$

Commented by jagoll last updated on 29/Jan/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented by MJS last updated on 29/Jan/20

$$y=\frac{ax+b}{cx+d}\Rightarrow \{\begin{array}{l}\mathrm{zero}\mathrm{at}x=-\frac{b}{a}\\ \mathrm{vertical}\mathrm{asymptote}\mathrm{at}x=-\frac{d}{c}\end{array}$$$$\mathrm{defined}\mathrm{for}x\in \mathbb{R}\mathrm{\setminus }\{-\frac{d}{c}\}$$$$y=\frac{ax+b}{cx+d}\iff x=-\frac{dy-b}{cy-a}\Rightarrow \mathrm{range}=\mathbb{R}\mathrm{\setminus }\left\{\frac{a}{c}\right\}$$$$\mathrm{horizontal}\mathrm{asymptote}\mathrm{at}y=\frac{a}{c}$$$$\mathrm{so}\mathrm{you}\mathrm{are}\mathrm{right}$$

Commented by jagoll last updated on 29/Jan/20

$$\mathrm{sir}\mathrm{your}\mathrm{notation}\mathrm{is}$$$$]14-4\sqrt{6};14+4\sqrt{6}[$$$$\mathrm{it}\mathrm{same}\mathrm{as}\mathrm{y}⩽14-4\sqrt{6}\vee \mathrm{y}⩾14+4\sqrt{6}$$$$?$$

Commented by MJS last updated on 29/Jan/20

$$\mathrm{depends}...$$$$y\in [a;b]\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}a⩽y⩽b$$$$\Rightarrow y\in \mathbb{R}\mathrm{\setminus }[a;b]\mathrm{means}y<a\vee y>b$$$$y\in ]a;b[\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}a<y<b$$$$\Rightarrow y\in \mathbb{R}\mathrm{\setminus }]a;b[\mathrm{means}y⩽a\vee y⩾b$$

Commented by jagoll last updated on 29/Jan/20

$$\mathrm{thanks}\mathrm{you}\mathrm{mister}$$

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