What is the tangent of f(x)= x/(x-2),(3,3) ?

The tangent of f(x)= x/(x-2),(3,3) is y=-2x+9

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tangent of f(x)= x/(x-2),(3,3)

tangent of

f (x) = \frac{x}{x - 2}, (3, 3)

y = - 2 x + 9

Solution steps

Find the slope of

f (x) = \frac{x}{x - 2} : \frac{df ( x )}{d x} = - \frac{2}{( x - 2 ) ^{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 = - 2

Find the line with slope m=

- 2

and passing through

(3, 3) : y = - 2 x + 9

y = - 2 x + 9

y^{^{''}} = y^{^{'}}

\int \frac{8}{z z ^{2} + 2} d z

\frac{d}{d x} (cos (x) - 7 tan (x))

y = e^{- x^{2}}

f (x) = x^{2} - 8, at x = 3

What is the tangent of f(x)= x/(x-2),(3,3) ?
The tangent of f(x)= x/(x-2),(3,3) is y=-2x+9