What is the tangent of f(x)=-4-2x^2,\at x=-3 ?

The tangent of f(x)=-4-2x^2,\at x=-3 is y=12x+14

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tangent of f(x)=-4-2x^2,\at x=-3

tangent of

f (x) = - 4 - 2 x^{2}, at x = - 3

y = 12 x + 14

Solution steps

Find the tangent point:

(- 3, - 22)

Find the slope of

f (x) = - 4 - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x

Compute the slope of

f (x) = - 4 - 2 x^{2}

x = - 3 : m = 12

Find the line with slope m=

12

and passing through

(- 3, - 22) : y = 12 x + 14

y = 12 x + 14

x \to \infty lim (x r^{x})

\int (x + 2) ln (x) d x

x = 9, x = y^{2}

\int \frac{6}{t ^{2}} d t

\int 12 x + 12 d x

What is the tangent of f(x)=-4-2x^2,\at x=-3 ?
The tangent of f(x)=-4-2x^2,\at x=-3 is y=12x+14