Which quadratic function has its vertex at (-2,7) and opens down?

\( The\ vertex\ form: y=a(x-h)^2+k\\\\vertex(h;\ k)\to(-2;\ 7)\\\\o pens\ down\ then\ a < 0\\\\Answer: y=a(x+2)^2+7\ where\ a < 0. \)

\( The \ vertex \ form \ of \ the \ function \ is: \\ \\y = a(x - h)^2 + k\\ \\vertex = \ is \ the point \\ \\(h, k) = (-2,7)\\\\h=-2\\k=7 \)

\( y=a(x-(-2))^{2} +7\\\\y=a(x+2)^{2} +7 \\\\ opens \ up \ down \ for : \ a < 0 \\ For \ instance, \ letting \ a = -1 \ gives \ this \ parabola: \\\\y= -(x+2)^{2} +7 \)

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