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What is the range of f(x)=(x+4)^2+7? 1 y>or equal to -4 2 y>or equal to 4 3 y=7 4 y>or equal to 7
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\[(x+4)^2\geq 0\] because it is a square that makes \[(x+4)^2+7\geq 7\]
the vertex is (-4,7) and the parabola is positive so there the vertex must be the minimum. so the range is the y>=7
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