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convert y^2-8x-4=0 to vertex form

Sagot :

The vertex form of a parabola that opens to the right (or left) is x = a (y-k)^2 + h , where
(h,k) is the vertex

y^2 - 8x - 4 = 0

y^2 = 8x - 4

(y-0)^2 = 8(x - 1/2)

1/8 (y-0)^2 = x - 1/2

x= 1/8 (y-0)^2 + 1/2

Therefore, the vertex of the parabola is at (1/2, 0).

Note: If the parabola opens upward or downward, the vertex form would look like:
y = a (x-h)^2 + k , where (h,k) is the vertex