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

\frac{d}{d x} (\frac{3 x - 10}{7 x + 13})

f (x) = - \frac{4}{3} x^{- \frac{5}{2}}

\int \frac{9 x ^{2} + 4}{x ^{2}} d x

2 x y - 9 x^{2} + (2 y + x^{2} + 1) \frac{d y}{d x} = 0, y (0) = 3

\int 6 t^{5} d t

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