If $y = \frac{\lvert 4p + 6\rvert}{-p^2 - 4}$, for what value of $p$ will the value of $y$ be greatest?

If $y = \frac{\lvert 4p + 6\rvert}{-p^2 - 4}$, for... | Free GMAT Practice Question | GMAT Panda

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If y = \frac{∣ 4 p + 6 ∣}{- p ^{2} - 4}, for what value of p will the value of y be greatest?

Explanation	Calculations	Help
$We factor out the negative sign from the denominator to simplify the expression.$	$y = \frac{∣ 4 p + 6 ∣}{- p ^{2} - 4} = - \frac{∣ 4 p + 6 ∣}{p ^{2} + 4}$
$We note that the denominator is always positive, so including the negative sign makes the entire fraction non-positive.$	$p^{2} + 4 > 0$
$Since y cannot be positive, its greatest value is zero, which occurs when the numerator equals zero.$	$∣ 4 p + 6 ∣ = 0 4 p + 6 = 0 4 p = - 6 p = - \frac{3}{2}$	Theory & Tactics Method Card ALG5-B Click to view full details

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B

If y = \frac{∣ 4 p + 6 ∣}{- p ^{2} - 4}, for what value of p will the value of y be greatest?

Explanation	Calculations	Help
$We factor out the negative sign from the denominator to simplify the expression.$	$y = \frac{∣ 4 p + 6 ∣}{- p ^{2} - 4} = - \frac{∣ 4 p + 6 ∣}{p ^{2} + 4}$
$We note that the denominator is always positive, so including the negative sign makes the entire fraction non-positive.$	$p^{2} + 4 > 0$
$Since y cannot be positive, its greatest value is zero, which occurs when the numerator equals zero.$	$∣ 4 p + 6 ∣ = 0 4 p + 6 = 0 4 p = - 6 p = - \frac{3}{2}$	Theory & Tactics Method Card ALG5-B Click to view full details

Scroll horizontally to view all columns

B