How do you write the vertex form equation of the parabola #y = -2(x+4)(x-2)#?

Question

Eli Assouline · Accepted Answer

Please see the explanation.

Given: #y = -2(x+4)(x-2)# Increase the multiplicities: #y = -2(x^2+4x-2x -8)# #y = -2(x^2+2x -8)# #y = -2x^2-4x +16" [1]"# Please observe that equation [1] is in the standard form #y = ax^2+bx+c# where #a = -2, b = -4 and c = 16# This kind of parabola's vertex form is: #y=a(x-h)^2+k" [2]"# Since the parabola's "a" in the standard form and "a" in the vertex form are the same, enter -2 in place of "a" in equation [2]: #y=-2(x-h)^2+k" [3]"# We know that #h = -b/(2a)#: #h = -(-4)/(2(-2)# #h = -1# Put a -1 in place of h in equation [3]: #y=-2(x--1)^2+k" [4]"# Evaluate equation [1] at #x = 0# #y = 16# After evaluating equation [4] and determining that x = 0 and y = 16, find k. #16=-2(1)^2+k# #k = 18# The form of the vertex is: #y=-2(x--1)^2+18" [5]"#

Eli Assouline · Answer

undefined To write the vertex form equation of the parabola $ y = -2(x+4)(x-2) $, first expand the expression and then complete the square to put it in the vertex form $ y = a(x-h)^2 + k $. The vertex form equation will be $ y = -2(x+1)^2 + 16 $.