Given: 3y = -(x-2)(x-1)
Vertex form is: y = a(x - h)^2 + k; where the vertex is (h, k) and a is a constant.
Distribute the two linear terms:" "3y = -( x^2 - 3x + 2)
Divide by 3 to get y by itself: y = -1/3(x^2 - 3x + 2)
One method is to use completing of the square to put in vertex form:
Only work with the x terms: " "y = -1/3(x^2 - 3x) -2/3
Half the coefficient of the x term: -3/2
Complete the square: y = -1/3(x - 3/2)^2 - 2/3 + 1/3(3/2)^2
Simplify: y = -1/3(x - 3/2)^2 - 2/3 + 1/3 * 9/4
y = -1/3(x - 3/2)^2 - 8/12 + 9/12
y = -1/3(x - 3/2)^2 + 1/12
A second method is to put the equation in y = Ax^2 + Bx + C:
Distribute the given equation: 3y = -x_2 + 3x - 2
Divide by 3: " "y = -1/3 x^2 + x -2/3
Find the vertex x = -B/(2A) = -1/(-2/3) = -1/1 * -3/2 = 3/2
Find the y of the vertex: y = -1/3 *(3/2)^2 + 3/2 - 2/3
y = -1/3 *9/4 + 9/6 - 4/6 = -9/12 + 5/6 = -9/12 + 10/12 = 1/12
Vertex form is: y = a(x - h)^2 + k; where the vertex is (h, k) and a is a constant.
y = a(x - 3/2)^2 + 1/12
Find a by inputting a point into the equation. Use the original equation to find this point:
Let x = 2, " " 3y = -(2-2)(2-1); " " 3y =0; " " y = 0
Use (2, 0) and substitute it into y = a(x - 3/2)^2 + 1/12:
0 = a(2 - 3/2)^2 + 1/12
-1/12 = a (1/2)^2
-1/12 = a 1/4
a =( -1/12)/(1/4) = -1/12 * 4/1 = -1/3
vertex form: y = -1/3(x-3/2)^2 + 1/12
