document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=138950>Question 138950</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have three quadratic equations that need to be solved and graphed on one graph to show the relationship between the size of three curves.\r<BR>\n" );
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document.write( "we choose at least 6 plots - I chose for X - o,1,2,3,-1,-2,-3. I have to solve for Y.\r<BR>\n" );
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document.write( "The equations are:<BR>\n" );
document.write( "1. y=2x^+10x+12\r<BR>\n" );
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document.write( "2. y= -1/2x^+4x+3\r<BR>\n" );
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document.write( "3. 2y=x^+4x+5\r<BR>\n" );
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document.write( "My problem has been how to solve these type of equations\r<BR>\n" );
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document.write( "Any help would be appreciated\r<BR>\n" );
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document.write( "Thank YOu\r<BR>\n" );
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document.write( "Sorry, my name is Leonora </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #101447 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=101447\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First, put number 3 in standard form by dividing by the coefficient on y.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%281%2F2%29x%2B2x%2B5%2F2 ALIGN=MIDDLE  ALT=\"y=%281%2F2%29x%2B2x%2B5%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Now it is in <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=ax%5E2%2Bbx%2Bc ALIGN=MIDDLE  ALT=\"y=ax%5E2%2Bbx%2Bc\"   DISABLED_style= \"max-width:90%; height:auto;\" > form.\r<BR>\n" );
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document.write( "Now, instead of just choosing values and trying to calculate the value of the function for each, you need to find the critical points and characteristics.\r<BR>\n" );
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document.write( "First thing to notice is that the sign on the lead coefficient tells you which way the curve opens, up or down.  Positive opens up, negative opens down.  So 1 and 3 open upward and 2 opens down.\r<BR>\n" );
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document.write( "Next thing to find is the axis of symmetry.  This is a vertical line about which the curve is symmetrical -- the two parts of the graph on either side of the axis of symmetry are mirror images of each other.  Once you have your equation in standard form, this is a simple calculation.  Take the negative of the coefficient on the 1st degree term (the 'b' in the standard form) and divide by 2 times the lead coefficient (the 'a').  So calculate <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-b%2F2a ALIGN=MIDDLE  ALT=\"-b%2F2a\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  For your first equation, you should get <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-10%29%2F4=%28-5%29%2F2 ALIGN=MIDDLE  ALT=\"%28-10%29%2F4=%28-5%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  The axis of symmetry is then the vertical line:  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=-b%2F2a ALIGN=MIDDLE  ALT=\"x=-b%2F2a\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=%28-5%29%2F2 ALIGN=MIDDLE  ALT=\"x=%28-5%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > for your first equation, for example.\r<BR>\n" );
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document.write( "The first point you want to find is the vertex of your parabola.  That is the point at the very bottom  (or top for #2) of the curve.  You already have the x-coordinate of this point -- it is the same as the value you calculated for the axis of symmetry.  The y-coordinate of this point is found by substituting the x-coordinate value into your function and calculating the result.\r<BR>\n" );
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document.write( "The next points to find are the x-intercepts.  This is where the curve intersects the x-axis.  Since the y-coordinate of any point on the x-axis is zero, just set the function equal to zero and solve the quadratic equation.  You can factor it or use the quadratic formula.  If you got a positive value for the y-coordinate when you calculated the vertex for a 'opens upward' curve, or you got a negative value when you did it for the 'opens downward' curve, then you can skip this step because the curve doesn't intersect the axis.  If you got a zero result when you calculated the y-coordinate of the vertex, then you can also skip this step because the vertex and the intersection of the curve with the x-axis are the same place.  \r<BR>\n" );
