What is the tangent of y=2x^3-x^2+3,(2,15) ?

The tangent of y=2x^3-x^2+3,(2,15) is y=20x-25

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tangent of y=2x^3-x^2+3,(2,15)

tangent of

y = 2 x^{3} - x^{2} + 3, (2, 15)

y = 20 x - 25

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 15) : m = 20

Find the line with slope m=

20

and passing through

(2, 15) : y = 20 x - 25

y = 20 x - 25

\frac{1}{2} y^{^{''}} + 2 y = 2 tan (2 t) - \frac{1}{2} e^{3 t}

(6, - 6), (8, 8)

x^{3} + y^{3} - 6 x y = 0, (\frac{4}{3}, \frac{8}{3})

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

\frac{d}{d x} (9 ln (4 x))

What is the tangent of y=2x^3-x^2+3,(2,15) ?
The tangent of y=2x^3-x^2+3,(2,15) is y=20x-25