Implicit differentiation results in no x I believe I understand what to do until a point, but I'm looking for a way to wrap my head around the last step.
Curve:
$x+y^2-y=1$
I'm asked to find a) All points where a vertical tangent line occurs, b) All points where a horizontal tangent line occurs.
I can differentiate it, resulting in
$y'(x)=-\frac{1}{2y-1}$
There would be a horizontal asymptote as the function approaches infinity, therefore SOMETHING equals zero, and a vertical asymptote at .5.
What I can't understand is if those points I listed above would count as x's or y's, and how I could actually obtain the coordinates (x,y). I know I should plug in the result into the original equation, but, again, I don't understand the result.
Therefore, I don't understand the function itself since there is no x.
Any help is greatly appreciated, thank you! 
 A: It can be dangerous to go all the way to solving for $y'$ without pausing at your division steps to see what is going on.   So,  rewinding a step and doing implicit differentiation, we have 
$1+2yy'-y'=0$,
$(2y-1)y'=-1$
Now,  if $y'=0$, this equation would have no solutions, as it leads to a contradiction, $0=-1$.   Therefore, no such points exist.  For vertical tangent lines, you look for when $y'$ doesn't exist,  which as you noted occurs at $y=\frac 1 2$.
Now, to find the points on the curve where this happens, we plug that value in for y and see that $x+{\frac 1 2}^2-\frac 1 2 =1$,  simplifying we have $x=\frac 5 4$,  so the point on the graph is $(\frac 5 4 ,\frac 1 2)$ we have a vertical tangent line
I actually just covered this in my own class 3 hours ago :)
A: You may view this curve as the graph of some function by writing the equation in the form $$x = f(y).$$
There are methods for general implicit equations $F(x, y) = 0$ involving the gradient of $F$ but no such technique is required here.
