What is the tangent of y=3x^3-x^2+3,(2,23) ?

The tangent of y=3x^3-x^2+3,(2,23) is y=32x-41

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tangent of y=3x^3-x^2+3,(2,23)

tangent of

y = 3 x^{3} - x^{2} + 3, (2, 23)

y = 32 x - 41

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 23) : m = 32

Find the line with slope m=

32

and passing through

(2, 23) : y = 32 x - 41

y = 32 x - 41

\frac{k}{( s - 4 ) ^{3}}

t^{3} e^{2 t}

\frac{d}{d x} (sin^{2} (3 x^{2} - 1))

\frac{3 ( x - 1 )}{4 x x}

t e^{- 2 t} cos (3 t)

What is the tangent of y=3x^3-x^2+3,(2,23) ?
The tangent of y=3x^3-x^2+3,(2,23) is y=32x-41