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

\int_{- 1}^{0} (x^{2} - 3) d x

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

(x + x) y^{^{'}} = y + y

n = 1 \sum \infty 2 \cdot 0. 2^{n}

3 x^{2} e^{y} + (x^{3} e^{y} - 1) y^{^{'}} = 0

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