The general equation for a hyperbola is
[tex]\frac{(y-h)^2}{b^2}\text{ + }\frac{(x-k)^2}{a^2}\text{ = 1}[/tex]
The equation we are considering is a hyperbola with the following parameters:
[tex]\begin{gathered} (k,\text{ h) = (-3 , 1)} \\ a^2\text{ = 9 } \\ \text{a = 3} \\ b^2\text{ = 16} \\ b\text{ = 4} \end{gathered}[/tex]
The co-vertices can be obtained by locating the vertices along the x-axis
[tex]\begin{gathered} (-3\text{ + 3, 1) and (-3 -3 , 1)} \\ (0,\text{ 1) and (-6 , 1)} \end{gathered}[/tex]
This corresponds to option B
