[tex]The\ vertex\ form:y=a(x-h)^2+k\\\\vertex\ (h;\ k)\to(-6;\ 0)\\\\then:y=a[x-(-6)]^2+0=a(x+6)^2\\\\graph\ is\ half\ as\ tall\ as\ y=x^2\ and\ o\ pens\ upside\ down\ then\ a=-2\\\\Answer:y=-2(x+6)^2[/tex]
If I understood correctly "graph is half as tall as y=x^2".
