document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=162673>Question 162673</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the problem is 3x^+x-5  I need to find the x intercept and the vertex . I am not sure how to set up the equation or what numbers go where. HELP PLEASE </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #119883 by </B><A HREF=/tutors/aboutme.mpl?userid=vleith><B>vleith(2983)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=vleith><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=119883\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> See this for a fairly detailed look at this issue --> http://www.analyzemath.com/quadratics/vertex_problems.html\r<BR>\n" );
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document.write( "I assume you meant <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2Bx-5 ALIGN=MIDDLE  ALT=\"3x%5E2%2Bx-5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Any point that is an x axis intercept must have a y coordinate value of 0. So to solve for those points. solve<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2Bx-5=0 ALIGN=MIDDLE  ALT=\"3x%5E2%2Bx-5=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > You can use quadratic equation for that.<BR>\n" );
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document.write( " <TR><TD  ALIGN=MIDDLE>Solved by <A  rel=nofollow HREF=http://wiki.algebra.com/index.php/Tutoring_-_pluggable_solvers>pluggable</A> solver: <A  rel=nofollow HREF=quadratic.solver>SOLVE quadratic equation with variable</A></TD></TR>\n" );
document.write( "<TR><TD  ALIGN=LEFT>Quadratic equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ax%5E2%2Bbx%2Bc=0 ALIGN=MIDDLE  ALT=\"ax%5E2%2Bbx%2Bc=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (in our case <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B1x%2B-5+=+0 ALIGN=MIDDLE  ALT=\"3x%5E2%2B1x%2B-5+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >) has the following solutons:<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  <BR>\n" );
document.write( "  For these solutions to exist, the <U>discriminant</U> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" > should not be a negative number.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  First, we need to compute the discriminant <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" >: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac=%281%29%5E2-4%2A3%2A-5=61 ALIGN=MIDDLE  ALT=\"b%5E2-4ac=%281%29%5E2-4%2A3%2A-5=61\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  Discriminant d=61 is greater than zero. That means that there are two solutions: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5B12%5D+=+%28-1%2B-sqrt%28+61+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"+x%5B12%5D+=+%28-1%2B-sqrt%28+61+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B1%5D+=+%28-%281%29%2Bsqrt%28+61+%29%29%2F2%5C3+=+1.13504161265111 ALIGN=MIDDLE  ALT=\"x%5B1%5D+=+%28-%281%29%2Bsqrt%28+61+%29%29%2F2%5C3+=+1.13504161265111\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B2%5D+=+%28-%281%29-sqrt%28+61+%29%29%2F2%5C3+=+-1.46837494598444 ALIGN=MIDDLE  ALT=\"x%5B2%5D+=+%28-%281%29-sqrt%28+61+%29%29%2F2%5C3+=+-1.46837494598444\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "    Quadratic expression <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B1x%2B-5 ALIGN=MIDDLE  ALT=\"3x%5E2%2B1x%2B-5\"   DISABLED_style= \"max-width:90%; height:auto;\" > can be factored:<BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B1x%2B-5+=+3%28x-1.13504161265111%29%2A%28x--1.46837494598444%29 ALIGN=MIDDLE  ALT=\"3x%5E2%2B1x%2B-5+=+3%28x-1.13504161265111%29%2A%28x--1.46837494598444%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  Again, the answer is: 1.13504161265111, -1.46837494598444.\n" );
document.write( "Here's your graph:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B1%2Ax%2B-5+%29 ALIGN=MIDDLE  ALT=\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B1%2Ax%2B-5+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ></TD></TR>\n" );
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document.write( "The URl above gives you a formula for finding the vertex. Here are two others ways<BR>\n" );
document.write( "1)if you know deriviatives, take the first derivative and set that to 0. Then solve for x. Use that x vlaue to find the corresponding y in the original equation.<BR>\n" );
document.write( "2) find the x value that is the average of the 2 x intercepts. That point is on the line that 'splits' the parabola (the line about which the parabola has symmetry). Find that x value and plug it into the original equation to find the y value.  </SPAN>\n" );
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