In the xycoordinate 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
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 ycoordinate of the vertex, 48. To find the the value of the xcoordinate 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 xcoordinate of the vertex, c, is 2. B.) a = 2, b = 6, and c = 2 Ask Algebra House Comments are closed.

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