What is the tangent of y=((x+2)/(x-2))^2,(6,4) ?

The tangent of y=((x+2)/(x-2))^2,(6,4) is y=-x+10

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tangent of y=((x+2)/(x-2))^2,(6,4)

tangent of

y = (\frac{x + 2}{x - 2})^{2}, (6, 4)

y = - x + 10

Solution steps

Find the slope of

y = (\frac{x + 2}{x - 2})^{2} : \frac{d y}{d x} = - \frac{8 ( x + 2 )}{( x - 2 ) ^{3}}

Compute the slope of

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

(6, 4) : m = - 1

Find the line with slope m=

- 1

and passing through

(6, 4) : y = - x + 10

y = - x + 10

\int_{0}^{\frac{π}{2}} sin (2 t) d t

\int_{0}^{4} x + 1 d x

x \to 9 lim (3 - x)

\frac{d}{d x} (4 x^{3} - 7 x^{2})

y = 6 - 2 x, y = 2, x = 0

What is the tangent of y=((x+2)/(x-2))^2,(6,4) ?
The tangent of y=((x+2)/(x-2))^2,(6,4) is y=-x+10