What is the tangent of y=3x^3-x^2+8,(2,28) ?

The tangent of y=3x^3-x^2+8,(2,28) is y=32x-36

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tangent of y=3x^3-x^2+8,(2,28)

tangent of

y = 3 x^{3} - x^{2} + 8, (2, 28)

y = 32 x - 36

Solution steps

Find the slope of

y = 3 x^{3} - x^{2} + 8 : \frac{d y}{d x} = 9 x^{2} - 2 x

Compute the slope of

y = 3 x^{3} - x^{2} + 8

(2, 28) : m = 32

Find the line with slope m=

32

and passing through

(2, 28) : y = 32 x - 36

y = 32 x - 36

\frac{d}{d x} (\frac{ln ( x )}{2})

\int a sin (θ) (1 + sin (θ)) d θ

\frac{3}{x}, 12 x, \frac{x}{3}

\int (t^{2} - 2 cos (t)) d t

f (x) = 2 x^{2} - 3 x + 1

What is the tangent of y=3x^3-x^2+8,(2,28) ?
The tangent of y=3x^3-x^2+8,(2,28) is y=32x-36