Does the curve h(x)=y=ax^2+b intersects the x-axis?

I think the answer should be E.

the question asks us to check if -4ab is negative.

(1): 16a + b > 0

16(1) + 2 > 0; 16(1) - 2 > 0; 16(-1) + 17 > 0

SO, Not sufficient

(2) a>0. Nothing about b. So, -4ab can be positive or negative. Not sufficient.

(1) + (2): Still not sufficient since b can be + or -.

Ans: E.

Bunuel, please let us know if this approach is correct.

chetan2u, Bunuel

Dear experts: Please check if C is the correct answer. I think E should be the answer.

Regards,
Arup

Statement 1: 16a + b > 0

Both a and b are unknown and thus, insufficient

Statement 2: a > 0

This tells us ax^2 is always positive, however b could be negative. Insufficient

Both 1 and 2 together:

16a + b is positive and that 16a is positive.

B could be positive, no intersection or B could be a large negative number which could give us 1 or 2 intersections. Therefore insufficient.

E

Hey GMATaspirant641

After combining when we use a>0 in 16a + b > 0, both cases will be valid, b can be <0 or b can be > 0

*Editing it*

Now if -4ab < 0

This means that determinant can be greater than 0 or less than 0

Giving answer as ,rightly mentioned in previous posts, E

Two posts above suggest that these inequalities:

16a + b > 0
a > 0

imply that b > 0. That is not the case, as you can see by plugging in numbers: we might have a = 1 and b = -1, for example.

If our parabola is y = ax^2 + b, then when a is positive, the parabola (a U shape) opens upwards, and if a is negative, the parabola opens downwards (an upside-down U shape). The value of b is the y-intercept of the parabola. Using both statements here, we know the parabola opens upwards, because a > 0, and we know that when x=4, the parabola is above the x-axis. But that still leaves two possibilities: maybe the parabola has a y-intercept that is above the origin (b > 0), and lies entirely above the x-axis (e.g. in the parabola y = x^2 + 1) or maybe it starts at or below the x-axis (b < 0), crosses the x-axis in one or two places, and becomes positive by the time x = 4 (e.g. in the parabola y = x^2 - 1, which clearly has two x-intercepts at 1 and -1). So the answer is E.

If the source is claiming the answer is C here, then I'd suggest working with more reliable study materials.