What is the tangent of 1/3 x^3-4x^2+17x+13 ?

The tangent of 1/3 x^3-4x^2+17x+13 is y=(a_{0}^2-8a_{0}+17)x-2/3 a_{0}^3+4a_{0}^2+13

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tangent of 1/3 x^3-4x^2+17x+13

tangent of

\frac{1}{3} x^{3} - 4 x^{2} + 17 x + 13

y = (a_{0}^{2} - 8 a_{0} + 17) x - \frac{2}{3} a_{0}^{3} + 4 a_{0}^{2} + 13

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3} a_{0}^{3} - 4 a_{0}^{2} + 17 a_{0} + 13)

Find the slope of

y = \frac{1}{3} x^{3} - 4 x^{2} + 17 x + 13 : \frac{d y}{d x} = x^{2} - 8 x + 17

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 = a_{0}^{2} - 8 a_{0} + 17

Find the line with slope m=

a_{0}^{2} - 8 a_{0} + 17

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} - 4 a_{0}^{2} + 17 a_{0} + 13) : y = (a_{0}^{2} - 8 a_{0} + 17) x - \frac{2}{3} a_{0}^{3} + 4 a_{0}^{2} + 13

y = (a_{0}^{2} - 8 a_{0} + 17) x - \frac{2}{3} a_{0}^{3} + 4 a_{0}^{2} + 13

What is the tangent of 1/3 x^3-4x^2+17x+13 ?
The tangent of 1/3 x^3-4x^2+17x+13 is y=(a_{0}^2-8a_{0}+17)x-2/3 a_{0}^3+4a_{0}^2+13