#color(blue)("Determine the general shape of the graph")#
If you multiply out the brackets you have #y=+1/2x^2+....#
As the coefficient of #x^2# is positive we have the general shape of #uu#. Thus the vertex is a minimum.
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#color(blue)("Determine the x-intercepts")#
The x-intercept is at #y=0#
This indicates that we have
#(x-4)=0=>x=+4#
#(x+2)=0=>x=-2#
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#color(blue)("Determine the vertex")#
#x_("vertex")# will be mid point between the x-intercepts so we have:
#x_("vertex")=(-2+4)/2=1#
Substitute for #x# to obtain #y_("vertex")#
#y_("vertex")=1/2(1-4)(1+2)=-9/2#
#y_("vertex")=-9/2#
Vertex#->(x,y)=(1,-9/2)#
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#color(blue)("Determine y-intercept")#
Consider the standard form #y=ax^2+bx+c#
The y-intercept is at #x=0# giving:
#y_("intercept")=c#
So from #y=1/2(x-4)(x+2) # we have:
# c=1/2xx(-4)xx2=1/2xx(-8) = -4#
#y_("intercept")->(x,y)=(0,-4)#
