What is the tangent of y=2x^3-x^2+1,(2,13) ?

The tangent of y=2x^3-x^2+1,(2,13) is y=20x-27

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tangent of y=2x^3-x^2+1,(2,13)

tangent of

y = 2 x^{3} - x^{2} + 1, (2, 13)

y = 20 x - 27

Solution steps

Find the slope of

y = 2 x^{3} - x^{2} + 1 : \frac{d y}{d x} = 6 x^{2} - 2 x

Compute the slope of

y = 2 x^{3} - x^{2} + 1

(2, 13) : m = 20

Find the line with slope m=

20

and passing through

(2, 13) : y = 20 x - 27

y = 20 x - 27

\frac{d}{d x} (\frac{π 8100 + x ^{2}}{x})

\frac{4}{x + 10}

x \to + 1 + lim (\frac{5}{x ^{3} - 1})

\frac{d}{d x} (\frac{1}{1 - sin ( x )})

f (x) = 8 x + 2 x^{2} - x^{3}, g (x) = 0

What is the tangent of y=2x^3-x^2+1,(2,13) ?
The tangent of y=2x^3-x^2+1,(2,13) is y=20x-27