What is the tangent of y=3+4x^2-2x^3,(2,3) ?

The tangent of y=3+4x^2-2x^3,(2,3) is y=-8x+19

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tangent of y=3+4x^2-2x^3,(2,3)

tangent of

y = 3 + 4 x^{2} - 2 x^{3}, (2, 3)

y = - 8 x + 19

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 3) : m = - 8

Find the line with slope m=

- 8

and passing through

(2, 3) : y = - 8 x + 19

y = - 8 x + 19

\frac{d}{d x} (\frac{4 + csc ( x )}{8 - csc ( x )})

\int 4 x^{3} ln (x) d x

\frac{d}{d x} (c x + ln (cos (x)))

y^{^{'}} + r y^{^{''}} = 0

\frac{d}{d x} ((x^{2} + 2)^{- 5} (cos (x)))

What is the tangent of y=3+4x^2-2x^3,(2,3) ?
The tangent of y=3+4x^2-2x^3,(2,3) is y=-8x+19