What is the tangent of y=3x^3-x^2+9,(2,29) ?

The tangent of y=3x^3-x^2+9,(2,29) is y=32x-35

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tangent of y=3x^3-x^2+9,(2,29)

tangent of

y = 3 x^{3} - x^{2} + 9, (2, 29)

y = 32 x - 35

Solution steps

Find the slope of

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

Compute the slope of

y = 3 x^{3} - x^{2} + 9

(2, 29) : m = 32

Find the line with slope m=

32

and passing through

(2, 29) : y = 32 x - 35

y = 32 x - 35

\int (x - 1) sin (π x) d x

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

\int 9 x^{5} e^{8 x^{2}} d x

\frac{\partial}{\partial y} (x^{3} y^{2})

\int cos (3 θ) d θ

What is the tangent of y=3x^3-x^2+9,(2,29) ?
The tangent of y=3x^3-x^2+9,(2,29) is y=32x-35