What is the tangent of y=((x+1)/(x-3))^2,(7,4) ?

The tangent of y=((x+1)/(x-3))^2,(7,4) is y=-x+11

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tangent of y=((x+1)/(x-3))^2,(7,4)

tangent of

y = (\frac{x + 1}{x - 3})^{2}, (7, 4)

y = - x + 11

Solution steps

Find the slope of

y = (\frac{x + 1}{x - 3})^{2} : \frac{d y}{d x} = - \frac{8 ( x + 1 )}{( x - 3 ) ^{3}}

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 = - 1

Find the line with slope m=

- 1

and passing through

(7, 4) : y = - x + 11

y = - x + 11

\int 2 cos^{2} (\frac{x}{2}) d x

\int \frac{1}{t ln ( t )} d t

y = 5 e^{x} + x, (0, 5)

\int \frac{1}{49 - 64 x ^{2}} d x

\int t^{7} e^{- t^{4}} d t

What is the tangent of y=((x+1)/(x-3))^2,(7,4) ?
The tangent of y=((x+1)/(x-3))^2,(7,4) is y=-x+11