Question 366485
Since the vertex occurs at ({{{-2}}},{{{1}}}), the line of symmetry is {{{x=-2}}}.
Since ({{{-4}}},{{{5}}}) is a point on the line, find the x value equidistant from the line of symmetry. 
The x-distance from ({{{-4}}},{{{5}}}) to the line of symmetry is {{{d[x]=-2-(-4)=2}}}.
Then add {{{d[x]=2}}} to the x position of the line of symmetry x position to get the point symmetric to ({{{-4}}},{{{5}}}).
{{{x[2]=-2+2=0}}}
{{{y[2]=5}}}
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({{{0}}},{{{5}}}) is another point on the parabola.
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{{{drawing(300,300,-5,5,-2,8,grid(1),circle(-4,5,0.2),circle(0,5,0.2),circle(-2,1,0.2),blue(line(-2,10,-2,-10)),graph(300,300,-5,5,-2,8,0))}}}