The figure shows the graph of the polynomial function y = f(x). For

which of the values k = 0, 1, 2, or 3 will the equation f(x) = k have

complex roots?

The choices are 0, 1, 2, or 3

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- Jun 16th 2007, 07:04 PMwvmcanelly@cableone.netAny help would be greatly appreciatedThe figure shows the graph of the polynomial function y = f(x). For

which of the values k = 0, 1, 2, or 3 will the equation f(x) = k have

complex roots?

The choices are 0, 1, 2, or 3 - Jun 16th 2007, 07:14 PMThePerfectHacker
For k = 3 because the curve hits it once. We need 2 more solutions. Which the curve does not show because they are not real rather complex.

- Jun 16th 2007, 07:15 PMJhevon
I'd say k = 3

then we would have:

$\displaystyle f(x) = 3$

$\displaystyle \Rightarrow f(x) - 3 = 0$

However, the graph of $\displaystyle f(x) - 3$ will cut the $\displaystyle x$-axis at in one place. Since it is a cubic polynomial, we should get 3 roots, since we would have only 1 real root if we shift the graph 3 units down, the other two roots will be complex

EDIT: Beaten yet again! - Jun 16th 2007, 07:24 PMwvmcanelly@cableone.net
thanks for the quick response. I had 3 but was not sure.

Greatly appreciated