What is the tangent of y=((x+3)/(x-1))^2,(5,4) ?

The tangent of y=((x+3)/(x-1))^2,(5,4) is y=-x+9

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tangent of y=((x+3)/(x-1))^2,(5,4)

tangent of

y = (\frac{x + 3}{x - 1})^{2}, (5, 4)

y = - x + 9

Solution steps

Find the slope of

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

(5, 4) : y = - x + 9

y = - x + 9

f (x) = sin (x) + arcsin (x)

\frac{d}{d x} (lo g_{2} (5 x))

\int \frac{3 + 2 x ^{2}}{x ^{2}} d x

\frac{d}{d x} (\frac{10}{x ^{3}})

\frac{7}{4 s - 12}

What is the tangent of y=((x+3)/(x-1))^2,(5,4) ?
The tangent of y=((x+3)/(x-1))^2,(5,4) is y=-x+9