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

The tangent of y=x^3-3x+1,(2,3) is y=9x-15

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

tangent of

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

y = 9 x - 15

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 3) : m = 9

Find the line with slope m=

9

and passing through

(2, 3) : y = 9 x - 15

y = 9 x - 15

\int (in t)^{2} d t

y = - 18

\frac{d}{d x} (2 sec^{2} (x^{7}))

\frac{1}{2 x}, at x = 4

x \to 8 + lim (x - 8)

What is the tangent of y=x^3-3x+1,(2,3) ?
The tangent of y=x^3-3x+1,(2,3) is y=9x-15