What is the tangent of y=2x^3-5x+1,(1,-2) ?

The tangent of y=2x^3-5x+1,(1,-2) is y=x-3

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tangent of y=2x^3-5x+1,(1,-2)

tangent of

y = 2 x^{3} - 5 x + 1, (1, - 2)

y = x - 3

Solution steps

Find the slope of

y = 2 x^{3} - 5 x + 1 : \frac{d y}{d x} = 6 x^{2} - 5

Compute the slope of

y = 2 x^{3} - 5 x + 1

(1, - 2) : m = 1

Find the line with slope m=

1

and passing through

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

y = x - 3

\int x 1 - x^{2} d x

\int \frac{x + 2}{x ^{2} + 25} d x

f (x) = \frac{10 x}{1 + x ^{2}}, (3, 3)

x^{5} + y^{4} x^{6} + 5 y^{^{'}} = 0

(- 5) (7.8)

What is the tangent of y=2x^3-5x+1,(1,-2) ?
The tangent of y=2x^3-5x+1,(1,-2) is y=x-3