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

The tangent of f(x)=3-4x^2,\at x=-2 is y=16x+19

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

tangent of

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

y = 16 x + 19

Solution steps

Find the tangent point:

(- 2, - 13)

Find the slope of

f (x) = 3 - 4 x^{2} : \frac{df ( x )}{d x} = - 8 x

Compute the slope of

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

x = - 2 : m = 16

Find the line with slope m=

16

and passing through

(- 2, - 13) : y = 16 x + 19

y = 16 x + 19

y = 3 x, (3, 3)

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

\int x sin (3 t - 3 x) d x

\frac{\partial}{\partial y} (3 x + 2 y)

\frac{d}{d x} (\frac{8 + x f ( x )}{x})

What is the tangent of f(x)=3-4x^2,\at x=-2 ?
The tangent of f(x)=3-4x^2,\at x=-2 is y=16x+19