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

The tangent of f(x)=3x^3+8x^2,\at x=-2 is y=4x+16

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

tangent of

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

y = 4 x + 16

Solution steps

Find the tangent point:

(- 2, 8)

Find the slope of

f (x) = 3 x^{3} + 8 x^{2} : \frac{df ( x )}{d x} = 9 x^{2} + 16 x

Compute the slope of

f (x) = 3 x^{3} + 8 x^{2}

x = - 2 : m = 4

Find the line with slope m=

4

and passing through

(- 2, 8) : y = 4 x + 16

y = 4 x + 16

x \to 5 lim (\frac{7}{x - 5})

x \to \infty lim (7 + \frac{4}{x})

n = 1 \sum \infty \frac{3 n}{2 n + 1}

\frac{\partial}{\partial x} (\frac{y}{x ^{2} + y ^{2} + 1})

6 x y^{^{'}} = 2 x e^{x} - 6 y + 6 x^{2}

What is the tangent of f(x)=3x^3+8x^2,\at x=-2 ?
The tangent of f(x)=3x^3+8x^2,\at x=-2 is y=4x+16