What is the tangent of x^3-9x^2+10x,\at =-2 ?

The tangent of x^3-9x^2+10x,\at =-2 is y=(3a_{0}^2-18a_{0}+10)x-2a_{0}^3+9a_{0}^2

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tangent of x^3-9x^2+10x,\at =-2

tangent of

x^{3} - 9 x^{2} + 10 x, at = - 2

y = (3 a_{0}^{2} - 18 a_{0} + 10) x - 2 a_{0}^{3} + 9 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} - 9 a_{0}^{2} + 10 a_{0})

Find the slope of

y = x^{3} - 9 x^{2} + 10 x : \frac{d y}{d x} = 3 x^{2} - 18 x + 10

Compute the slope of

y = x^{3} - 9 x^{2} + 10 x

x = a_{0} : m = 3 a_{0}^{2} - 18 a_{0} + 10

Find the line with slope m=

3 a_{0}^{2} - 18 a_{0} + 10

and passing through

(a_{0}, a_{0}^{3} - 9 a_{0}^{2} + 10 a_{0}) : y = (3 a_{0}^{2} - 18 a_{0} + 10) x - 2 a_{0}^{3} + 9 a_{0}^{2}

y = (3 a_{0}^{2} - 18 a_{0} + 10) x - 2 a_{0}^{3} + 9 a_{0}^{2}

n = 0 \sum \infty \frac{x ^{n}}{n 3 ^{n}}

\frac{d}{d x} (\frac{e ^{x}}{2 - tan ( x )})

\frac{\partial}{\partial x} (sin (3 x) cos (3 y))

f (x) = 5 x - 3 x^{2}, at x = 1

\frac{d}{d x} (\frac{( x ^{2} - 6 ) ^{3}}{x ^{2} + 4})

What is the tangent of x^3-9x^2+10x,\at =-2 ?
The tangent of x^3-9x^2+10x,\at =-2 is y=(3a_{0}^2-18a_{0}+10)x-2a_{0}^3+9a_{0}^2