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I have this function. I noticed that it can be written as $$xy^2 +y(1-x^2) +(x-2) = 0,$$ so this is a quadratic in $y$. Thus $$y=\frac{(x^2-1) \mp \sqrt{(x^2-1)^2 -4x(x-2)}}{2x}$$ So I notice that this is not actually a function since it maps two $y$ values given one $x$ value. Anyway, the domain is $x \neq 0$, and from the expression inside the square root I find: $x^4 -6x^2 +8x +1 \geqslant 0$ which I can't solve. I don't know how to find the range either. From graphing this on WolframAlpha it gives the domain but not the range. How did it find the domain? And what about the range? Can somebody help? Thanks in advance.

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Actually, $x$ can equal zero. Just substitute $x=0$ in your original equation and you get $y=2$. In your quadratic expression for $y$, setting $x=0$ give you the indeterminate $\frac 00$: that's what confused you.

I used my TI-Nspire CX calculator to get the same approximate domain that WolframAlpha gave you:

$$x\le -2.93432\quad\text{or}\quad x\ge -0.115088$$

Note that $x=0$ is included in that domain. This graph agrees with that domain.

enter image description here

To find the range, do the same trick you did with $x$ for $y$: change your equation into a quadratic in $x$. We get

$$(-y)x^2+(y^2+1)x+(y-2)=0$$

The discriminant for this quadratic is

$$(y^2+1)^2-4(-y)(y-2)\ge 0$$

This equation has one easy exact answer, $y=1$. My TI-Nspire CX gives as the values of $y$ as

$$y\le 0.139681\quad\text{or}\quad y\ge 1$$

That also agrees with the graph. Note that $y=0,x=2$ is also included, though the quadratic again would have $y$ in the denominator.

Note also that exact values for the domain and range are possible. SageMath gives me formulas for the exact values, but they are long formulas with square roots and cube roots. The exact values do not look useful.

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