SOLUTION: f(x)=x^2-6x+12 g(x)=k In the equation above, f(x)>=g(x) for all real numbers x. If k is a constant, what is the maximum value of k?

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Question 1043305: f(x)=x^2-6x+12
g(x)=k
In the equation above, f(x)>=g(x) for all real numbers x. If k is a constant, what is the maximum value of k?

Answer by josgarithmetic(39623)   (Show Source): You can put this solution on YOUR website!




Maybe better, expect k to be the vertex, minimum value for f(x).

to complete-the-square;

Vertex is (3,3).
The line g(x)=k must be no higher than y=3. The MAXIMUM value for k is 3.

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