Axis of Symmetry of a Generalised Quartic

Question

How can one determine the axis of symmetry of a generalised quartic in $x,y$ such as :

$$y-2x=(x+2y-1)(x+2y-2)(x+2y-3)(x+2y-4)$$

in particular if it is given in expanded form? This equation has been constructed from a nice upright symmetrical curve (which can be easily guessed by inspection) and then rotated. The objective is to see how the axis/axes of symmetry can be worked out from the equation.

Answer 1

$t=y-2x,u=x+2y$ define a similarity transform which restores the presumably original equation:

$$t=(u-1)(u-2)(u-3)(u-4).$$

By a further shift of $u$, on gets an even function

$$t=\left(v-\frac32\right)\left(v-\frac12\right)\left(v+\frac12\right)\left(v+\frac32\right)=\left(v^2-\frac14\right)\left(v^2-\frac94\right)=v^4-\frac52v^2+\frac9{16}.$$

Answer 2

Your curve could be parametrised by

$$x=\frac{-1}{4}((t-1)(t-2)(t-3)(t-4)-t)$$

write $\frac{dx}{dt}=0$

to find $t=x+2y$.