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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5x6=30
x2≤64
The possible values for k are -2 and -14 because in order for the line to be tangent to the curve the discriminant must be equal to 0 as follows:- -2x-2 = x2-8x+7 => 6-x2-9 = 0 -14x-2 = x2-8x+7 => -6-x2-9 = 0 Discriminant: 62-4*-1*-9 = 0
30
x2 + 5x =6 x2+5x -6 = 0 (x+6)(x-1) = 0 x+ 6 = 0 x = -6 x-1 = 0 x = 1