$f(x) = \dfrac{(x-1)(x-3)}{(x-2)^2}$

Considering the function above, what is the intersection point of the graph's asymptotes?

$(3 , 0)$
$(2 , 1)$
$(3 , 1)$
$(2 , 0)$
$(1 , 1)$
$(1 , 0)$

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$f(x) = \dfrac{(x-1)(x-3)}{(x-2)^2} = \dfrac{x^2 - 4x +3}{x^2 - 4x +4} = 1 - \dfrac{1}{x^2 - 4x +4}$ .

From the graph above, this function (red) is hyperbolic with $x = 2$ (blue) as its vertical asymptote, for $x$ can't equal to 2.

Also, if $y$ = $f(x)$ = 1, then 0 = - $\dfrac{1}{x^2 - 4x +4}$ .

Then $f(x)$ can only approach $y = 1$ once $x$ reaches +/- infinity. Therefore, $y = 1$ (green) is the graph's horizontal asymptote.

Hence, these two asymptotes will intersect each other at the coordinates

(2 , 1).