What is the tangent of y=x^3,\at x=-3 ?

The tangent of y=x^3,\at x=-3 is y=27x+54

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tangent of y=x^3,\at x=-3

tangent of

y = x^{3}, at x = - 3

y = 27 x + 54

Solution steps

Find the tangent point:

(- 3, - 27)

Find the slope of

y = x^{3} : \frac{d y}{d x} = 3 x^{2}

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 = 27

Find the line with slope m=

27

and passing through

(- 3, - 27) : y = 27 x + 54

y = 27 x + 54

y = \frac{x ^{5}}{4 - x ^{4}}

\frac{2}{s ( 5 s + 1 )}

e^{- y^{2}}

\int + 3 x^{3} e^{- x^{2}} d x

\int (8) d x

What is the tangent of y=x^3,\at x=-3 ?
The tangent of y=x^3,\at x=-3 is y=27x+54