This is the vertex form of a quadratic. standard form #y=a(x+b/(2a))^2+c+k# where #k# is a correction due to the error produced by #a(b/(2a))^2# so we have #a(b/(2a))^2+k=0#
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Given: #y=2/3(xcolor(red)(-2))^2color(green)(+0.5)# where #0.5=k+c#
#color(blue)("General shape")#
Note that #2/3xx x^2>0# so the graph is of form #uu#
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#color(blue)("Determine the vertex")#
#x_("vertex")=(-1)xxcolor(red)(-2)= +2#
#y_("vertex")=color(green)(+5)#
#=>"Verex"->(x,y)=(2,5)#
Note that as the graph is of form #uu# and the vertex is above the x-axis then the graph does NOT cross the x-axis.
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#color(blue)("Determine "y" intercept")#
#y_("intercept")=2/3(color(red)(-2))^2color(green)(+0.5) = 19/6#
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#color(blue)("Method demonstration: determine "x" intercpts")#
Set #y=0=2/3(x-2)^2+0.5#
#(x-2)^2=3/2(-1/2) = -3/4#
#sqrt((x-2)^2)=sqrt(-3/4)#
#x=2+-sqrt(-3/4)#
You can go further with this using complex numbers. However, you do not need to as you are only plotting/sketching the graph.
