# What are the asymptotes for (x^2 - 2x - 3 )/(-4x) ?

Jun 3, 2016

Vertical asymptote at $x = 0$
Slant asymptote given by $y = - \frac{x}{4} + \frac{1}{2}$

#### Explanation:

In a polynomial fraction $f \left(x\right) = \frac{{p}_{n} \left(x\right)}{{p}_{m} \left(x\right)}$ we have:

1) vertical asymptotes for ${x}_{v}$ such that ${p}_{m} \left({x}_{v}\right) = 0$
2) horizontal asymptotes when $n \le m$
3) slant asymptotes when $n = m + 1$
In the present case we have ${x}_{v} = 0$ and $n = m + 1$ with $n = 2$ and $m = 1$

Slant asymptotes are obtained considering (p_n(x))/(p_{n-1}(x)) approx y = a x+b  for large values of $\left\mid x \right\mid$

In the present case we have

$\frac{{p}_{n} \left(x\right)}{{p}_{n - 1} \left(x\right)} = \frac{{x}^{2} - 2 x - 3}{- 4 x}$
${p}_{n} \left(x\right) = {p}_{n - 1} \left(x\right) \left(a x + b\right) + {r}_{n - 2} \left(x\right)$
${x}^{2} - 2 x - 3 = \left(- 4 x\right) \left(a x + b\right) + c$

equating coefficients

{ (-3 - c=0), (-2 + 4 b=0), (1 + 4 a=0) :}

solving for $a , b , c$ we get $\left\{a = - \left(\frac{1}{4}\right) , b = \frac{1}{2} , c = - 3\right\}$
and substituting

$y = - \frac{x}{4} + \frac{1}{2}$