What is the tangent of y=2x^3-x^2+5,(2,17) ?

The tangent of y=2x^3-x^2+5,(2,17) is y=20x-23

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tangent of y=2x^3-x^2+5,(2,17)

tangent of

y = 2 x^{3} - x^{2} + 5, (2, 17)

y = 20 x - 23

Solution steps

Find the slope of

y = 2 x^{3} - x^{2} + 5 : \frac{d y}{d x} = 6 x^{2} - 2 x

Compute the slope of

y = 2 x^{3} - x^{2} + 5

(2, 17) : m = 20

Find the line with slope m=

20

and passing through

(2, 17) : y = 20 x - 23

y = 20 x - 23

f (x) = 3 x^{2} + 4 x + 1, at x = 0

8 cos^{2} (y) d x + csc^{2} (x) d y = 0, y (\frac{π}{12}) = \frac{π}{4}

f (x) = \frac{x + 3}{x}

\frac{d}{d x} (\frac{a + b x + c x ^{2}}{x})

f (x) = x - 1, (17, 4)

What is the tangent of y=2x^3-x^2+5,(2,17) ?
The tangent of y=2x^3-x^2+5,(2,17) is y=20x-23