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

Question

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

GMATBusters · Accepted Answer

I. f(4) > 0
It means a part of the curve is above the x axis at x = 4. But, No info whether
a) the curve is above the x axis

or

b) intersect it in more points.

So, NOT SUFFICIENT.

II. a < 0
It means the parabola, extends downwards but 2 cases are possible.
a) it can intersects

or
b) does not intersect the x axis.

So, NOT SUFFICIENT.

Combing Statement 1 & 2, We get a part of the curve is above the x axis and it extends downwards.
So, It must intersects the x axis.

Hence SUFFICIENT

So, Answer is C

II. a < 0

kiran120680

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

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

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eabhgoy · Answer

Could anyone help with the solution?

Thanks in advance.

eabhgoy · Answer

Thank you  gmatbusters !

GMATBusters · Answer

Algebraic approach

Gladiator59 · Answer

Algebraic solution attached. Similar to what  gmatbusters  has outlined.

pielkay · Answer

chetan2u   VeritasKarishma   Bunuel , could you help with the reasoning for this, please?

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Siddharthjain07 · Answer

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

Posted from my mobile device

KarishmaB · Answer

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. IMG_1843.JPG 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. IMG_3399.JPG 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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Does the curve f(x)=y=ax^2+c intersect with the x - axis?

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

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