document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1054657>Question 1054657</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Write a polynomial function that has the following characteristics?<BR>\n" );
document.write( "Extends from quadrant II to IV\r<BR>\n" );
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document.write( "Y-intercept of 6\r<BR>\n" );
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document.write( "One x-intercept is at 2\r<BR>\n" );
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document.write( "2 turning points </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #669988 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=669988\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Since the function must have two turning points, it must be at least of degree 3.<BR>\n" );
document.write( "It crosses the x-axis at x=2, which means one of the factors must be (x-2)<BR>\n" );
document.write( "Also, the constant term, when x=0, is equal to 6.<BR>\n" );
document.write( "From this information we can write:<BR>\n" );
document.write( "f(x) = (x-2)(ax^2+bx-3)<BR>\n" );
document.write( "The function starts out in quadrant II and ends in quadrant IV, <BR>\n" );
document.write( "so for negative values of x, it must be >0 and for positive values of x, it must be <0. <BR>\n" );
document.write( "This implies the coefficient on the leading term is negative.<BR>\n" );
document.write( "Let us assume that the quadratic term of f(x) = 0<BR>\n" );
document.write( "Thus we need to find a and b such that x-2 is a factor of ax^3+bx+6<BR>\n" );
document.write( "If we perform the division, we are left with the requirement that b = -4a - 3<BR>\n" );
document.write( "At the turning points, the df/dx = 0, and there must be two.<BR>\n" );
document.write( "df/dx = 3ax^2+b = 0 -> x = +- sqrt(-b/3a) = +- sqrt((4a+3)/3a)) = +- sqrt(4/3+1/a), which must be >0 for there to be two real roots.<BR>\n" );
document.write( "Therefore a must be less than -3/4.<BR>\n" );
document.write( "Let a = -1, then b = -4(-1) - 3 = 1<BR>\n" );
document.write( "So a function that satisfies the requirements is<BR>\n" );
document.write( "f(x) = -x^3 + x + 6<BR>\n" );
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