text: first you have to put the hyperbola in standard form. Divide everything by 4.\n\n9x^2 - y^2 = 1\nWhich is the same as\n(x^2)/(1/9) - (y^2)/1 = 1\nHere, a^2 = 1/9, so a = 1/3,\nand b^2 = 1, so b = 1\n\nIn this case, the center is (0,0)\nSince the "x" term is first, then it opens left and right. The vertices will be left and right from the center by "a" units, so the vertices are (1/3, 0) and (-1/3, 0).\n\nThe foci are left and right of the center by "c" units. In a hyperbola, a^2 + b^2 = c^2, so c^2 = (1/3)^2 + (1^2) = 1/9 + 1 = 10/9. c = root(10)/3. The foci will be (root10/3, 0) and (-root10/3, 0).\n\nThe asymptotes have slopes that are +b/a and -b/a, and pass through the center, in this case, (0,0). So these slopes are +1/(1/3) and -1/(1/3) or 3 and -3, and the equations are y = 3x and y = -3x.