If y=|3 x-5|-x^2-3, for what value of x will the value of y be greatest?

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5/3 We factor out the negative sign from the denominator to simplify the expression. y = 3x - 5/-x^2 - 3 = - 3x - 5/x^2 + 3 We note that the denominator is always positive, so including the negative sign makes it negative. x^2 + 3 > 0 -(x^2 + 3) < 0 We see that y cannot be positive, so its highest value is zero, which happens when the numerator equals zero. 3x - 5 = 0 3x - 5 = 0 3x = 5 x = 5/3

Explanation	Calculations	Help
$We factor out the negative sign from the denominator to simplify the expression.$	$y = \frac{∣ 3 x - 5 ∣}{- x ^{2} - 3} = - \frac{∣ 3 x - 5 ∣}{x ^{2} + 3}$
$We note that the denominator is always positive, so including the negative sign makes it negative.$	$x^{2} + 3 > 0 - (x^{2} + 3) < 0$
$We see that y cannot be positive, so its highest value is zero, which happens when the numerator equals zero.$	$∣ 3 x - 5 ∣ = 0 3 x - 5 = 0 3 x = 5 x = \frac{5}{3}$

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