What is the tangent of x^3-6x^2-34x+40,\at x=-2 ?

The tangent of x^3-6x^2-34x+40,\at x=-2 is y=2x+80

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tangent of x^3-6x^2-34x+40,\at x=-2

tangent of

x^{3} - 6 x^{2} - 34 x + 40, at x = - 2

y = 2 x + 80

Solution steps

Find the tangent point:

(- 2, 76)

Find the slope of

y = x^{3} - 6 x^{2} - 34 x + 40 : \frac{d y}{d x} = 3 x^{2} - 12 x - 34

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2

Find the line with slope m=

2

and passing through

(- 2, 76) : y = 2 x + 80

y = 2 x + 80

\int arcsec (2 x) d x

x \to - 4 lim (x + 3)

\int \frac{x}{x ( x - 2 ) ( x + 3 )} d x

\int cos (3 y) d y

x \to \infty lim ((cos (\frac{1}{x}))^{\frac{1}{x ^{2}}})

What is the tangent of x^3-6x^2-34x+40,\at x=-2 ?
The tangent of x^3-6x^2-34x+40,\at x=-2 is y=2x+80