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Question Number 167567 by pete last updated on 19/Mar/22

Find the coordinates of the point on the  curve y=3x^2 −2x−5, where the tangent  is parallel to the line y−5=8x.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{curve}\:\mathrm{y}=\mathrm{3x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{5},\:\mathrm{where}\:\mathrm{the}\:\mathrm{tangent} \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}−\mathrm{5}=\mathrm{8x}. \\ $$

Commented by greogoury55 last updated on 19/Mar/22

 y′= 8    6x−2=8⇒x=(5/3)     y=3(((25)/9))−2((5/3))−5     y=((25−10−15)/3)=0     the point ((5/3),0)

$$\:{y}'=\:\mathrm{8} \\ $$$$\:\:\mathrm{6}{x}−\mathrm{2}=\mathrm{8}\Rightarrow{x}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\:\:{y}=\mathrm{3}\left(\frac{\mathrm{25}}{\mathrm{9}}\right)−\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)−\mathrm{5} \\ $$$$\:\:\:{y}=\frac{\mathrm{25}−\mathrm{10}−\mathrm{15}}{\mathrm{3}}=\mathrm{0} \\ $$$$\:\:\:{the}\:{point}\:\left(\frac{\mathrm{5}}{\mathrm{3}},\mathrm{0}\right) \\ $$

Commented by pete last updated on 19/Mar/22

Thanks for your time, sir

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time},\:\mathrm{sir} \\ $$

Commented by otchereabdullai@gmail.com last updated on 21/Mar/22

nice!

$$\mathrm{nice}! \\ $$

Answered by som(math1967) last updated on 19/Mar/22

slope of  y=8x+5 is 8  ∴slope of tangent also 8  now y=3x^2 −2x−5   (dy/dx)=6x−2 ∴6x−2=8⇒x=((10)/6)=(5/3)  ∴ y=3×((25)/9) −((10)/3) −5=5−5=0  ∴ point is ((5/3),0)

$${slope}\:{of}\:\:{y}=\mathrm{8}{x}+\mathrm{5}\:{is}\:\mathrm{8} \\ $$$$\therefore{slope}\:{of}\:{tangent}\:{also}\:\mathrm{8} \\ $$$${now}\:{y}=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{5} \\ $$$$\:\frac{{dy}}{{dx}}=\mathrm{6}{x}−\mathrm{2}\:\therefore\mathrm{6}{x}−\mathrm{2}=\mathrm{8}\Rightarrow{x}=\frac{\mathrm{10}}{\mathrm{6}}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\therefore\:{y}=\mathrm{3}×\frac{\mathrm{25}}{\mathrm{9}}\:−\frac{\mathrm{10}}{\mathrm{3}}\:−\mathrm{5}=\mathrm{5}−\mathrm{5}=\mathrm{0} \\ $$$$\therefore\:{point}\:{is}\:\left(\frac{\mathrm{5}}{\mathrm{3}},\mathrm{0}\right) \\ $$$$ \\ $$$$ \\ $$

Commented by pete last updated on 19/Mar/22

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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