Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Ex 6.3, 7 Find points at which the tangent to the curve π¦=π₯^3β3π₯^2β9π₯+7 is parallel to the x-axisEquation of Curve is π¦=π₯^3β3π₯^2β9π₯+7 Differentiating w.r.t. π₯ ππ¦/ππ₯=π(π₯^3β 3π₯^2 β 9π₯ + 7)/ππ₯ ππ¦/ππ₯=3π₯^2β6π₯β9+0 ππ¦/ππ₯=3π₯^2β6π₯β9 ππ¦/ππ₯=3(π₯^2β2π₯β3) ππ¦/ππ₯=3(π₯^2β3π₯+π₯β3) ππ¦/ππ₯=3(π₯(π₯β3)+1(π₯β3)) ππ¦/ππ₯=3(π₯+1)(π₯β3) Given tangent to the curve is parallel to the π₯βππ₯ππ  i.e. the Slope of tangent = Slope of π₯βππ₯ππ  ππ/ππ=π 3(π₯+1)(π₯β3)=0 (π₯+1)(π₯β3)=0 Thus π₯=β1 & π₯=3 When π=βπ π¦=π₯^3β3π₯^2β9π₯+7 =(β1)^3β3(β1)^2β9(β1)+7 =β1β3+9+7 =12 Point is (β1 , 12) When π=π π¦=π₯^3β3π₯^2β9π₯+7 =(3)^3β3(3)^2β9(3)+7 =27β27β27+7 =β 20 Point is (3 , β20) Hence , the tangent to the Curve is parallel to the π₯βππ₯ππ  at (βπ , ππ) & (π , βππ)