# 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!

• Would the y, in this case, simply count as x? That would make sense to me because the vertical asymptote is at a point x in all other cases. Oct 6, 2014 at 22:42

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 :)

• Awesome answer! I did exactly what you explained but simply thought it was wrong because of my x and y confusions. Thank you very much! Oct 6, 2014 at 23:19
• not a problem, I could see you were on the right track. Always stop before dividing by variables in these equations to check for problems with 0 :)
– Alan
Oct 6, 2014 at 23:20

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.