What is the tangent of y=3x^3-x^2+6x+4 ?

The tangent of y=3x^3-x^2+6x+4 is y=(9a_{0}^2-2a_{0}+6)x-6a_{0}^3+a_{0}^2+4

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tangent of y=3x^3-x^2+6x+4

tangent of

y = 3 x^{3} - x^{2} + 6 x + 4

y = (9 a_{0}^{2} - 2 a_{0} + 6) x - 6 a_{0}^{3} + a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 3 x^{3} - x^{2} + 6 x + 4 : \frac{d y}{d x} = 9 x^{2} - 2 x + 6

Compute the slope of

y = 3 x^{3} - x^{2} + 6 x + 4

x = a_{0} : m = 9 a_{0}^{2} - 2 a_{0} + 6

Find the line with slope m=

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

and passing through

(a_{0}, 3 a_{0}^{3} - a_{0}^{2} + 6 a_{0} + 4) : y = (9 a_{0}^{2} - 2 a_{0} + 6) x - 6 a_{0}^{3} + a_{0}^{2} + 4

y = (9 a_{0}^{2} - 2 a_{0} + 6) x - 6 a_{0}^{3} + a_{0}^{2} + 4

\int 2 x + 1 d x

(8 x^{3} + 2 x^{2})^{4}

\int_{- 4}^{4} 16 - x^{2} d x

\frac{d y}{d x} = 5 y

21 x y^{^{'}} - ln (x^{7}) = 0

What is the tangent of y=3x^3-x^2+6x+4 ?
The tangent of y=3x^3-x^2+6x+4 is y=(9a_{0}^2-2a_{0}+6)x-6a_{0}^3+a_{0}^2+4