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

The tangent of y=x^3,\at x=-4 is y=48x+128

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tangent of y=x^3,\at x=-4

tangent of

y = x^{3}, at x = - 4

y = 48 x + 128

Solution steps

Find the tangent point:

(- 4, - 64)

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 = 48

Find the line with slope m=

48

and passing through

(- 4, - 64) : y = 48 x + 128

y = 48 x + 128

\int_{- 4}^{0} (4 + 16 - x^{2}) d x

\int - 5 e^{6 x} d x

\int \frac{1}{x ^{2} 100 - x ^{2}} d x

x \to - \infty lim (sin (x))

arctan (x)

What is the tangent of y=x^3,\at x=-4 ?
The tangent of y=x^3,\at x=-4 is y=48x+128