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

The tangent of y=2x^3-x^2+1,(1,2) is y=4x-2

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

tangent of

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

y = 4 x - 2

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

(1, 2) : m = 4

Find the line with slope m=

4

and passing through

(1, 2) : y = 4 x - 2

y = 4 x - 2

x \to 0 + lim (\frac{x}{x})

\frac{\partial}{\partial x} ((2 - x)^{6})

\frac{\partial}{\partial x} (2 x y - 4 x)

n = 0 \sum \infty e^{- n}

\int 2 y^{2} d y

What is the tangent of y=2x^3-x^2+1,(1,2) ?
The tangent of y=2x^3-x^2+1,(1,2) is y=4x-2