What is the tangent of y=x^3-3x^2-9x+27,\at x=4 ?

The tangent of y=x^3-3x^2-9x+27,\at x=4 is y=15x-53

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tangent of y=x^3-3x^2-9x+27,\at x=4

tangent of

y = x^{3} - 3 x^{2} - 9 x + 27, at x = 4

y = 15 x - 53

Solution steps

Find the tangent point:

(4, 7)

Find the slope of

y = x^{3} - 3 x^{2} - 9 x + 27 : \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 = 15

Find the line with slope m=

15

and passing through

(4, 7) : y = 15 x - 53

y = 15 x - 53

What is the tangent of y=x^3-3x^2-9x+27,\at x=4 ?
The tangent of y=x^3-3x^2-9x+27,\at x=4 is y=15x-53