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

The tangent of f(x)=4x^2+2x,\at x=-2 is y=-14x-16

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

tangent of

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

y = - 14 x - 16

Solution steps

Find the tangent point:

(- 2, 12)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = - 14

Find the line with slope m=

- 14

and passing through

(- 2, 12) : y = - 14 x - 16

y = - 14 x - 16

\frac{1}{s ^{2} + 6 s + 8}

\frac{d}{d x} (\frac{2 - x ^{2}}{( 2 + x ^{2} ) ^{2}})

- x^{2} + 4 x, x^{2} - 2 x

f (x) = 8 e^{x} cos (x)

\frac{s}{( s ^{2} + 2 s + 2 ) ( s ^{2} + 4 )}

What is the tangent of f(x)=4x^2+2x,\at x=-2 ?
The tangent of f(x)=4x^2+2x,\at x=-2 is y=-14x-16