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#
