How do you find the coordinates of one of the vertices for the parabola ysquared-6x plus 4y plus 16 equals 0?

Answer 1

The quadratic is "sideways" which has the form: x = ay2 + by + c,

or the vertex form: x = a(y - k)2 + h where (h, k) is the vertex.

y

2

- 6x + 4y + 16 = 0

y

2

+ 4y + 16 = 6x

y

2

+ 4y + 4 - 4 + 16 = 6x

(y + 2)2 + 12 = 6x

(1/6)(y + 2)2+ 2 = x

(1/6)[y -(-2)]2+ 2 = x

Thus, the vertex is (2, -2).