How do you find the asymptotes for y= (x+1)^2 / ((x-1)(x-3))?

Mar 26, 2016

Vertical
$x = 1$
$x = 3$

Horizontal
$x = 1$ (for both $\pm \infty$)

Oblique
Don't exist

Explanation:

Let $y = f \left(x\right)$

Find the limits of the function as it tends to the limits of its domain except infinity. If their result is infinity, than that $x$ line is an asymptote. Here, the domain is:

$x \in \left(- \infty , 1\right) \cup \left(1 , 3\right) \cup \left(3 , + \infty\right)$

So the 4 possible vertical asymptotes are:

${\lim}_{x \to {1}^{-}} f \left(x\right)$

${\lim}_{x \to {1}^{+}} f \left(x\right)$

${\lim}_{x \to {3}^{-}} f \left(x\right)$

${\lim}_{x \to {3}^{+}} f \left(x\right)$

Asymptote $x \to {1}^{-}$

${\lim}_{x \to {1}^{-}} f \left(x\right) = {\lim}_{x \to {1}^{-}} {\left(x + 1\right)}^{2} / \left(\left(x - 1\right) \left(x - 3\right)\right) = {2}^{2} / \left({0}^{-} \cdot \left(- 2\right)\right) =$

$= - {2}^{2} / \left(0 \cdot \left(- 2\right)\right) = \frac{4}{0 \cdot 2} = \frac{4}{0} = + \infty$ Vertical asymptote for $x = 1$

Note: for $x - 1$ since $x$ is slightly lower than 1 the result will be something a bit lower than 0, so the sign will be negative, hence the note ${0}^{-}$ which later translates to a negative sign.

Confirmation for asymptote $x \to {1}^{+}$

${\lim}_{x \to {1}^{+}} f \left(x\right) = {\lim}_{x \to {1}^{+}} {\left(x + 1\right)}^{2} / \left(\left(x - 1\right) \left(x - 3\right)\right) = {2}^{2} / \left({0}^{+} \cdot \left(- 2\right)\right) =$

$= {2}^{2} / \left(0 \cdot \left(- 2\right)\right) = - \frac{4}{0 \cdot 2} = - \frac{4}{0} = - \infty$ Confirmed

Asymptote $x \to {3}^{-}$

${\lim}_{x \to {3}^{-}} f \left(x\right) = {\lim}_{x \to {3}^{-}} {\left(x + 1\right)}^{2} / \left(\left(x - 1\right) \left(x - 3\right)\right) = {3}^{2} / \left(2 \cdot {0}^{-}\right) =$

$= - {3}^{2} / \left(2 \cdot 0\right) = - \frac{9}{0} = - \infty$ Vertical asymptote for $x = 3$

Confirmation for asymptote $x \to {3}^{+}$

${\lim}_{x \to {3}^{+}} f \left(x\right) = {\lim}_{x \to {3}^{+}} {\left(x + 1\right)}^{2} / \left(\left(x - 1\right) \left(x - 3\right)\right) = {3}^{2} / \left(2 \cdot {0}^{+}\right) =$

$= {3}^{2} / \left(2 \cdot 0\right) = \frac{9}{0} = + \infty$ Confirmed

• Horizontal asymptotes

Find both limits as the function tends to $\pm \infty$

Minus infinity $x \to - \infty$

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} {\left(x + 1\right)}^{2} / \left(\left(x - 1\right) \left(x - 3\right)\right) =$

$= {\lim}_{x \to - \infty} \frac{{x}^{2} + 2 x + 1}{{x}^{2} - 4 x - 3} = {\lim}_{x \to - \infty} \frac{{x}^{2} \left(1 + \frac{2}{x} + \frac{1}{x} ^ 2\right)}{{x}^{2} \left(1 - \frac{4}{x} - \frac{3}{x} ^ 2\right)} =$

$= {\lim}_{x \to - \infty} \frac{\cancel{{x}^{2}} \left(1 + \frac{2}{x} + \frac{1}{x} ^ 2\right)}{\cancel{{x}^{2}} \left(1 - \frac{4}{x} - \frac{3}{x} ^ 2\right)} = {\lim}_{x \to - \infty} \frac{1 + \frac{2}{x} + \frac{1}{x} ^ 2}{1 - \frac{4}{x} - \frac{3}{x} ^ 2} =$

$= \frac{1 + 0 + 0}{1 - 0 - 0} = 1$ Horizontal asymptote for $y = 1$

Plus infinity $x \to + \infty$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} {\left(x + 1\right)}^{2} / \left(\left(x - 1\right) \left(x - 3\right)\right) =$

$= {\lim}_{x \to + \infty} \frac{{x}^{2} + 2 x + 1}{{x}^{2} - 4 x - 3} = {\lim}_{x \to + \infty} \frac{{x}^{2} \left(1 + \frac{2}{x} + \frac{1}{x} ^ 2\right)}{{x}^{2} \left(1 - \frac{4}{x} - \frac{3}{x} ^ 2\right)} =$

$= {\lim}_{x \to + \infty} \frac{\cancel{{x}^{2}} \left(1 + \frac{2}{x} + \frac{1}{x} ^ 2\right)}{\cancel{{x}^{2}} \left(1 - \frac{4}{x} - \frac{3}{x} ^ 2\right)} = {\lim}_{x \to + \infty} \frac{1 + \frac{2}{x} + \frac{1}{x} ^ 2}{1 - \frac{4}{x} - \frac{3}{x} ^ 2} =$

$= \frac{1 + 0 + 0}{1 - 0 - 0} = 1$ Horizontal asymptote for $y = 1$

Note: it just so happens that this function has a common horizontal for both $- \infty$ and $+ \infty$. You should always check both.

• Oblique asymptotes

You must first find both limits:

${\lim}_{x \to \pm \infty} f \frac{x}{x}$

For each, if this limit is a real number, then the asymptote exists and the limit is its slope. The $y$ intercept of each is the limit:

${\lim}_{x \to \pm \infty} \left(f \left(x\right) - m \cdot x\right)$

However, to save us the trouble, you can use some function "knowledge" to avoid this. Since we know $f \left(x\right)$ has horizontal asymptote for both $\pm \infty$ the only way of having an oblique is having another line as $x \to \pm \infty$. However, $f \left(x\right)$ is a $1 - 1$ function so there can't be two $y$ values for one $x$, hence a second line is impossible, so it's impossible to have oblique asymptotes.