Question 435518
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
y= .5x^2 -5  
1) Parabola 
2)Parabola with vertex Pt(0,-5)
3)  y-axis the line of symmetry
4) Parabola Opens Upward a = .5 >0
5) x-intercepts(when y = 0) are x =  ± {{{sqrt(10)}}}
{{{drawing(300,300,   -6, 6, -6, 6, grid(1),
circle(3.16, 0,0.3),
circle(-3.16, 0,0.3),
circle(0, -5,0.3),
graph( 300, 300, -6, 6, -6, 6,0,.5x^2-5))}}}