Given:#" "y=-2x^2+2x-3#......................(1)
Standard form:#" "y=ax^2+bx+c#
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#color(blue)("Determine y intercept")#
#color(blue)(y_("intercept")=c=-3#
(note: y intercept is at x=0 so #-2x^2=0" and "2x=0" leaving " y=-3#
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#color(blue)("Determine x-vertex")#
The coefficient of #x^2# is negative so the graph is of general shape #nn# Thus the vertex is a maximum.
Write #" "y=-2x^2+2x-3" as "y=-2(x^2color(red)( -x)) -3#
Using the coefficient of #color(red)(-x -> -1)#
Apply: #color(blue)(x_("vertex")=(-1/2)xx(color(red)(-1))=+1/2)#
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#color(blue)("Determine y-vertex")#
Substitute #x=1/2# into equation (1) giving
#y_("vertex")=-2(1/2)^2+2(1/2)-3#
#color(blue)(y_("vertex")=-2/4+2/2-3 = -5/2#
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#color(blue)("Vertex"->(x,y)->(1/2,-5/2))#
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#color(red)("As the graph is of shape "nn" and the vertex")#
#color(red)("is below the x-axis there are no x-intercepts")#
