What is the tangent of y=x^3-2x+1,(3,22) ?

The tangent of y=x^3-2x+1,(3,22) is y=25x-53

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tangent of y=x^3-2x+1,(3,22)

tangent of

y = x^{3} - 2 x + 1, (3, 22)

y = 25 x - 53

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 22) : m = 25

Find the line with slope m=

25

and passing through

(3, 22) : y = 25 x - 53

y = 25 x - 53

(- 3 x + 14)^{\frac{4}{3}}, 2

\frac{\partial}{\partial x} (x^{5} + x^{2} y + x + 1)

\int \frac{13}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

(x^{2} + 1) \frac{d y}{d x} = x (y^{2} + 1)

x^{2} - 6 x - 7, - x^{2} + 2 x + 3

What is the tangent of y=x^3-2x+1,(3,22) ?
The tangent of y=x^3-2x+1,(3,22) is y=25x-53