Diffirential Cal (rush pls help)

Question

locate the point of inflection (x,y), of the curve y=x^3-3x^2+4x-3 and find the equation of the inflectional tangent at the point of inflection. the point of inflection is _____ and the equation of inflectional tangent is ____

Answer 1

x^3 -3x^2 +4x -3

y' = 3x^2 -6x +4

inflection points are where the 2nd derivative = 0

y" = 6x -6 =0

x-1 = 0

x = 1

plug that x value into the original equation to find the y coordinate of the inflection point

y = 1-3+4-3 = -1

inflection point is (1,-1)

slope at that point is y'(1) = 3-6+4 = 1

line tangent at the inflection point is:

y =x +b, plug in the inflection point to solve for b

-1 = 1 +b

b =-2

tangent line is y=x-2