How to calculate the asymptotes to $y^2+2y-4x^2=0$ Calculate the center and asymptotes to the hyperbola $y^2+2y-4x^2=0$, aswell as the intersections of the actual coordinate axis. 
The hyperbola takes the form $\left ( \frac{x}{\frac{1}{\sqrt{4}}}\right )^{2}-\left ( {y+1}\right )^{2}=-1$ and I understand how to complete all of the steps except how to find the asymptotes to the hyperbola. My course textbook states that they should be given by the formula $y=\pm \frac{b}{a}x$ if $(\frac {y}{b}^{2})=(\frac{x}{a})^{2}\pm1\approx (\frac{x}{a})^2$. But if I use this formula I get that $y+1=\pm \sqrt4 x $. What did I do wrong? 

I think your hyperbola has the form
\begin{align*}y^2 + 2y {\color{red}+\color{red}1\color{red}-\color{red}1}-4x^2 =0 &\Leftrightarrow (y+1)^2 -1- 4x^2= 0 \\&\Leftrightarrow (y+1)^2 - 4x^2=1
\end{align*}
So $a = \frac{1}{2}$ and $b= 1$.
Then you should get the asymptotes 
$$y = \pm 2x-1$$
