Did I write the equation in standard form of a parabola correctly?

Question

Put the equation into the standard form of a parabola by completing the square.

Given equation: 12y=5x^2-60x+264

12y/12=5x^2/12-60x/12+264/12

y=5x^2/12-5x+22

Answer 1

12y = 5x^2 - 60x + 264

y = 5/12 x^2 - 5x + 22

y = 5/12 ( x^2 - 12x ) + 22

y = 5/12( x^2 - 12x + 36) + 22 - 15

y = 5/12 ( x-6)^2 + 7 <---- the axis of symmetry is x=6; the vertex is (6,7)

STANDARD FORM:

y = (5/12)x^2 - 5x + 22

GOOD JOB