Given: #y = 1/2 x^2 + 2x -8#
Put the function in standard/vertex form: #y = a(x-h)^2 + k; " where the vertex is "(h, k)# and #a# is a constant.
Use completing of the square :
First put the #x# terms together and factor out the #1/2#, knowing that #4/2 = 2#:
#y = 1/2(x^2 + 4x) - 8#
Half the #x#-term: #1/2(4) = 2# to complete the square and subtract #1/2(2)^2 = 1/2(4)# because this was added when the square was completed, since #1/2(x+2)^2 = 1/2(x^2+4x+4) #:
#y = 1/2(x + 2)^2 - 8 - 1/2(2)^2#
Simplify:
#y = 1/2(x + 2)^2 -10; " vertex"(-2, -10)#
Find #x#-intercepts using the quadratic formula,
where #x = (-B +-sqrt(B^2 - 4AC))/(2A); "when " y = Ax^2 + Bx + C = 0#
#x = (-2 +-sqrt(4 - 4(1/2)(-8)))/(2*1/2) = -2 +- sqrt(20) = -2 +- 2 sqrt(5)#
Find other points such as the #y#-intercept by setting #x = 0#:
#y = -8; " "y"-intercept":(0, -8)#
