What is the tangent of f(x)=x^3-2x^2-5x+6,\at x=2 ?

The tangent of f(x)=x^3-2x^2-5x+6,\at x=2 is y=-x-2

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tangent of f(x)=x^3-2x^2-5x+6,\at x=2

tangent of

f (x) = x^{3} - 2 x^{2} - 5 x + 6, at x = 2

y = - x - 2

Solution steps

Find the tangent point:

(2, - 4)

Find the slope of

f (x) = x^{3} - 2 x^{2} - 5 x + 6 : \frac{df ( x )}{d x} = 3 x^{2} - 4 x - 5

Compute the slope of

f (x) = x^{3} - 2 x^{2} - 5 x + 6

x = 2 : m = - 1

Find the line with slope m=

- 1

and passing through

(2, - 4) : y = - x - 2

y = - x - 2

\int_{0}^{3} 15 8 x + 1 d x

\int g e^{- c t} d t

\frac{d}{d x} (a x sin (4 x) + b x cos (4 x))

f^{^{'}} (t) - f (t) = 2 t

\int e^{2 x + 1} d x

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