How do you find all the asymptotes for $\frac{x - 2}{x^{2} - 12 x + 12}$ $(x-2)/(x^2-12x+12)$ ?

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Answer 1 · 2016-03-12T13:18:39

Two vertical asymptotes $x = 10.899$ $x=10.899$ or $x = 1.101$ $x=1.101$

Using quadratic formulahe denominator $x^{2} - 12 x + 12$ $x^2-12x+12$ will be zero at

$x = \frac{12 \pm \sqrt{12^{2} - 4 \times 1 \times 12}}{2} = \frac{12 \pm \sqrt{96}}{2}$ $x=(12+-sqrt(12^2-4xx1xx12))/2=(12+-sqrt96)/2$ or

$x = \frac{12 \pm 4 \sqrt{6}}{2} = 6 \pm 2 \times 2.4495$ $x=(12+-4sqrt6)/2=6+-2xx2.4495$ i.e.

$x = 10.899$ $x=10.899$ or $x = 1.101$ $x=1.101$

Hence we have two vertical asymptotes $x = 10.899$ $x=10.899$ or $x = 1.101$ $x=1.101$ .

As the degree of numerator is less than that of denominator, there is no horizontal or slanting asymptote.

graph{(x-2)/(x^2-12x+12) [-5, 20, -5, 5]}

How do you find all the asymptotes for (x-2)/(x^2-12x+12) x−2x2−12x+12 (x-2)/(x^2-12x+12) ?