What is the tangent of y= 1/6 x^3+2/5 x^2 ?

The tangent of y= 1/6 x^3+2/5 x^2 is y=((a_{0}^2)/2+(4a_{0})/5)x-(a_{0}^2(5a_{0}+6))/(15)

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tangent of y= 1/6 x^3+2/5 x^2

tangent of

y = \frac{1}{6} x^{3} + \frac{2}{5} x^{2}

y = (\frac{a _{0}^{2}}{2} + \frac{4 a _{0}}{5}) x - \frac{a _{0}^{2} ( 5 a _{0} + 6 )}{15}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{6} a_{0}^{3} + \frac{2}{5} a_{0}^{2})

Find the slope of

y = \frac{1}{6} x^{3} + \frac{2}{5} x^{2} : \frac{d y}{d x} = \frac{x ^{2}}{2} + \frac{4 x}{5}

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 = \frac{a _{0}^{2}}{2} + \frac{4 a _{0}}{5}

Find the line with slope m=

\frac{a _{0}^{2}}{2} + \frac{4 a _{0}}{5}

and passing through

(a_{0}, \frac{1}{6} a_{0}^{3} + \frac{2}{5} a_{0}^{2}) : y = (\frac{a _{0}^{2}}{2} + \frac{4 a _{0}}{5}) x - \frac{a _{0}^{2} ( 5 a _{0} + 6 )}{15}

y = (\frac{a _{0}^{2}}{2} + \frac{4 a _{0}}{5}) x - \frac{a _{0}^{2} ( 5 a _{0} + 6 )}{15}

What is the tangent of y= 1/6 x^3+2/5 x^2 ?
The tangent of y= 1/6 x^3+2/5 x^2 is y=((a_{0}^2)/2+(4a_{0})/5)x-(a_{0}^2(5a_{0}+6))/(15)