What is the tangent of y=2x^3-x^2+4,(2,16) ?

The tangent of y=2x^3-x^2+4,(2,16) is y=20x-24

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tangent of y=2x^3-x^2+4,(2,16)

tangent of

y = 2 x^{3} - x^{2} + 4, (2, 16)

y = 20 x - 24

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 16) : m = 20

Find the line with slope m=

20

and passing through

(2, 16) : y = 20 x - 24

y = 20 x - 24

x = \frac{y ^{2} - 1}{3}, x = 3 + y

\frac{d}{d x} (- 2 ln (x) + x^{2} - 4 + e^{x} + x^{e})

\frac{\partial}{\partial x} (2 yz)

x \to 6 - lim (e^{\frac{3}{6 - x}})

\frac{d y}{d x} = y x

What is the tangent of y=2x^3-x^2+4,(2,16) ?
The tangent of y=2x^3-x^2+4,(2,16) is y=20x-24