1, 2, 3, Go!

Algebra Level 3

f ( x ) = ( x − 1 ) ( x − 3 ) ( x − 2 ) 2 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 ) (3 , 0) ( 2 , 1 ) (2 , 1) ( 3 , 1 ) (3 , 1) ( 2 , 0 ) (2 , 0) ( 1 , 1 ) (1 , 1) ( 1 , 0 ) (1 , 0)

Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

f ( x ) = ( x − 1 ) ( x − 3 ) ( x − 2 ) 2 = x 2 − 4 x + 3 x 2 − 4 x + 4 = 1 − 1 x 2 − 4 x + 4 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 x = 2 (blue) as its vertical asymptote, for x x can't equal to 2.

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

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

Hence, these two asymptotes will intersect each other at the coordinates (2 , 1) .

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