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Thank you gmatbusters!
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Algebraic approach
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Algebraic solution attached. Similar to what gmatbusters has outlined.
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chetan2u VeritasKarishma Bunuel, could you help with the reasoning for this, please?

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gmatbusters can u help in clarifying how did u get c as some positive value?

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kiran120680
Does the curve f(x)=y=ax^2+c intersect with the x - axis?

I. f(4) > 0
II. a < 0

y = ax^2 + c is a quadratic so its graph will be a parabola.

I. f(4) > 0
This means when x = 4, y > 0. So at x = 4, the graph lies above the x axis. It can be done in many ways as shown.

Attachment:
IMG_1843.JPG
IMG_1843.JPG [ 2.5 MiB | Viewed 5662 times ]

II. a < 0
This means that ax^2 is negative so graph is downward opening. Again, it can be done in many ways as shown.
Attachment:
IMG_3399.JPG
IMG_3399.JPG [ 2.82 MiB | Viewed 5646 times ]

Using both, it should lie above x axis at x = 4 but should be downward opening. Hence, only one case is possible. (No 3 in first diagram and no 1 in second diagram, both same)

So the graph will cut x axis.

Answer (C)
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