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

The tangent of f(x)=2x^2+2x,\at x=-2 is y=-6x-8

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

tangent of

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

y = - 6 x - 8

Solution steps

Find the tangent point:

(- 2, 4)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 6

Find the line with slope m=

- 6

and passing through

(- 2, 4) : y = - 6 x - 8

y = - 6 x - 8

\int_{- 3}^{3} (e^{x} - e^{- x})^{2} d x

\frac{d}{d t} (cos (t) + t sin (t))

t a y l or ln (\frac{z + \frac{1}{2}}{z})

n = 5 \sum \infty \frac{1}{4 ^{n} + 1}

in v erse l a pl a ce \frac{z}{( z + 1 ) ^{2}}

What is the tangent of f(x)=2x^2+2x,\at x=-2 ?
The tangent of f(x)=2x^2+2x,\at x=-2 is y=-6x-8