# How do you find the vertical, horizontal or slant asymptotes for f(x)= (x^2-5x+6)/ (x^2-8x+15)?

Mar 6, 2016

$\text{Asymptote "->" } x = 5$

${\lim}_{x \to \pm \infty} f \left(x\right) = 1$

#### Explanation:

Given:$\text{ } f \left(x\right) = \frac{{x}^{2} - 5 x + 6}{{x}^{2} - 8 x + 15}$

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Investigate potential simplification by factoring

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

At $x = 5$ the denominator becomes zero so the expression is undefined. Thus the asymptote is at $x = 5$
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As $x$ tends to $\infty$ then the constants are of no consequence.

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{x - 2}{x - 5} \to 1$