#y=-1/2x^2# is a quadratic equation in standard form:
#y=ax^2+bx+c#,
where:
#a=-1/2#, #b=0#, and #c=0#.
You need the vertex, which is the maximum or minimum point of the parabola, and other points. Hint, since #a<0#, the vertex will be the maximum point and the parabola will open downwards.
The axis of symmetry is the #x#-value of the vertex. For a quadratic equation in standard form, the formula for determining the axis of symmetry is #x=(-b)/(2*a)#.
#x=(-0)/(-2 1/2)=0#
To find the #y#-value for the vertex, substitute #0# for the value of #x# and solve for #y#.
#y=-1/2(0)^2=0#
Vertex: #(0,0)#
Since the vertex is at the origin, and the parabola opens downward, the parabola will not cross the x- or y- axes. So we will need to choose values for #x# and solve for #y# to determine several points.
#x=1,##y=-1/2(1)^2=-1/2#
Point: #(1,-1/2)#
#x=-1,##y=-1/2(-1)^2=-1/2#
Point: #(-1,-1/2)#
#x=2,##y=-1/2(2)^2=-1/2(4)=-2#
Point: #(2,-2)#
#x=-2,##y=-1/2(-2)^2=-1/2(4)=-2#
Point: #(-2,-2)#
#x=4,##y=-1/2(4)^2=-1/2(16)=-8#
Point: #(4,-8)#
#x=-4,##y=-1/2(-4)^2=-1/2(16)=-8#
Point: #(-4,-8)#
Plot the points and sketch a parabola through them. Do not connect the dots.
graph{y=-1/2x^2 [-10, 10, -5, 5]}
