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Question Number 43923 by pieroo last updated on 17/Sep/18

$$\mathrm{A}\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(1,-11)\mathrm{and}\mathrm{its}$$$$\mathrm{gradient}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{is}\mathbf{a}{\mathrm{x}}^2+\mathbf{b},\mathrm{where}\mathbf{a}\mathrm{and}\mathbf{b}\mathrm{are}$$$$\mathrm{constants}.\mathrm{The}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{curve}\mathrm{at}\mathrm{the}\mathrm{point}$$$$(2,-16)\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathbf{x}-\mathrm{axis}.\mathrm{Find}$$$$\mathrm{i}.\mathrm{the}\mathrm{values}\mathrm{of}\mathbf{a}\mathrm{and}\mathbf{b}$$$$\mathrm{ii}.\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18

$$\frac{dy}{dx}={ax}^2+b$$$$y=\frac{{ax}^3}{3}+bx+c$$$$-11=\frac{a}{3}+b+c$$$$-16=\frac{8a}{3}+2b+c$$$$given4a+b=0$$$$5=\frac{a-8a}{3}+(-b)$$$$5=\frac{-7a}{3}+4a$$$$5=\frac{5a}{3}a=3b=4\times -3=-12$$$$-11=\frac{a}{3}+b+c$$$$-11=1-12+cc=0$$$$soa=3b=-12$$$$curvey=\frac{{ax}^3}{3}+bx+c$$$$y=x^3-12x$$$$$$$$$$

Commented by pieroo last updated on 18/Sep/18

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

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