What is the tangent of f(x)=4x^2+4x-2,\at x=-1 ?

The tangent of f(x)=4x^2+4x-2,\at x=-1 is y=-4x-6

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tangent of f(x)=4x^2+4x-2,\at x=-1

tangent of

f (x) = 4 x^{2} + 4 x - 2, at x = - 1

y = - 4 x - 6

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, - 2) : y = - 4 x - 6

y = - 4 x - 6

\frac{d}{d x} ((x^{3} + 5) e^{x})

\int cos (4 x) 2 - sin (4 x) d x

\int \frac{1}{( 9 - x ^{2} )} d x

\frac{d}{d x} (4 (x + 4)^{3})

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

What is the tangent of f(x)=4x^2+4x-2,\at x=-1 ?
The tangent of f(x)=4x^2+4x-2,\at x=-1 is y=-4x-6