What is the tangent of y=(4x)/(x^2+1),(-1,-2) ?

The tangent of y=(4x)/(x^2+1),(-1,-2) is y=-2

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tangent of y=(4x)/(x^2+1),(-1,-2)

tangent of

y = \frac{4 x}{x ^{2} + 1}, (- 1, - 2)

y = - 2

Solution steps

Find the slope of

y = \frac{4 x}{x ^{2} + 1} : \frac{d y}{d x} = \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{4 x}{x ^{2} + 1}

(- 1, - 2) : m = 0

Find the line with slope m=

0

and passing through

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

y = - 2

x^{3} - 13 x^{2} + 30 x, - x^{3} + 13 x^{2} - 30 x

- \frac{e ^{- 3 (t - 1)}}{3} - \frac{1}{3}

f (x) = \frac{6 x ^{3}}{4 - x}

x \to 0 lim (\frac{3 ^{x^{2}} - 1}{x})

\frac{1}{x + 6}

What is the tangent of y=(4x)/(x^2+1),(-1,-2) ?
The tangent of y=(4x)/(x^2+1),(-1,-2) is y=-2