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

The tangent of 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 2/3 x^3+4x^2+6x-5

tangent of

\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

y = \frac{2}{3} x^{3} + 4 x^{2} + 6 x - 5 : \frac{d y}{d x} = 2 x^{2} + 8 x + 6

Compute the slope of

y = \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

\int \frac{196}{w ^{2} 49 - w ^{2}} d w

\int \frac{1}{2 x + 3} d x

\frac{\partial}{\partial y} (2 x (5 + y)^{- 1})

\frac{sec ( 9 x ^{2} )}{tan ( 2 x ) \cdot cos ( x )}

y^{^{'}} = 3

What is the tangent of 2/3 x^3+4x^2+6x-5 ?
The tangent of 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