How do you find the vertical, horizontal or slant asymptotes for $(x^2 - 5x + 6)/( x - 3)$ ?

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Answer 1 · 2016-11-22T17:53:23

We have a hole when $x=2$

Let's factorise the numerator.

$x^2-5x+6=(x-3)(x-2)$

Therefore,

$(x^2-5x+6)/(x-3)=(cancel(x-3)(x-2))/cancel(x-3)$

$=(x-2)$

We have a hole at $x=2$

$y=(x-2)$ is the equation of a straight line.

graph{x-2 [-10, 10, -5, 5]}

How do you find the vertical, horizontal or slant asymptotes for (x^2 - 5x + 6)/( x - 3)(x^2 - 5x + 6)/( x - 3)?