What is the tangent of y=x^3-3x^2+25 ?

The tangent of y=x^3-3x^2+25 is y=(3a_{0}^2-6a_{0})x-2a_{0}^3+3a_{0}^2+25

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tangent of y=x^3-3x^2+25

tangent of

y = x^{3} - 3 x^{2} + 25

y = (3 a_{0}^{2} - 6 a_{0}) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 25

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} + 25)

Find the slope of

y = x^{3} - 3 x^{2} + 25 : \frac{d y}{d x} = 3 x^{2} - 6 x

Compute the slope of

y = x^{3} - 3 x^{2} + 25

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} + 25) : y = (3 a_{0}^{2} - 6 a_{0}) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 25

y = (3 a_{0}^{2} - 6 a_{0}) x - 2 a_{0}^{3} + 3 a_{0}^{2} + 25

\int_{0}^{π} 9 e^{x} sin (x) d x

\frac{d}{d x} (arctan (sec (x) + tan (x)))

y = \frac{sin ( x )}{x ^{2}}

\frac{d}{d x} (\frac{5}{x + 1})

y^{^{'}} x y = x^{2} + 2 y^{2}

What is the tangent of y=x^3-3x^2+25 ?
The tangent of y=x^3-3x^2+25 is y=(3a_{0}^2-6a_{0})x-2a_{0}^3+3a_{0}^2+25