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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-val




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
Acurvepassesthroughthepoint(1,11)anditsgradientatanypointisax2+b,whereaandbareconstants.Thetangenttothecurveatthepoint(2,16)isparalleltothexaxis.Findi.thevaluesofaandbii.theequationofthecurve
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
dydx=ax2+by=ax33+bx+c11=a3+b+c16=8a3+2b+cgiven4a+b=05=a8a3+(b)5=7a3+4a5=5a3a=3b=4×3=1211=a3+b+c11=112+cc=0soa=3b=12curvey=ax33+bx+cy=x312x
Commented by pieroo last updated on 18/Sep/18
thanks sir
thankssir

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