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

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.