document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=885061>Question 885061</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find x-intercepts of the graph of the equation<BR>\n" );
document.write( "y=x^2-3x+2<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #534878 by </B><A HREF=/tutors/aboutme.mpl?userid=algebrapro18><B>algebrapro18(249)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=algebrapro18><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=534878\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To find the x-intercepts of a graph you need to set the equation of the graph equal to 0 and solve.  NOTE:  While there are two ways to solve this equation I would suggest solving by factoring as its much quicker than the quadratic formula. To factor <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-3X%2B2 ALIGN=MIDDLE  ALT=\"x%5E2-3X%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" > we need to find two factors of 2 that multiply to 2 and add to -3.  Well the factors of 2 are 1 and 2 so the only way these add to -3 is if they are both negative.  After we have the equation factored we can set each factor equal to 0 and solve for x.  Doing so we get:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-3x%2B2 ALIGN=MIDDLE  ALT=\"x%5E2-3x%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" > = 0<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-1%29%28x-2%29 ALIGN=MIDDLE  ALT=\"%28x-1%29%28x-2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > = 0 <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-1+=+0 ALIGN=MIDDLE  ALT=\"x-1+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-2=0 ALIGN=MIDDLE  ALT=\"x-2=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "x = 1 or x = 2 \r<BR>\n" );
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document.write( "So the x-intercepts of the graph are at x = 1 and x = 2.   </SPAN>\n" );
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