[tex]f(x)=ax^2+bx+c[/tex] and a ≠ 0
Has Maximum Value at the vertex, ok so.
Let's talk about Parabola a bit.
Given [tex]y=ax^2+bx+c[/tex] when a≠0 and a, b, c ∈ R (a, b and c are part of real numbers.)
This is just basic of Parabola, let's move to the question.
The problem gives the standard form. However, it says maximum value at its vertex (Meaning that a<0)
So for the choice A. a<0 which is right because the maximum value is at vertex and not minimum value (that'd be a>0 if it's minimum.)
For the choice B. -b/2a<0 which is wrong and doesn't have anything to do with it. -b/2a is the axis of symmetry. If -b/2a<0 then the value of h would be in negative. (Doesn't have anything to do with form of Parabola, it just changes the value of h.)
For the choice C. -b/2a>0 which is similar to the choice B because if -b/2a>0 then the value of h would be positive and doesn't have anything to do with it.
For the choice D. a>0 which is wrong because that'd be minimum value at vertex, not maximum. (maximum value at vertex when a<0 and minimum value at vertex when a>0)
So the answer is choice A.
