What is the tangent of x^3-8x^2 ?

The tangent of x^3-8x^2 is y=(3a_{0}^2-16a_{0})x-2a_{0}^3+8a_{0}^2

tangent of x^3-8x^2

tangent of

x^{3} - 8 x^{2}

y = (3 a_{0}^{2} - 16 a_{0}) x - 2 a_{0}^{3} + 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 8 a_{0}^{2})

Find the slope of

y = x^{3} - 8 x^{2} : \frac{d y}{d x} = 3 x^{2} - 16 x

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 = 3 a_{0}^{2} - 16 a_{0}

Find the line with slope m=

3 a_{0}^{2} - 16 a_{0}

and passing through

(a_{0}, a_{0}^{3} - 8 a_{0}^{2}) : y = (3 a_{0}^{2} - 16 a_{0}) x - 2 a_{0}^{3} + 8 a_{0}^{2}

y = (3 a_{0}^{2} - 16 a_{0}) x - 2 a_{0}^{3} + 8 a_{0}^{2}

\frac{d}{d u} (\frac{( 2 u ^{3} + 7 ) ( 3 u ^{2} - 5 )}{u ^{2} + 1})

\int \frac{x}{( 4 + x ^{4} ) ^{2}} d x

\frac{d}{d x} (- arctan (\frac{x}{y}))

\frac{d}{d x} (sin (x^{2} - x))

t an g e n t y = (1 + 3 x)^{8}, (0, 1)

What is the tangent of x^3-8x^2 ?
The tangent of x^3-8x^2 is y=(3a_{0}^2-16a_{0})x-2a_{0}^3+8a_{0}^2