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y>=-1/4

Restate the question: "What is the range of the function f(x) =x2+5x+6?".

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Solve the quadratic x2+5x+6=0 <=> (x+2)(x+3) = 0 <=> x=-2 or x=-3. The minimum value is at x = (-2+-3)/2 = -5/2. The minimum value will be f(-5/2) = (-5/2)2+5(-5/2)+6 = 25/4-25/2+6 = (25-50+24)/4 = -1/4.

The range (or set of possible values) is f(x)>= -1/4 ... all numbers greater than or equal to -1/4.

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Q: What are the possible values for x2 5x6?
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