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Question Number 107766 by mohammad17 last updated on 12/Aug/20

Answered by bobhans last updated on 12/Aug/20

$$\frac{\mathcal{B}\mathrm{obhans}}{\mathrm{\Pi }}$$$$\mathrm{y}=\frac{{\mathrm{x}}^2-4}{{\mathrm{x}}^2-5\mathrm{x}+6}=\frac{{\mathrm{x}}^2-5\mathrm{x}+6+5\mathrm{x}-10}{{\mathrm{x}}^2-5\mathrm{x}+6}$$$$\mathrm{y}=\frac{{\mathrm{x}}^2-5\mathrm{x}+6}{{\mathrm{x}}^2-5\mathrm{x}+6}+\frac{5\mathrm{x}-10}{{\mathrm{x}}^2-5\mathrm{x}+6}$$$$\mathrm{y}=1+\frac{5\mathrm{x}-10}{{\mathrm{x}}^2-5\mathrm{x}+6}$$$$\mathrm{oblique}\mathrm{asymptotes}\mathrm{is}\mathrm{y}=5\mathrm{x}-10$$

Commented by mohammad17 last updated on 12/Aug/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 12/Aug/20

$$\mathrm{no}\mathrm{correct}\mathrm{sir}.$$

Answered by ajfour last updated on 12/Aug/20

$$noslantasymptote.$$

Answered by mathmax by abdo last updated on 12/Aug/20

$${\lim }_{\mathrm{x}\to \mathrm{\infty }}\mathrm{y}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2}=1\Rightarrow \mathrm{y}=1\mathrm{is}\mathrm{assymtote}\mathrm{to}\mathrm{graph}$$$$\mathrm{but}\mathrm{not}\mathrm{oblique}$$

Answered by mathmax by abdo last updated on 12/Aug/20

$${\mathrm{x}}^2-5\mathrm{x}+6=0\to \mathrm{\Delta }=25-24=1\Rightarrow {\mathrm{x}}_1=\frac{5+1}{2}=3\mathrm{and}{\mathrm{x}}_2=\frac{5-1}{2}=2$$$$\mathrm{so}\mathrm{the}\mathrm{line}\mathrm{x}=3\mathrm{and}\mathrm{x}=2\mathrm{are}\mathrm{assyptote}\mathrm{to}{\mathrm{C}}_{\mathrm{f}}\mathrm{but}\mathrm{not}\mathrm{oblique}$$

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