Question

find the equation of the tangent and normal to the curve y=3x^2-x^3 where the curve meets the x axis​

Answer 1

Given : curve y=3x^2-x^3

To Find  :  equation of the tangent and normal to the curve

Solution:

y  = 3x²  - x³

curve meet x axis

=> y = 0

3x²  - x³ = 0

=> x² (3 - x) = 0

=> x = 0 , 0 , x = 3

Hence curve touches x axis at x = 0

x = 0  then    y = 0

dy/dx = 6x - 3x² = 0

Hence tangent is  y = 0

and normal is x  = 0

at x = 3   it cuts x axis

y = 0

(3 , 0)

y  = 3x²  - x³

dy/dx = 6x - 3x²  =   18 - 27 = 9

Tangent = y - 0 = 9(x - 3)

=> y = 9x - 27

Normal  y - 0 = (-1/9)(x - 3)

=> 9y = -x + 3

=> x + 9y = 3

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