SOLUTION: if f(x)= 3x^2, what are all real values of a and b for which the graph of g(x)= ax^2+b os below the graph of f(x) for all values of x

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Question 1078221: if f(x)= 3x^2, what are all real values of a and b for which the graph of g(x)= ax^2+b os below the graph of f(x) for all values of x
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
if f(x)= 3x^2, what are all real values of a and b for which the graph of g(x)= ax^2+b is below the graph of f(x) for all values of x
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Since the vertex of f(x) is (0,0) and f(x) opens upward,
a must be negative to force g(x) to open down,
and b must be negative to assure the vertex is below (0,0)
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

For g%28x%29 to be below the graph of f%28x%29, f%28x%29 must be "bigger" than g%28x%29.
g%28x%29+=+ax%5E2+%2B+b+is below the graph of f%28x%29 if shifted+down and it will be for any b%3C0 ( b is any negative number)
and g%28x%29+=+ax%5E2+%2B+b is below the graph of f%28x%29 if a++%3C=+3 , multiplying by negative number will flip it upside down AND stretch it in the y-direction
so, answer is: for any b%3C0 , and for any a++%3C=+3