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

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. 4) sketch the graph of the function.

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

Answered by bobhans last updated on 11/Apr/21

(1) intercept to y−axis ⇒ x=0 , y=−6 have no intercept x−axis (2) vertical asymptotes x=(1/3) oblique asymptotes y=(1/3)x+((10)/3)

$$\left(\mathrm{1}\right)\:{intercept}\:{to}\:{y}−{axis}\: \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:,\:{y}=−\mathrm{6} \\ $$$${have}\:{no}\:{intercept}\:{x}−{axis}\: \\ $$$$\left(\mathrm{2}\right)\:{vertical}\:{asymptotes}\:{x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${oblique}\:{asymptotes}\: \\ $$$$\:{y}=\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{10}}{\mathrm{3}} \\ $$

Commented by bobhans last updated on 11/Apr/21

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

thank you sir! sir left option( 3)

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

Answered by EDWIN88 last updated on 11/Apr/21

y = ((x^2 +3x+6)/(3x−1)) ⇒y′=(((2x+3)(3x−1)−3(x^2 +3x+6))/((3x−1)^2 )) y′=((3x^2 −2x−21)/((3x−1)^2 )) =0 ; 3x^2 −2x−21=0 x=((2±16)/6) ; → { ((x= 3)),((x=−(7/3))) :} decreasing for −(7/3)<x<3 and increasing for x<−(7/3) ∪ x > 4. → { ((max value for x=−(7/3))),((min value for x= 3)) :} { ((y_(max) = f(−(7/3)))),((y_(min) = f(3))) :} .

$${y}\:=\:\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{6}}{\mathrm{3}{x}−\mathrm{1}}\:\Rightarrow{y}'=\frac{\left(\mathrm{2}{x}+\mathrm{3}\right)\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{6}\right)}{\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${y}'=\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{21}}{\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{0}\:;\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{21}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\pm\mathrm{16}}{\mathrm{6}}\:;\:\rightarrow\begin{cases}{{x}=\:\mathrm{3}}\\{{x}=−\frac{\mathrm{7}}{\mathrm{3}}}\end{cases} \\ $$$${decreasing}\:{for}\:−\frac{\mathrm{7}}{\mathrm{3}}<{x}<\mathrm{3}\:{and}\:{increasing} \\ $$$${for}\:{x}<−\frac{\mathrm{7}}{\mathrm{3}}\:\cup\:{x}\:>\:\mathrm{4}.\:\:\rightarrow\begin{cases}{{max}\:{value}\:{for}\:{x}=−\frac{\mathrm{7}}{\mathrm{3}}}\\{{min}\:{value}\:{for}\:{x}=\:\mathrm{3}}\end{cases} \\ $$$$\:\begin{cases}{{y}_{{max}} \:=\:{f}\left(−\frac{\mathrm{7}}{\mathrm{3}}\right)}\\{{y}_{{min}} =\:{f}\left(\mathrm{3}\right)}\end{cases}\:. \\ $$

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