What is the tangent of f(x)= 2/3 x^3+4x^2+6x-5 ?

The tangent of f(x)= 2/3 x^3+4x^2+6x-5 is y=(2a_{0}^2+8a_{0}+6)x-4/3 a_{0}^3-4a_{0}^2-5

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tangent of f(x)= 2/3 x^3+4x^2+6x-5

tangent of

f (x) = \frac{2}{3} x^{3} + 4 x^{2} + 6 x - 5

y = (2 a_{0}^{2} + 8 a_{0} + 6) x - \frac{4}{3} a_{0}^{3} - 4 a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2}{3} a_{0}^{3} + 4 a_{0}^{2} + 6 a_{0} - 5)

Find the slope of

f (x) = \frac{2}{3} x^{3} + 4 x^{2} + 6 x - 5 : \frac{df ( x )}{d x} = 2 x^{2} + 8 x + 6

Compute the slope of

f (x) = \frac{2}{3} x^{3} + 4 x^{2} + 6 x - 5

x = a_{0} : m = 2 a_{0}^{2} + 8 a_{0} + 6

Find the line with slope m=

2 a_{0}^{2} + 8 a_{0} + 6

and passing through

(a_{0}, \frac{2}{3} a_{0}^{3} + 4 a_{0}^{2} + 6 a_{0} - 5) : y = (2 a_{0}^{2} + 8 a_{0} + 6) x - \frac{4}{3} a_{0}^{3} - 4 a_{0}^{2} - 5

y = (2 a_{0}^{2} + 8 a_{0} + 6) x - \frac{4}{3} a_{0}^{3} - 4 a_{0}^{2} - 5

x \to 4 + lim (x + 1)

\int x^{2} x d x

\int e^{3 x} \cdot sin (2 x) d x

\frac{d y}{d t} = k y^{2}

\int_{0}^{64} \frac{3}{x ^{\frac{2}{3}}} d x

What is the tangent of f(x)= 2/3 x^3+4x^2+6x-5 ?
The tangent of f(x)= 2/3 x^3+4x^2+6x-5 is y=(2a_{0}^2+8a_{0}+6)x-4/3 a_{0}^3-4a_{0}^2-5