MathRevolution wrote:

[GMAT math practice question]

For which value of x will \(y=ax^2+100x+b\) have a maximum in the \(xy\)-plane?

1) \(b=100\)

2) \(a=-4\)

The maximum or minimum value of a quadratic function \(ax^2+bx+c\) occurs at

\(x=-\frac{b}{2a}\)

so if \(a>0\) then minima will occur and the parabola will open upwards

and if \(a<0\) then maxima will occur and the parabola will open downwards

so for \(y=ax^2+100x+b\)

maxima will be at \(x=-\frac{100}{2a}\) where \(a<0\)

Statement 1: no information about \(a\).

InsufficientStatement 2: provides the value of \(a\).

SufficientOption

BAlternatively, we can use derivative

(out of scope for GMAT)\(y=ax^2+100x+b\)

so \(\frac{dy}{dx} = 2ax+100\)

maximum point will be \(\frac{dy}{dx}=0 => 2ax+100=0\)

so \(x=-\frac{100}{2a}\). We need the value of \(a\) to get the maximum point.