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

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

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

tangent of

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

y = - 4 x - 8

Solution steps

Find the tangent point:

(- 2, 0)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 0) : y = - 4 x - 8

y = - 4 x - 8

\frac{d}{d x} (4 (3 x^{2} - 2))

\int sin (x) cos (x) + x d x

x \to \infty lim (x sin (\frac{4 π}{x}))

\frac{\partial}{\partial z} (x^{2})

sin^{2} (t)

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