QuestionConvert the hyperbola equation $-16x^{2}+y^{2}+128x-272=0$ to standard form and graph it.

Studdy Solution
To graph the hyperbola, we need to find the center, vertices, and asymptotes. The center is at (h, k), which is (4,0) in this case. The distance from the center to each vertex is a, which is $\sqrt{33}$ in this case. The equations of the asymptotes are y = ±(b/a)(x - h) + k, which simplifies to y = ±$\sqrt{528/33}$(x -4).
The hyperbola opens to the left and right because the x term is positive in the standard form of the equation. The vertices are at (4 - $\sqrt{33}$,0) and (4 + $\sqrt{33}$,0). The asymptotes are the lines y = ±$\sqrt{528/33}$(x -4).