Quoted By:
1(×^2 + 2x + 1) -8 -1
Complete the square
(× + 1)^2 -9
Vertex at (-1, -9), and a positive primary coefficient so ×->inf as y -> inf, x-> -inf as y-> -inf, which makes -9 the minimum value for the range.
Range is [-9, inf), domain is all reals.
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