Question

find the equation of tangent to the curve y = 4x^3 -3x+5 which are perpendicular to the line 9y + x + 3 = 0

Answer 1

y = 4x^3 -3x+5 
=> dy/dx = 12x^2 - 3 

and 9y + x + 3 = 0
     => 9dy/dx + 1 = 0;
     => dy/dx = -1/9
so perpendicular slope would be 9
so 12x^2 - 3 = 9
=> 12x^2 = 12;
=> x^2 = 1;
=> x = 1,-1
therefore y = 9x + c
putting x = 1 in y = 4x^3 -3x+5 
we get y = 6 and y = 4

so equations are y = 9x -3 
and y = 9x +13

Answer 2

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