Does $y = \dfrac{ax^2 + bx + c}{dx + e}$ have any lines of symmetry. If it does, what are they, and if not, how would one prove that it doesn't have any lines of symmetry?
(Please consider the general case. And please ignore trivial cases where the expression reduces to a quadratic expression in $x$, linear expression in $x$, or a constant.)
I have written this solution independently of the solution of @Steven Gregory, with which it shares many common points . The differences are that
1) I have tried to concentrate on the main ideas, escaping computational details.
2) I am guided by the common general shape of the curves, an example of which I give below
3) I consider the exceptional case where the function is in fact affine.
Assuming $a \neq 0$ and $c \neq 0$, Euclidean division of the numerator by the denominator gives :
$$ax^2+bx+c=(Ax+B)(dx+e)+C \ \ \text{for certain real numbers} \ \ A,B,C.$$
Thus
$$f(x)=y=Ax+B+\dfrac{C}{dx+e} \ \ \text{with} \ \ A:=\frac{a}{d} \neq 0 \ \ (1)$$
Two cases:
Either: $C=0$ (which happens when $dx+e$ exactly divides $ax^2+bx+c$) in this case, the graphical representation of $f$ is a straight line which is its own symmetry axis and has an infinite number of axes of symmetry in the perpendicular direction...
Or $C\neq0$: we recognize here the equation of a hyperbola having a vertical asymptote with equation $x=-e/d$ and oblique (slant) asymptote with equation $y=Ax+B$. See graphics below. Thus the graphical representation has two axis of symmetry.
Explanation about the fact that the graphical representation of (1) is a hyperbola : let us write (1) under the form $(y-Ax-B)(cx+d)=C$ and make the affine change of variables $X=y-Ax-B, Y=cx+d$ (this does not modify the nature of the conical section) thus we obtain $XY=C$, which is one of the canonical forms for the equation of a hyperbola.
The example of $y=0.5x-1+\epsilon(x)$ with $\epsilon(x)=\frac{2}{x-1}\rightarrow 0 \ $ when $x \rightarrow \infty$, proving that $y=0.5x-1$ is the equation of the oblique asymptote (vertical asymptote at $x=1$):
$y=\dfrac{ax^2 + bx + c}{dx+e}$
can be rewritten as
$ax^2 - dxy + 0y^2 + bx + ey + c = 0$
which is known to be the equation of a conic section.
Comparing this to $Ax^2 + Bxy + Cy^2 + Dx + Ey + F$
we compute the descriminant to be $B^2 - 4AC = d^2 \gt 0$.
This implies that you have the equation of a hyperbola. Hence it will have two axis of symmetry.
$y=\dfrac{ax^2 + bx +c}{dx+e}$
\begin{array}{ccccccc} &&&& \dfrac adx &+& \dfrac{bd-ae}{d^2}\\ & &--&--&---&-&-----\\ dx + e&|&ax^2 &+& bx &+& c\\ & &ax^2 &+& \dfrac{ae}{d}x \\ & &---&-& -----\\ & && &\dfrac{bd-ae}{d}x &+& c \\ & && &\dfrac{bd-ae}{d}x &+& \dfrac{bde-ae^2}{d^2} \\ & && &-----&-&-------\\ & &&&&& \dfrac{cd^2 - bde + ae^2}{d^2} \\ \end{array}
We find $y = \dfrac adx + \dfrac{bd-ae}{d^2} + \dfrac{cd^2 - bde + ae^2}{d^2(dx+e)}$
Which admits to the vertical asymptote $dx + e = 0$ and the slant asymptote $adx-d^2y + (bd - ae) = 0$
The two asymptotes intersect at the point P = $\left(-\dfrac ed, \dfrac{bd - 2ae}{d^2}\right)$.
The change of coordinates \begin{array}{l} x = u - \dfrac ed\\ y = v + \dfrac{bd - 2ae}{d^2} \end{array}
translates the point $(x,y) = P$ to the point $(u,v) = (0,0)$
The asymptotes become $u = 0$ and $au -dv = 0$
The two axis of symmetry are the two lines that bisect the vertical angles formed by the two asymptotes. Their equations are described by
\begin{align} |u| &= \dfrac{|au-dv|}{\sqrt{a^2+d^2}} \\ au - dv &= \mp \sqrt{a^2 + d^2}\;u \\ (a \pm \sqrt{a^2 + d^2})u - dv &= 0 \\ v &= \dfrac{a \pm \sqrt{a^2 + d^2}}{d} u \end{align}
Translating back into $(x,y)$ coordinates, we get
\begin{align} y - \dfrac{bd - 2ae}{d^2} &= \dfrac{a \pm \sqrt{a^2 + d^2}}{d}(x + \dfrac ed) \\ y- \dfrac{bd - 2ae}{d^2} &= \dfrac{a \pm \sqrt{a^2 + d^2}}{d}x + \dfrac{e(a \pm \sqrt{a^2 + d^2})}{d^2}\\ y &= \dfrac{a \pm \sqrt{a^2 + d^2}}{d}x + \dfrac{-ea + bd \pm e\sqrt{a^2 + d^2})}{d^2} \end{align}
So the two axis of symmetry are $$y = \dfrac{a + \sqrt{a^2 + d^2}}{d}x + \dfrac{-ea + bd + e\sqrt{a^2 + d^2})}{d^2}$$
and
$$y = \dfrac{a - \sqrt{a^2 + d^2}}{d}x + \dfrac{-ea + bd - e\sqrt{a^2 + d^2})}{d^2}$$
There does exist a line of symmetry when $a =1, b = c = d = 0, e = 1$, and the line is $x = 0$.
