#### Answer

$y=−2x+9$

#### Work Step by Step

The question asks us to find the equation of the tangent of the function
$f(x)=\frac{x}{x-2}$ at the point (3,3)
Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function.
Using the power rule and quotient rule:
$f′(x)=\frac{1}{x-2}=\frac{-x}{(x-2)^2}=\frac{-2}{(x-2)^2}$
$f′(3)=−2$
Thus, the gradient of that function at the value x=3 is -2.
Using this information, we can find the intercept value of the tangent line.
The equation of the gradient is given by:
$\frac{y_{1}−y_{0}}{x_{1}−x_{0}}=m$
We can now use the points (3,3) and (0,c) (where c is the y-intercept).
$\frac{3−c}{3-0}=-2$
Solving for c gives us c=9
Thus, the equation of the tangent at (3,3) is $y=−2x+9$