How do you find the vertical, horizontal or slant asymptotes for f(x)= (x+3)/(x-3)?

Jun 19, 2016

Vertical asymptote is $x = 3$
Horizontal asymptote is $y = 1$

Explanation:

The equation becomes undefined at $x = 3$ in the denominator as $3 - 3 = 0$. Basically this means that mathematically you are not allowed to divide by 0.

$\textcolor{b l u e}{\text{Vertical asymptotes}}$

${\lim}_{x \to {3}^{+}} \frac{x + 3}{x - 3} \to \frac{x + 3}{{0}^{+}} = + \infty$

${\lim}_{X \to {3}^{-}} \frac{x + 3}{x - 3} \to \frac{x + 3}{{0}^{-}} = - \infty$

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$\textcolor{b l u e}{\text{Horizontal asymptotes}}$

As $x$ becomes bigger and bigger then the addition or subtraction of 3 becomes insignificant. Consequently we end up with basically $\frac{x}{x}$

${\lim}_{x \to {\infty}^{+}} \frac{x + 3}{x - 3} \to \frac{+ \infty}{+ \infty} = + 1$

${\lim}_{x \to {\infty}^{-}} \frac{x + 3}{x - 3} \to \frac{- \infty}{- \infty} = + 1$

So the horizontal asymptote is +1
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Check: using polynomial division.

$\left(x + 3\right) \div \left(x - 3\right) = 1 + \frac{6}{x - 3}$

${\lim}_{x \to {3}^{\pm}} = 1 \pm \infty$

${\lim}_{x \to {\infty}^{\pm}} = 1 \pm 0 = 1$