What is the tangent of f(x)=(6x^2-1)/(-3x+1),\at x=0 ?

The tangent of f(x)=(6x^2-1)/(-3x+1),\at x=0 is y=-3x-1

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tangent of f(x)=(6x^2-1)/(-3x+1),\at x=0

tangent of

f (x) = \frac{6 x ^{2} - 1}{- 3 x + 1}, at x = 0

y = - 3 x - 1

Solution steps

Find the tangent point:

(0, - 1)

Find the slope of

f (x) = \frac{6 x ^{2} - 1}{- 3 x + 1} : \frac{df ( x )}{d x} = \frac{- 18 x ^{2} + 12 x - 3}{( - 3 x + 1 ) ^{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 = - 3

Find the line with slope m=

- 3

and passing through

(0, - 1) : y = - 3 x - 1

y = - 3 x - 1

\int_{2}^{4} \frac{1}{16 - x ^{2}} d x

\frac{\partial}{\partial y} (e^{x^{2}})

\int 4 x^{- 5} d x

\int (x - 1) e^{- x - 1} d x

n = 0 \sum \infty e

What is the tangent of f(x)=(6x^2-1)/(-3x+1),\at x=0 ?
The tangent of f(x)=(6x^2-1)/(-3x+1),\at x=0 is y=-3x-1