Zeros and the minimum of the graph of a quadratic equation

In the xy-coordinate plane, the graph of the equation y = 3x² - 12x - 36 has zeros at x = a and x = b, 
where a < b.  The graph has a minimum at (c , -48).  What are the values of a, b, and c?
A.)  a = 2, b = 4, and c = 2
B.)  a = -2, b = 6, and c = 2
C.)  a = -3, b = 3, and c = 0
D.)  a = 3, b = 6, and c = 2

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y = 3x² - 12x - 36
3x² - 12x - 36 = 0   {substituted 0 for y}
3(x² - 4x - 12) = 0   {factored out the greatest common factor, 3}
x² - 4x - 12 = 0   {divided each side by 3}
(x - 6)(x + 2) = 0   {factored into two binomials}
x - 6 = 0   or   x + 2 = 0   {set each factor equal to 0}
x = 6   or   x = -2   {added 6 and subtracted 2, respectively}

Since a < b
a = -2 and b = 6

The minimum, is the y-coordinate of the vertex, -48.  To find the the value of the x-coordinate of the
vertex, c, substitute -48 in for y into the equation y = 3x² - 12x - 36 and solve for x.

-48 = 3x² - 12x - 36
3x² - 12x + 12 = 0   {added 48 to each side}
3(x² - 4x + 4) = 0   {factored out the greatest common factor, 3}
x² - 4x + 4 = 0   {divided each side by 3}
(x - 2)(x - 2) = 0   {factored into two binomials}
x - 2 = 0   {set the factor equal to 0}
x = 2   {added 2 to each side}

Therefore, the x-coordinate of the vertex, c, is 2.

B.)  a = -2, b = 6, and c = 2

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