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How do you find the coordinates of one of the vertices for the parabola ysquared-6x plus 4y plus 16 equals 0?
JeroGeronimo
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).
Wiki User
∙ 11y ago
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