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Consider the following, a polynomial $P(x)$ is given by $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0x^0$$if we then have another polynomial, say $$G(x)=b_mx^m+b_{m-1}x^{m-1}+\ldots+b_0x^0$$and then a function defined by$$H(x)=\frac{P(x)}{G(x)}$$then $H(x)$ is a rational function. Furthermore, if $m=n$ then the graph of $H(x)$ has a horizontal asymptote at $$y=\frac{a_n}{b_m}$$conversely, if $n<m$ then the graph of $H(x)$ will have a horizontal asymptote at $y=0$, but if $n>m$ then there is no horizontal asymptote. All of this in mind, consider the rational function $$f(x)=\frac{x}{x^2-8x+15}$$since the degree of the denominator is greater than the degree of the numerator, there must be a horizontal asymptote at $y=0$, however, this is not true for the entire domain, given that at $x=0$, $y=0$. The question therefore is why is this the case? I've speculated that it's because the degree of the denominator can be reduced to $1$, by factoring the expression to $(x-3)(x-5)$, but I'd appreciate some confirmation of this.

[Note: for values of $x$ greater than $5$, $y=0$ is a horizontal asymptote, but for the prior mentioned rule to apply, obviously this must be the case $\forall x$]

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Horizontal asymptotes are finite limits of your function as $x$ approaches $\infty$ or $-\infty.$ The behavior of your function around $x=0$ will not affect the horizontal asymptotes.

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If a graph has a horizontal asymptote, it means that $$\lim_{x\to \infty}f(x)=a$$ or $$\lim_{x\to -\infty}f(x)=b$$ for some finite numbers $a$ and $b$.

The function doesn't necessarily approach this limiting value at all points of the rational function. It only tends to the limit as $x$ takes on increasingly negative or increasingly positive values. It can attain this limiting value elsewhere, as long as it still approaches the asymptote as I have desribed above.

As a side note, the product $(x - 3)(x - 5)$ is still a second degree polynomial. Factoring a polynomial has no effect on its degree, as it only simplifies the equation without changing its value.

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