Find all asymptotes of y=dfrac1ax²+bx+c, for real

Find all asymptotes of $y=\dfrac1{ax^2+bx+c}$, for real constants $a,b,c$, in every case.

Step 1: Horizontal asymptote

Whenever the denominator is non-constant ($a\neq0$, or $a=0,\ b\neq0$), it grows without bound as $x\to\pm\infty$, so $y\to0$; also $y/x\to0$, so there is no oblique asymptote in any case:

\[ y = 0. \tag{1} \]

Step 2: Vertical asymptotes — case analysis on $D=b^2-4ac$

$a\neq0,\ D\gt 0$: two distinct real roots $x_{1,2}=\dfrac{-b\pm\sqrt D}{2a}$, each a vertical asymptote (the sign of $y$ flips across each root).

$a\neq0,\ D=0$: a double root $x_0=-b/(2a)$; since $ax^2+bx+c=a(x-x_0)^2\to0$ with one fixed sign from both sides, $y\to\pm\infty$ (same sign on both sides) — one vertical asymptote.

$a\neq0,\ D\lt 0$: no real roots; the denominator never vanishes — no vertical asymptote.

$a=0,\ b\neq0$: $y=1/(bx+c)$, one vertical asymptote at $x=-c/b$.

$a=b=0,\ c\neq0$: $y=1/c=\text{const}$ — a constant function, with no asymptotes at all.

Step 3: Verification

\[a=1,b=-3,c=2\ (D=1\gt 0)\text{: roots }x=1,2\text{, and }y\to0\text{ as }x\to\pm500\ (y\approx4\times10^{-6})\ \checkmark\]

\[a=1,b=-2,c=1\ (D=0)\text{: double root }x=1,\ \lim_{x\to1}y=+\infty\text{ from both sides}\ \checkmark\]

\[a=1,b=0,c=1\ (D=-4\lt 0)\text{: no real root, }y(x)\text{ finite for all tested }x\text{, and }y\to0\text{ at }\pm\infty\ \checkmark\]