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 Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 43923 by pieroo last updated on 17/Sep/18 Acurvepassesthroughthepoint(1,−11)anditsgradientatanypointisax2+b,whereaandbareconstants.Thetangenttothecurveatthepoint(2,−16)isparalleltothex−axis.Findi.thevaluesofaandbii.theequationofthecurve Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18 dydx=ax2+by=ax33+bx+c−11=a3+b+c−16=8a3+2b+cgiven4a+b=05=a−8a3+(−b)5=−7a3+4a5=5a3a=3b=4×−3=−12−11=a3+b+c−11=1−12+cc=0soa=3b=−12curvey=ax33+bx+cy=x3−12x Commented by pieroo last updated on 18/Sep/18 thankssir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-109459Next Next post: Question-109460 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.