PLEASE HELP WITH MY TRIG HW: Graph the hyperbola *BEST ANSWER AWARDED*?

-4x² + y² - 10y - 11 = 0

(y²-10y) - 4x² = 11

(y²-10y+5²) - 4x² = 5²+11

(y-5)² - 4x² = 36

(y-5)²/36 - x²/9 = 1

(y-5)²/6² - x²/3² = 1

This is a vertical hyperbola.

Center-radius equation for vertical hyperbola:

 (y-k)²/a² - (x-k)²/b² = 1

with

 center (h,k)

 vertices (h,k±a)

 co-vertices (h±b,k)

 asymptotes y = ±(a/b)x + k∓(a/b)h

Apply your equation.

h = 0

k = 5

vertices (0,5±6) = (0,11) and (0,-1)

co-vertices (0±3,5) = (3,5) and (-3,5)

asymptotes

 y = ±(6/3)x + 5∓(6/3)0 = 2x + 5

 y = -2x + 5

(see graph below)