Consider-the-function-y-x-2-3x-6-3x-1-1-Determine-the-intercept-of-the-function-2-Find-the-asymptotes-if-they-exist-3-find-the-turning-point-and-determine-the-type-of-turnin-point-they-are

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Question Number 138219 by otchereabdullai@gmail.com last updated on 11/Apr/21

$$\mathrm{Consider}\mathrm{the}\mathrm{function}\mathrm{y}=\frac{{\mathrm{x}}^2+3\mathrm{x}+6}{3\mathrm{x}-1}$$$$1)\mathrm{Determine}\mathrm{the}\mathrm{intercept}\mathrm{of}\mathrm{the}\mathrm{function}$$$$2)\mathrm{Find}\mathrm{the}\mathrm{asymptotes}\mathrm{if}\mathrm{they}\mathrm{exist}$$$$3)\mathrm{find}\mathrm{the}\mathrm{turning}\mathrm{point}\mathrm{and}$$$$\mathrm{determine}\mathrm{the}\mathrm{type}\mathrm{of}\mathrm{turnin}\mathrm{point}\mathrm{they}$$$$\mathrm{are}.$$$$4)\mathrm{sketch}\mathrm{the}\mathrm{graph}\mathrm{of}\mathrm{the}\mathrm{function}.$$

Answered by bobhans last updated on 11/Apr/21

$$\left(1\right)intercepttoy-axis$$$$\Rightarrow x=0,y=-6$$$$havenointerceptx-axis$$$$\left(2\right)verticalasymptotesx=\frac{1}{3}$$$$obliqueasymptotes$$$$y=\frac{1}{3}x+\frac{10}{3}$$

Commented by bobhans last updated on 11/Apr/21

Commented by otchereabdullai@gmail.com last updated on 11/Apr/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!\mathrm{sir}\mathrm{left}\mathrm{option}\left(3\right)$$

Answered by EDWIN88 last updated on 11/Apr/21

$$y=\frac{x^2+3x+6}{3x-1}\Rightarrow y^{\prime }=\frac{(2x+3)(3x-1)-3(x^2+3x+6)}{{(3x-1)}^2}$$$$y^{\prime }=\frac{3x^2-2x-21}{{(3x-1)}^2}=0;3x^2-2x-21=0$$$$x=\frac{2\pm 16}{6};\to \{\begin{array}{l}x=3\\ x=-\frac{7}{3}\end{array}$$$$decreasingfor-\frac{7}{3}<x<3andincreasing$$$$forx<-\frac{7}{3}\cup x>4.\to \{\begin{array}{l}maxvalueforx=-\frac{7}{3}\\ minvalueforx=3\end{array}$$$$\{\begin{array}{l}{y}_{max}=f(-\frac{7}{3})\\ y_{min}=f\left(3\right)\end{array}.$$

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