Finding the $x$-coordinate of the vertex of a hyperbola?

Question

How would I go about finding the $x$-coordinates for the vertex of the hyperbola $y^2 + y = x^2 + 4$?

Thank you!

Answer 1

Go about completing that square:

$$\frac{\left(y+\frac12\right)^2}{17/4}-\frac{x^2}{17/4}=1$$

We then know that hyperbolas of the form

$$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1$$

have vertexes at $x=k$ (because that is where the graph is symmetric) so for our case, the point in interest lies on $x=0$.