Sketch the graph of the function. Show vertical and horizontal asymptotes and x-intercepts.
$y=\frac{x^2-16}{x^2-6 x+8}$

Solution

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$y = \lim\limits_{x \to \infty}\dfrac{x^2 - 16}{x^2 - 6x + 8} = \lim\limits_{x \to \infty}\dfrac{x^2 }{x^2} = \lim\limits_{x \to \infty}1 =1$ is the HA.

For horizontal asymptote (HA), take the limit of the rational function as $x$ approaches $\infty$ . If the limit is a finite constant $k$ , then $y = k$ is the HA, otherwise, there is no HA.

(Note: Technically we need to take the limit of the function as $x$ approaches $-\infty$ also, however, since a rational function can have at most one HA, we can omit taking the limit as $x$ approaches $-\infty$ .)