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

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

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

tangent of

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

y = - 4 x - 2

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

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

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 = - 4

Find the line with slope m=

- 4

and passing through

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

y = - 4 x - 2

\frac{d}{d x} (\frac{- 10}{3 x})

\int_{2}^{6} (\frac{1}{x - 2}) d x

x \to 1 lim (ln ∣ ln (x) ∣)

\frac{d y}{d t} = 8 y

\frac{d y}{d x} = x^{3} + 2 x + π

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