What is the tangent of y=x^3-2x+2,(3,23) ?

The tangent of y=x^3-2x+2,(3,23) is y=25x-52

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tangent of y=x^3-2x+2,(3,23)

tangent of

y = x^{3} - 2 x + 2, (3, 23)

y = 25 x - 52

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 23) : m = 25

Find the line with slope m=

25

and passing through

(3, 23) : y = 25 x - 52

y = 25 x - 52

y^{^{''}} + 6 y^{^{'}} + 25 y = 0

\frac{d}{d x} (1 + 7 sin (x) - tan (x))

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

f (x) = - 8 x^{2} + 10, (0.5, 8)

\frac{d}{d x} ((x^{2} - 9)^{\frac{1}{2}})

What is the tangent of y=x^3-2x+2,(3,23) ?
The tangent of y=x^3-2x+2,(3,23) is y=25x-52