What is the tangent of y=x^{3-5x^2}+7x+3 ?

The tangent of y=x^{3-5x^2}+7x+3 is

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tangent of y=x^{3-5x^2}+7x+3

tangent of

y = x^{3 - 5 x^{2}} + 7 x + 3

y = (a_{0}^{- 5 a_{0}^{2} + 3} (- 10 a_{0} ln (a_{0}) + \frac{- 5 a _{0}^{2} + 3}{a _{0}}) + 7) x + 10 a_{0}^{- 5 a_{0}^{2} + 5} ln (a_{0}) + 5 a_{0}^{- 5 a_{0}^{2} + 5} - 2 a_{0}^{- 5 a_{0}^{2} + 3} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3 - 5 a_{0}^{2}} + 7 a_{0} + 3)

Find the slope of

y = x^{3 - 5 x^{2}} + 7 x + 3 : \frac{d y}{d x} = x^{- 5 x^{2} + 3} (- 10 x ln (x) + \frac{- 5 x ^{2} + 3}{x}) + 7

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = a_{0}^{- 5 a_{0}^{2} + 3} (- 10 a_{0} ln (a_{0}) + \frac{- 5 a _{0}^{2} + 3}{a _{0}}) + 7

Find the line with slope m=

a_{0}^{- 5 a_{0}^{2} + 3} (- 10 a_{0} ln (a_{0}) + \frac{- 5 a _{0}^{2} + 3}{a _{0}}) + 7

and passing through

(a_{0}, a_{0}^{3 - 5 a_{0}^{2}} + 7 a_{0} + 3) : y = (a_{0}^{- 5 a_{0}^{2} + 3} (- 10 a_{0} ln (a_{0}) + \frac{- 5 a _{0}^{2} + 3}{a _{0}}) + 7) x + 10 a_{0}^{- 5 a_{0}^{2} + 5} ln (a_{0}) + 5 a_{0}^{- 5 a_{0}^{2} + 5} - 2 a_{0}^{- 5 a_{0}^{2} + 3} + 3

y = (a_{0}^{- 5 a_{0}^{2} + 3} (- 10 a_{0} ln (a_{0}) + \frac{- 5 a _{0}^{2} + 3}{a _{0}}) + 7) x + 10 a_{0}^{- 5 a_{0}^{2} + 5} ln (a_{0}) + 5 a_{0}^{- 5 a_{0}^{2} + 5} - 2 a_{0}^{- 5 a_{0}^{2} + 3} + 3