What is the tangent of x^3+8,\at x=-2 ?

The tangent of x^3+8,\at x=-2 is y=12x+24

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tangent of x^3+8,\at x=-2

tangent of

x^{3} + 8, at x = - 2

y = 12 x + 24

Solution steps

Find the tangent point:

(- 2, 0)

Find the slope of

y = x^{3} + 8 : \frac{d y}{d x} = 3 x^{2}

Compute the slope of

y = x^{3} + 8

x = - 2 : m = 12

Find the line with slope m=

12

and passing through

(- 2, 0) : y = 12 x + 24

y = 12 x + 24

\int \frac{3 x ^{3} - 5 x ^{2} + 10 x - 3}{3 x + 1} d x

\frac{\partial}{\partial y} (x \cdot y)

(2 (t - 2)^{2}) + 10 (t - 2) + 15

\int \frac{1}{x ( ln ( x ) ) ^{8}} d x

\int \frac{2 + x + x}{x} d x

What is the tangent of x^3+8,\at x=-2 ?
The tangent of x^3+8,\at x=-2 is y=12x+24