Final answer:
Given the constraints of the quartic polynomial, the leading coefficient 'a' is determined to be positive because the y-intercept is provided as y = 4, which equates to a = 1 after substituting x = 0 into the function.
Step-by-step explanation:
The question asks for information about a quartic polynomial given certain constraints: it has a given form f(x) = ax^4 + bx^3 + cx^2 + dx + e, two real zeros with multiplicity 2 (which means they are each repeated twice), and a known y-intercept.
Knowing that the x-intercepts are x = -2 and x = 1, we can deduce that f(x) has factors (x + 2)^2 and (x - 1)^2. This gives us the expanded form of f(x) = a(x + 2)^2(x - 1)^2. The y-intercept occurs when x = 0, which means e = f(0) = a(2)^2(-1)^2 = 4a. Since we are given that the y-intercept is y = 4, this means that 4a = 4, from which we find that a = 1, which is a positive number. Thus, the leading coefficient a is positive.