Home Competitions AHSME 1951 Problem 16
If in applying the quadratic formula to a quadratic equation
f ( x ) ≡ a x 2 + b x + c = 0 , f(x) \equiv ax^2 + bx + c = 0, f ( x ) ≡ a x 2 + b x + c = 0 ,
it happens that c = b 2 4 a c = \frac{b^2}{4a} c = 4 a b 2 , then the graph of y = f ( x ) y = f(x) y = f ( x ) will certainly:
( A ) h a v e a m a x i m u m ( B ) h a v e a m i n i m u m \mathrm{(A) \ have\ a\ maximum } \mathrm{(B) \ have\ a\ minimum} ( A ) have a maximum ( B ) have a minimum ( C ) b e t a n g e n t t o t h e x − a x i s \mathrm{(C) \ be\ tangent\ to\ the\ x-axis} ( C ) be tangent to the x − axis ( D ) b e t a n g e n t t o t h e y − a x i s \mathrm{(D) \ be\ tangent\ to\ the\ y-axis} ( D ) be tangent to the y − axis ( E ) l i e i n o n e q u a d r a n t o n l y \mathrm{(E) \ lie\ in\ one\ quadrant\ only} ( E ) lie in one quadrant only
Dataset text NuminaMath-1.5 . Quoted with source.
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