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vipulgoyal
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If y = a(x - b)² + p intersects x-axis, then the intersection point will be of the form (k, 0)
So, 0 = a(k - b)² + p
So, (k - b)² = -p/a
Now, (k - b)² cannot be negative.
So, if -p/a is non-negative, then only the graph of the given equation can have intersection with x-axis.
So, the problem is basically asking "Is -p/a ≥ 0?"
Clearly, none of the statements are individually sufficient but both of them together is sufficient.
Answer : C
