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Is $$y$$ a real function of $$x$$ in the equation $$|y + 2008| = x^2 + 2$$?
Correct Answer: 0
PLACEHOLDER ANSWER - the answer is "No". Since $$x^2 + 2 > 0$$, the equation gives $$y = -2008 + (x^2 + 2)$$ or $$y = -2008 - (x^2 + 2)$$. Every real $$x$$ therefore corresponds to two different values of $$y$$, so $$y$$ is not a function of $$x$$.
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