How to find the equation of the tangent line to the curve f(x)=(x3−3x+1)(x+2) at the point (1, -3)?

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Bucharly9 · Accepted Answer

Explanation:First, expand the given function so that we can derive it with ease:f(x)=(x3−3x+1)(x+2)f(x)=x4+2x3−3x2−5x+2Then, we differentiate the function to find its gradient/derivative(s):f′(x)=4x3+6x2−6x−5Since we have been given the coordinates of a point (1, -3) in the curve / tangent line, we know that we need to find the gradient of the tangent line at x=1:f′(1)=4⋅13+6⋅12−6⋅1−5f&#039;(1)=4+6-6-5f&#039;(1)=1Finally, we use the gradient-intercept formula y2−y1=m(x2−x1) to find the equation of the tangent line:y- -3=1(x-1)y+3=x-1y=x-4

How to find the equation of the tangent line to

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