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

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

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tangent of y=(-6x)/(x^2+1),(1,-3)

tangent of

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

y = - 3

Solution steps

Find the slope of

y = \frac{- 6 x}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 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 = 0

Find the line with slope m=

0

and passing through

(1, - 3) : y = - 3

y = - 3

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