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

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

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

tangent of

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

y = 8 x + 8

Solution steps

Find the tangent point:

(- 2, - 8)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = 8

Find the line with slope m=

8

and passing through

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

y = 8 x + 8

4 x - 16

y^{^{''}} + 10 y^{^{'}} + 20 y = 0

\frac{1}{( s - 2 ) ( s + 2 )}

n = 0 \sum \infty 2 (\frac{2}{3})^{n}

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

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