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

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

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

tangent of

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

y = - 8 x - 8

Solution steps

Find the tangent point:

(- 2, 8)

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 = - 8

Find the line with slope m=

- 8

and passing through

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

y = - 8 x - 8

\frac{\partial}{\partial y} (ln (x + y) - ln (x - y))

y = 4 x + 48, (4, 8)

\frac{74}{s ^{2} + 1}

\int 3 cos^{3} (3 x) d x

\int_{4}^{5} x x - 4 d x

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