SOLUTION: The graph of y=ax^2+bx+c cannot have points in the third or fourth quadrants if
a) b^2-4ac>0 and b>a
b) b^2-4ac>0 and b<0
c) b^2-4ac=0 and a<0
d) b^2-4ac<0 and a<0
e) b^2-4ac
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: The graph of y=ax^2+bx+c cannot have points in the third or fourth quadrants if
a) b^2-4ac>0 and b>a
b) b^2-4ac>0 and b<0
c) b^2-4ac=0 and a<0
d) b^2-4ac<0 and a<0
e) b^2-4ac
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Question 625837: The graph of y=ax^2+bx+c cannot have points in the third or fourth quadrants if
a) b^2-4ac>0 and b>a
b) b^2-4ac>0 and b<0
c) b^2-4ac=0 and a<0
d) b^2-4ac<0 and a<0
e) b^2-4ac<0 and a>0 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The graph of y=ax^2+bx+c cannot have points in the third or fourth quadrants if
a) b^2-4ac>0 and b>a
b) b^2-4ac>0 and b<0
c) b^2-4ac=0 and a<0
d) b^2-4ac<0 and a<0
e) b^2-4ac<0 and a>0
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Ans: e because no x-intercepts and the parabola opens up
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Cheers
Stan H.
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