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

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

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

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

(dy/dx)=ax^2 +b y=((ax^3 )/3)+bx+c −11=(a/3)+b+c −16=((8a)/3)+2b+c given 4a+b=0 5=((a−8a)/3)+(−b) 5=((−7a)/3)+4a 5=((5a)/3) a=3 b=4×−3=−12 −11=(a/3)+b+c −11=1−12+c c=0 so a=3 b=−12 curve y=((ax^3 )/3)+bx+c y=x^3 −12x

$$\frac{{dy}}{{dx}}={ax}^{\mathrm{2}} +{b}\: \\ $$$${y}=\frac{{ax}^{\mathrm{3}} }{\mathrm{3}}+{bx}+{c} \\ $$$$−\mathrm{11}=\frac{{a}}{\mathrm{3}}+{b}+{c} \\ $$$$−\mathrm{16}=\frac{\mathrm{8}{a}}{\mathrm{3}}+\mathrm{2}{b}+{c} \\ $$$${given}\:\:\mathrm{4}{a}+{b}=\mathrm{0} \\ $$$$\mathrm{5}=\frac{{a}−\mathrm{8}{a}}{\mathrm{3}}+\left(−{b}\right) \\ $$$$\mathrm{5}=\frac{−\mathrm{7}{a}}{\mathrm{3}}+\mathrm{4}{a} \\ $$$$\mathrm{5}=\frac{\mathrm{5}{a}}{\mathrm{3}}\:\:\:\:{a}=\mathrm{3}\:\:\:\:\:\:{b}=\mathrm{4}×−\mathrm{3}=−\mathrm{12} \\ $$$$−\mathrm{11}=\frac{{a}}{\mathrm{3}}+{b}+{c} \\ $$$$−\mathrm{11}=\mathrm{1}−\mathrm{12}+{c}\:\:\:\:{c}=\mathrm{0} \\ $$$${so}\:{a}=\mathrm{3}\:\:\:\:{b}=−\mathrm{12} \\ $$$${curve}\:\:\:{y}=\frac{{ax}^{\mathrm{3}} }{\mathrm{3}}+{bx}+{c} \\ $$$${y}={x}^{\mathrm{3}} −\mathrm{12}{x} \\ $$$$ \\ $$$$ \\ $$

Commented by pieroo last updated on 18/Sep/18

thanks sir

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

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