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

The tangent of f(x)=x^2-2x,(-2,8) is y=-6x-4

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tangent of f(x)=x^2-2x,(-2,8)

tangent of

f (x) = x^{2} - 2 x, (- 2, 8)

y = - 6 x - 4

Solution steps

Find the slope of

f (x) = x^{2} - 2 x : \frac{df ( x )}{d x} = 2 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, 8) : y = - 6 x - 4

y = - 6 x - 4

\frac{x ^{2} - 1}{x}

(3 x^{2} - x - 4) (- x^{2} + 2)

x \to 0 lim (6 x - ln (6 x + 1))

y^{^{''}} + 16 y = t e^{t} cos (4 t)

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

What is the tangent of f(x)=x^2-2x,(-2,8) ?
The tangent of f(x)=x^2-2x,(-2,8) is y=-6x-4