Precalculus
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How do you find vertical, horizontal and oblique asymptotes for f(x)= (x^2-5x+6)/ (x^2-8x+15)?

May 23, 2016

vertical asymptote x = 5
horizontal asymptote y = 1

Explanation:

The first step here is to factorise and simplify f(x).

$\Rightarrow f \left(x\right) = \frac{\cancel{\left(x - 3\right)} \left(x - 2\right)}{\cancel{\left(x - 3\right)} \left(x - 5\right)} = \frac{x - 2}{x - 5}$

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 5 = 0 → x = 5 is the asymptote.

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , f \left(x\right) \to 0$

divide terms on numerator/denominator by x.

$\Rightarrow \frac{\frac{x}{x} - \frac{2}{x}}{\frac{x}{x} - \frac{5}{x}} = \frac{1 - \frac{2}{x}}{1 - \frac{5}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{1 - 0}$

$\Rightarrow y = 1 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1). Hence there are no oblique asymptotes.
graph{(x-2)/(x-5) [-20, 20, -10, 10]}

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