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the equation of a tangent to the curve of f(x)=ax3 + bx is y-x-4=0 if the point of contact is (-1;3). calculate a and b
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im not sure i just want to know how to get to the answer what are the steps
first find first derivative of f(x) i.e f'(x)=3ax^2 +b and since the tangent is at (-1,3) therfore the f'(x)=3a+b at(-1,3) now slope of the tangent line is 1 thus 3a+b=1 further the curve passes through (-1,3) then 3=-a-b thus solving the two eq. as obtained we have a=2 and b=-5
the slope of the tangent to a curve is given by f'(x)
sorry can you please explain how you got to the 2 equations
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