What is the tangent of y=2x^3-6x^2+5x+1,(2,3) ?

The tangent of y=2x^3-6x^2+5x+1,(2,3) is y=5x-7

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tangent of y=2x^3-6x^2+5x+1,(2,3)

tangent of

y = 2 x^{3} - 6 x^{2} + 5 x + 1, (2, 3)

y = 5 x - 7

Solution steps

Find the slope of

y = 2 x^{3} - 6 x^{2} + 5 x + 1 : \frac{d y}{d x} = 6 x^{2} - 12 x + 5

Compute the slope of

y = 2 x^{3} - 6 x^{2} + 5 x + 1

(2, 3) : m = 5

Find the line with slope m=

5

and passing through

(2, 3) : y = 5 x - 7

y = 5 x - 7

h^{^{'}} = - h

\frac{d}{d x} (\frac{4}{3})

\frac{d}{d x} (sin^{3} (x) cos (x))

(7 - 4) (73)

x \to - 5 + lim (\frac{x - 3}{x + 5})

What is the tangent of y=2x^3-6x^2+5x+1,(2,3) ?
The tangent of y=2x^3-6x^2+5x+1,(2,3) is y=5x-7