document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=48318>Question 48318</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given a table of x and F(x) values how would you find the quadratic equation that describes it?<BR>\n" );
document.write( "x -4, -3, -2, -1, 0<BR>\n" );
document.write( "y -1,   0,     3,    8,   15\r<BR>\n" );
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document.write( "Since when x=0, y=15; I know that the constant term in the quadratic equation must be 15.  Also I know that one of the roots is -3 because when y=0, x=-3.  Additionally I have found that the second difference beteen the \"y\" values is always +2. (pattern of change is 2).  I however, do not know how to use this piece of information to solve the problem.  I have to write the quadratic equation that describes the relationship.  I know that the relationship is quadratic because the second difference is a constant term - 2. Please help.  Thank you for your time and help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #31959 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><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=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=31959\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> well done with what you said and have done so far.\r<BR>\n" );
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document.write( "Right then, my method:\r<BR>\n" );
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document.write( "assuming it is a quadratic (and as you say, the second difference being constant implies a quadratic) then the equation will have the form of f(x) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+ax%5E2+%2B+bx+%2B+c+ ALIGN=MIDDLE  ALT=\"+ax%5E2+%2B+bx+%2B+c+\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "We know:<BR>\n" );
document.write( "when x = 0 that y=15, so:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+f%28x%29+=+ax%5E2+%2B+bx+%2B+c+ ALIGN=MIDDLE  ALT=\"+f%28x%29+=+ax%5E2+%2B+bx+%2B+c+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+15+=+a%280%29%5E2+%2B+b%280%29+%2B+c+ ALIGN=MIDDLE  ALT=\"+15+=+a%280%29%5E2+%2B+b%280%29+%2B+c+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "--> c = 15\r<BR>\n" );
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document.write( "So we have <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+f%28x%29+=+ax%5E2+%2B+bx+%2B+15+ ALIGN=MIDDLE  ALT=\"+f%28x%29+=+ax%5E2+%2B+bx+%2B+15+\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Knowing one of the roots is fine but it doesn't help us since there are many quadratic that will have that as a root. What we need to do now (seeing that we only have 2 unknowns to find --> a and b) is pick two of the coordinates and solve them simultaneously. Pick easier ones.\r<BR>\n" );
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document.write( "Using (-3,0) gives <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+0+=+a%28-3%29%5E2+-+3b+%2B+15+ ALIGN=MIDDLE  ALT=\"+0+=+a%28-3%29%5E2+-+3b+%2B+15+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+0+=+9a+-+3b+%2B+15+ ALIGN=MIDDLE  ALT=\"+0+=+9a+-+3b+%2B+15+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "--> 9a - 3b = -15\r<BR>\n" );
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document.write( "Using (-1,8) gives <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+8+=+a%28-1%29%5E2+-+b+%2B+15+ ALIGN=MIDDLE  ALT=\"+8+=+a%28-1%29%5E2+-+b+%2B+15+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+8+=+a+-+b+%2B+15+ ALIGN=MIDDLE  ALT=\"+8+=+a+-+b+%2B+15+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "--> a - b = -7\r<BR>\n" );
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document.write( "So we have 2 equations now:<BR>\n" );
document.write( "9a - 3b = -15<BR>\n" );
document.write( "a - b = -7\r<BR>\n" );
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document.write( "scale up the second one:<BR>\n" );
document.write( "9a - 3b = -15<BR>\n" );
document.write( "3a - 3b = -21\r<BR>\n" );
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document.write( "and subtract them:<BR>\n" );
document.write( "6a = 6<BR>\n" );
document.write( "--> a = 1\r<BR>\n" );
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document.write( "and from a - b = -7 we get<BR>\n" );
document.write( "1 - b = -7<BR>\n" );
document.write( "--> b = 8\r<BR>\n" );
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document.write( "So the quadratic is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+f%28x%29+=+x%5E2+%2B+8x+%2B+15+ ALIGN=MIDDLE  ALT=\"+f%28x%29+=+x%5E2+%2B+8x+%2B+15+\"   DISABLED_style= \"max-width:90%; height:auto;\" > which can be factorised to f(x) = (x+3)(x+5) meaning that its two roots are at -3 and -5, in agreement with the data quoted.\r<BR>\n" );
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document.write( "jon.<BR>\n" );
document.write( " </SPAN>\n" );
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