What is the tangent of y=x^3-3x^2+1 ?

The tangent of y=x^3-3x^2+1 is y=(3a_{0}^2-6a_{0})x-2a_{0}^3+3a_{0}^2+1

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tangent of y=x^3-3x^2+1

tangent of

y = x^{3} - 3 x^{2} + 1

y = (3 a_{0}^{2} - 6 a_{0}) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 3 a_{0}^{2} + 1)

Find the slope of

y = x^{3} - 3 x^{2} + 1 : \frac{d y}{d x} = 3 x^{2} - 6 x

Compute the slope of

y = x^{3} - 3 x^{2} + 1

x = a_{0} : m = 3 a_{0}^{2} - 6 a_{0}

Find the line with slope m=

3 a_{0}^{2} - 6 a_{0}

and passing through

(a_{0}, a_{0}^{3} - 3 a_{0}^{2} + 1) : y = (3 a_{0}^{2} - 6 a_{0}) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 1

y = (3 a_{0}^{2} - 6 a_{0}) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 1

f (x) = - 7

y^{^{'}} - \frac{1}{x} y = 9

x^{2} + 2, at x = - 1

x \to 0 lim (\frac{1 - 2 cos ( x ) + cos ( 2 x )}{x ^{2}})

y = - \frac{2}{5} x - 3

What is the tangent of y=x^3-3x^2+1 ?
The tangent of y=x^3-3x^2+1 is y=(3a_{0}^2-6a_{0})x-2a_{0}^3+3a_{0}^2+1