Answer :

Given equation:


y = x3 + 2x - 4


On differentiating with respect to x


…(1)


It is to the line x + 14y + 3 = 0


The slope of the line is


The slope of the tangent is


m1 = 14


…(2)


Equating equ(1) and equ(2)


3x + 2 = 14


3x = 12


x = 4


y = (4)3 + 2(4) - 4


= 64 + 8 - 4


= 68.


Point of contact is (x,y) = (4,68)


Equation of the tangent is y−y1 = m(x−x1)


y−68 = 14(x−4)


y−68 = 14x−64


14x - y + 4 = 0


This is the required equation.


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