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document.write( "You can also find the y-intercept.  This is the one place where the curve intersects the y-axis.  Points on the y-axis have x-coordinates that are all 0, so just substitute 0 for x in your function and solve (you just get the value of the constant term).  Since the curve is symmetrical, take the value of the x-coordinate of the vertex and double it, then you will have a point at this new x-coordinate and the value of the y-intercept.\r<BR>\n" );
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document.write( "If you think you need a couple more points to get a smooth curve, go to the nearest whole unit one side or the other of the vertex, then calculate the value and plot your point.  Because of symmetry, you will have a point with the same y value the same distance on the other side of the vertex.\r<BR>\n" );
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document.write( "Let's do the first one.  We have already established that the axis of symmetry is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=%28-5%29%2F2 ALIGN=MIDDLE  ALT=\"x=%28-5%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so the x-coordinate of the vertex is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-5%29%2F2 ALIGN=MIDDLE  ALT=\"%28-5%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  The y-coordinate of the vertex is then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=2%28%28-5%29%2F2%29%5E2%2B10%28%28-5%29%2F2%29%2B12=25%2F2-50%2F2%2B24%2F2=-1%2F2 ALIGN=MIDDLE  ALT=\"y=2%28%28-5%29%2F2%29%5E2%2B10%28%28-5%29%2F2%29%2B12=25%2F2-50%2F2%2B24%2F2=-1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  So, the vertex is at <big><b> (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-5%29%2F2 ALIGN=MIDDLE  ALT=\"%28-5%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%29%2F2 ALIGN=MIDDLE  ALT=\"%28-1%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >)</big></b>\r<BR>\n" );
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document.write( "This is an \"opens upward\" with a negative y-coordinate on the vertex, meaning that the curve will intersect the x-axis in two points, so let's solve the quadratic to find them:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x%5E2%2B10x%2B12=0 ALIGN=MIDDLE  ALT=\"2x%5E2%2B10x%2B12=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B2%29%28x%2B3%29=0 ALIGN=MIDDLE  ALT=\"%28x%2B2%29%28x%2B3%29=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "So the function has zeros at -2 and -3, and the x-intercepts are at (-2,0) and (-3,0)\r<BR>\n" );
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document.write( "The y-intercept:  Substitute 0 for x:   <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=2%280%29%5E2%2B10%280%29%2B12=12 ALIGN=MIDDLE  ALT=\"y=2%280%29%5E2%2B10%280%29%2B12=12\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so the y-intercept is (0,12).  The mirror image point is at 2 times the x-coordinate of the vertex, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%28-5%2F2%29=-5 ALIGN=MIDDLE  ALT=\"2%28-5%2F2%29=-5\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so the point is (-5,12).\r<BR>\n" );
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document.write( "The next whole number from the x-coordinate of the vertex is -2 or -3, but we already have those points as x-intercepts, so lets pick -1.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%28-1%29%5E2%2B10%28-1%29%2B12=2-10%2B12=4 ALIGN=MIDDLE  ALT=\"2%28-1%29%5E2%2B10%28-1%29%2B12=2-10%2B12=4\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so we have a point at (-1,4).  By symmetry, we also have a point at (-4,4).\r<BR>\n" );
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document.write( "Here's a picture of what we have so far (the green line is the axis of symmetry):\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28600%2C600%2C-8%2C8%2C-2%2C14%2C%0D%0Agrid%281%29%2C%0D%0Acircle%28-5%2F2%2C-1%2F2%2C.2%29%2C%0D%0Acircle%28-3%2C0%2C.2%29%2C%0D%0Acircle%28-2%2C0%2C.2%29%2C%0D%0Acircle%280%2C12%2C.2%29%2C%0D%0Acircle%28-5%2C12%2C.2%29%2C%0D%0Acircle%28-1%2C4%2C.2%29%2C%0D%0Acircle%28-4%2C4%2C.2%29%2C%0D%0Agreen%28line%28-5%2F2%2C-2%2C-5%2F2%2C14%29%29%2C%0D%0Agraph%28600%2C600%2C-8%2C8%2C-2%2C14%2C2x%5E2%2B10x%2B12%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Now, go through this whole process again with the other two functions. </SPAN>\n" );
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