In Desmos, function fills entire screen unless $y$ is used in some denominator In Desmos, I am graphing the equation
$$(g^2 - 4x)y^2 + (2g - 4x - 4)y + 1 = 0$$
which uses the helper function
$$g = x + 1 - \frac{x^2 - 2x + 1}{x + 1 + 2\sqrt x}$$
which fills up the entire screen, even though it should be a smooth curve. Strangely, when I add $y$ in any denominator of the main function, say setting the function equal to $0/y$ instead of $0$, the function works just fine and is displayed correctly.
Buggy Curve:

Intended Curve:

I would think that having $y$ in the denominator like this would change things when $y=0$, but without it, the whole screen is filled up (every $y$ value.) I can't figure out the reason that this might be.

Here are the equation sources to paste into desmos:

Helper Function: g=x+1-\frac{x^{2}-2x+1}{x+1+2\sqrt{x}}
Buggy Equation: \left(g^{2}-4x\right)\cdot y^{2}+\left(2g-4x-4\right)\cdot y+1=0
Fixed Equation: \left(g^{2}-4x\right)\cdot y^{2}+\left(2g-4x-4\right)\cdot y+1=\frac{0}{y}


There are quite some redundancy in the way your implicit function was formulated. Note that the rational function in $g$ can be simplified.
$$
\frac{x^2 - 2x + 1}{x + 1 + 2\sqrt x}= \left(\frac{x-1}{\sqrt{x} + 1} \right)^2 = \left(\frac{(\sqrt{x} + 1)(\sqrt{x} - 1)}{\sqrt{x} + 1} \right)^2 = (\sqrt{x} - 1)^2
$$
Put this back into $g$ we have
$$g=x+1-(x-2\sqrt{x} + 1) = 2 \sqrt{x}$$
The curve $(g^2 - 4x)y^2 + (2g - 4x - 4)y + 1 = 0$ therefore goes from an implicit function to an explicit function.
$$\begin{align}
&\implies 0 \cdot y^2 +(4\sqrt{x}-4x-4)y+1=0 \\
&\implies y = \frac14 \frac1{x-\sqrt{x}+1}
\end{align}$$
