What is the tangent of x^3-3x^2-9x+11,\at x=3 ?

The tangent of x^3-3x^2-9x+11,\at x=3 is y=-16

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tangent of x^3-3x^2-9x+11,\at x=3

tangent of

x^{3} - 3 x^{2} - 9 x + 11, at x = 3

y = - 16

Solution steps

Find the tangent point:

(3, - 16)

Find the slope of

y = x^{3} - 3 x^{2} - 9 x + 11 : \frac{d y}{d x} = 3 x^{2} - 6 x - 9

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 = 0

Find the line with slope m=

0

and passing through

(3, - 16) : y = - 16

y = - 16

\int (cos (θ))^{2} d θ

\int (ln (t))^{2} d t

x \to 6 lim (x^{3} + 3 x)

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

\frac{d}{d x} ((x^{2} + 8)^{4})

What is the tangent of x^3-3x^2-9x+11,\at x=3 ?
The tangent of x^3-3x^2-9x+11,\at x=3 is y=-16