What is the tangent of y=2x^3-3x+1,(1,0) ?

The tangent of y=2x^3-3x+1,(1,0) is y=3x-3

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tangent of y=2x^3-3x+1,(1,0)

tangent of

y = 2 x^{3} - 3 x + 1, (1, 0)

y = 3 x - 3

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 0) : m = 3

Find the line with slope m=

3

and passing through

(1, 0) : y = 3 x - 3

y = 3 x - 3

x^{2} + 6 x + 9

4 ln (9 - 5 x^{2})

(x^{2} + 1) d y = (x - 4 x y) d x

\int 22 tan^{3} (x) d x

\int \frac{1}{1 + e ^{x}} d x

What is the tangent of y=2x^3-3x+1,(1,0) ?
The tangent of y=2x^3-3x+1,(1,0) is y=3x-3