# Any help would be greatly appreciated

• Jun 16th 2007, 08:04 PM
wvmcanelly@cableone.net
Any help would be greatly appreciated
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
• Jun 16th 2007, 08:14 PM
ThePerfectHacker
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, 08:15 PM
Jhevon
Quote:

Originally Posted by wvmcanelly@cableone.net
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

I'd say k = 3

then we would have:

$f(x) = 3$

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

However, the graph of $f(x) - 3$ will cut the $x$-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, 08:24 PM
wvmcanelly@cableone.net
thanks for the quick response. I had 3 but was not sure.

Greatly appreciated