document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209686>Question 1209686</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose the polynomial f(x) is of degree 3 and satisfies f(3) = 2, f(4) = -4, f(5) = 6, and f(8) = 8.\r<BR>\n" );
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document.write( "Determine the value of f(0). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849804 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849804\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $f(x)$ be a polynomial of degree 3. We are given the following values:<BR>\n" );
document.write( "$f(3) = 2$<BR>\n" );
document.write( "$f(4) = -4$<BR>\n" );
document.write( "$f(5) = 6$<BR>\n" );
document.write( "$f(8) = 8$\r<BR>\n" );
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document.write( "We can use the Lagrange interpolation formula to find the polynomial $f(x)$:<BR>\n" );
document.write( "\[f(x) = \sum_{i=0}^3 y_i L_i(x)\]<BR>\n" );
document.write( "where $y_i$ are the given function values and $L_i(x)$ are the Lagrange basis polynomials.\r<BR>\n" );
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document.write( "In our case, the points are $(3, 2), (4, -4), (5, 6), (8, 8)$.\r<BR>\n" );
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document.write( "\[L_0(x) = \frac{(x-4)(x-5)(x-8)}{(3-4)(3-5)(3-8)} = \frac{(x-4)(x-5)(x-8)}{(-1)(-2)(-5)} = -\frac{1}{10}(x-4)(x-5)(x-8)\]<BR>\n" );
document.write( "\[L_1(x) = \frac{(x-3)(x-5)(x-8)}{(4-3)(4-5)(4-8)} = \frac{(x-3)(x-5)(x-8)}{(1)(-1)(-4)} = \frac{1}{4}(x-3)(x-5)(x-8)\]<BR>\n" );
document.write( "\[L_2(x) = \frac{(x-3)(x-4)(x-8)}{(5-3)(5-4)(5-8)} = \frac{(x-3)(x-4)(x-8)}{(2)(1)(-3)} = -\frac{1}{6}(x-3)(x-4)(x-8)\]<BR>\n" );
document.write( "\[L_3(x) = \frac{(x-3)(x-4)(x-5)}{(8-3)(8-4)(8-5)} = \frac{(x-3)(x-4)(x-5)}{(5)(4)(3)} = \frac{1}{60}(x-3)(x-4)(x-5)\]\r<BR>\n" );
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document.write( "Then,<BR>\n" );
document.write( "\[f(x) = 2L_0(x) - 4L_1(x) + 6L_2(x) + 8L_3(x)\]\r<BR>\n" );
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document.write( "We want to find $f(0)$:<BR>\n" );
document.write( "\[f(0) = 2L_0(0) - 4L_1(0) + 6L_2(0) + 8L_3(0)\]<BR>\n" );
document.write( "\[L_0(0) = -\frac{1}{10}(-4)(-5)(-8) = -\frac{1}{10}(-160) = 16\]<BR>\n" );
document.write( "\[L_1(0) = \frac{1}{4}(-3)(-5)(-8) = \frac{1}{4}(-120) = -30\]<BR>\n" );
document.write( "\[L_2(0) = -\frac{1}{6}(-3)(-4)(-8) = -\frac{1}{6}(-96) = 16\]<BR>\n" );
document.write( "\[L_3(0) = \frac{1}{60}(-3)(-4)(-5) = \frac{1}{60}(-60) = -1\]\r<BR>\n" );
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document.write( "\[f(0) = 2(16) - 4(-30) + 6(16) + 8(-1) = 32 + 120 + 96 - 8 = 240\]\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{240}$<BR>\n" );
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