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

The tangent of 2x^3+4x^2-x+2 is y=(6a_{0}^2+8a_{0}-1)x-4a_{0}^3-4a_{0}^2+2

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tangent of 2x^3+4x^2-x+2

tangent of

2 x^{3} + 4 x^{2} - x + 2

y = (6 a_{0}^{2} + 8 a_{0} - 1) x - 4 a_{0}^{3} - 4 a_{0}^{2} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 2 x^{3} + 4 x^{2} - x + 2 : \frac{d y}{d x} = 6 x^{2} + 8 x - 1

Compute the slope of

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

x = a_{0} : m = 6 a_{0}^{2} + 8 a_{0} - 1

Find the line with slope m=

6 a_{0}^{2} + 8 a_{0} - 1

and passing through

(a_{0}, 2 a_{0}^{3} + 4 a_{0}^{2} - a_{0} + 2) : y = (6 a_{0}^{2} + 8 a_{0} - 1) x - 4 a_{0}^{3} - 4 a_{0}^{2} + 2

y = (6 a_{0}^{2} + 8 a_{0} - 1) x - 4 a_{0}^{3} - 4 a_{0}^{2} + 2

y = \frac{x - 4}{x}

x \to - 10 lim (f (x))

x \to 3 - lim (\frac{x + 3}{x ^{2} - 9})

\int \frac{4}{e ^{t}} d t

\int \frac{1}{4 sin ( x ) + 3 cos ( x )} d x

What is the tangent of 2x^3+4x^2-x+2 ?
The tangent of 2x^3+4x^2-x+2 is y=(6a_{0}^2+8a_{0}-1)x-4a_{0}^3-4a_{0}^2+2