What is the vertices of the hyperbola with the equation (16x^2) - (4y^2) = 64?

recall that x^2/a^2 - y^2/b^2 = 1

has vertices at (±a,0)
Your hyperbola can be written as
x^2/4 - y^2/16 = 1
so its vertices are at (±2,0)

Did you know?

Did you know that the vertices of a hyperbola can be found by manipulating its equation? In the equation (16x^2) - (4y^2) = 64, the vertices can be identified by first rearranging the terms to isolate the x or y variable. By dividing both sides by 64, we obtain the equation x^2/4 - y^2/16 = 1. The standard form for a hyperbola equation is (x - h)^2/a^2 - (y - k)^2/b^2 = 1, where (h, k) represents the coordinates of the center and a and b are the distances from the center to the vertices. Comparing this with our equation, we can conclude that the center is at (0, 0). By taking the square root of the denominators, we find that a = 2 and b = 4. Therefore, the vertices of this hyperbola are located at (0 ± a, 0), which gives us the points (2, 0) and (-2, 0).