What is the tangent of f(x)=2x^2+2x,\at x=-3 ?

The tangent of f(x)=2x^2+2x,\at x=-3 is y=-10x-18

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tangent of f(x)=2x^2+2x,\at x=-3

tangent of

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

y = - 10 x - 18

Solution steps

Find the tangent point:

(- 3, 12)

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 = - 10

Find the line with slope m=

- 10

and passing through

(- 3, 12) : y = - 10 x - 18

y = - 10 x - 18

\int x e - x d x

s l o p e (- 4, 2), (- 9, - 10)

n = 0 \sum \infty - \frac{1}{3 ^{n}}

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

d er i v a t i v e arcsin (\frac{x}{5})

What is the tangent of f(x)=2x^2+2x,\at x=-3 ?
The tangent of f(x)=2x^2+2x,\at x=-3 is y=-10x-18