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

The tangent of 2x^3-x^2 is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2

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tangent of 2x^3-x^2

tangent of

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

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

(a_{0}, 2 a_{0}^{3} - a_{0}^{2}) : y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2}

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2}

e^{3 t} sin (4 t)

\int x^{2} + x d x

n = 0 \sum \infty 9^{n}

f (x) = \frac{x ^{2}}{( x ^{3} ) ( x ^{2} )}

\int \frac{1}{x x ^{2} - 81} d x

What is the tangent of 2x^3-x^2 ?
The tangent of 2x^3-x^2 is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2