document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=6520>Question 6520</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Dear Sir/Madam,\r<BR>\n" );
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document.write( "in finding the vertex of a parabola, I have been taught the following:\r<BR>\n" );
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document.write( "Rewrite the formula in point slope form, which means y - a = m(x - b), where the vertex is (b,a).\r<BR>\n" );
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document.write( "I tried to use this procedure for the following parabola: x^2 - 4x - 2y - 4 = 0. Thus, I rewrote it as follows: 2(y + 2) = x(x - 4), which would in theory mean that the vertex is (4,-2). However, this is not the case. Instead, the vertex is (2,-4). What did I do wrong?\r<BR>\n" );
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document.write( "Thanks in advance.<BR>\n" );
document.write( "Regards,<BR>\n" );
document.write( "-Mike </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #3602 by </B><A HREF=/tutors/aboutme.mpl?userid=prince_abubu><B>prince_abubu(198)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=prince_abubu><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=3602\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I'm wondering where you learned that formula y - a = m(x - b) as the vertex-point form for a quadratic function. I'm also particularly curious as to what the m actually is, and I don't think it's slope in this case.\r<BR>\n" );
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document.write( "A quadratic function can be written in the form y = a(x - h)^2 + k where (h,k) is the vertex of the parabola. We're gonna go through a long process called completing the square to get to the answer (2,-4) for our example.\r<BR>\n" );
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document.write( "The equation as thrown to you is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+-+4x+-+2y+-+4+=+0+ ALIGN=MIDDLE  ALT=\"+x%5E2+-+4x+-+2y+-+4+=+0+\"   DISABLED_style= \"max-width:90%; height:auto;\" >. First of all, we need to put this in standard quadratic form so that we can write it as \"y = \".\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+-+4x+-+4+=+2y+ ALIGN=MIDDLE  ALT=\"+x%5E2+-+4x+-+4+=+2y+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <---- Moved the 2y to the other side of equation\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29x%5E2+-+2x+-+2 ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29x%5E2+-+2x+-+2\"   DISABLED_style= \"max-width:90%; height:auto;\" > <---- Divided both sides by 2 and put the y in front.\r<BR>\n" );
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document.write( "We haven't \"completed the square yet.\" This is a tricky process, and we must go through some basics first before going on.\r<BR>\n" );
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document.write( "First up, let's say that we'll begin with a number a. If we say a + 3 - 3, we really didn't change the value of a. This may seem like a \"duh\" but we'll need to use this trick later on.\r<BR>\n" );
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document.write( "Say that we have a*(b) to begin with. Suppose that we are to add a \"+ 3\" to the b inside the parentheses, so that a(b + 3). Now, how do we undo that so that we'll get back to the original a(b)? Since a(b + 3) = ab + 3a, we'll need to subtract the 3a from ab + 3a to get back to a(b). In other words a(b) = ab + 3a - 3a = a(b + 3) - 3a. This is just a step above the trick mentioned in the first paragraph. This is the trick you'll be using for this example because you have an a-value that isn't 1.\r<BR>\n" );
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document.write( "The second thing we need to establish is the recognition of trinomial squares and backwards thinking with them. Say that we have (x + 3)(x + 3). We know that that's equal to x^2 + 6x + 9. When we see a quadratic expression in the form ax^2 + bx + c, we must know how to tell whether it's a perfect trinomial square, aka, whether it can be written as (x + n)(x + n). The way to do that is to take the b term, in this case, 6. Divide that by two (which brings it to 3) and square it, which brings it to 9. Does the result equal the c term of the expression? If yes, then the expression is a trinomial square. We will use this fact heavily in completing the square.\r<BR>\n" );
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document.write( "---------------------------------------------------<BR>\n" );
document.write( "On with the game.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29x%5E2+-+2x+-+2+ ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29x%5E2+-+2x+-+2+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <----- Start with quadratic function in standard form.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29%28x%5E2+-+4x%29+-+2+ ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29%28x%5E2+-+4x%29+-+2+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <----- We factored 1/2 from the first two terms and left the -2 untouched. (If you performed distributive property on this equation, you'll end up with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29x%5E2+-+2x+-+2+ ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29x%5E2+-+2x+-+2+\"   DISABLED_style= \"max-width:90%; height:auto;\" >. The \"increase\" of the -2x to the -4x confuses most people because factoring usually cuts down a number, but we're factoring out a fraction less than 1, so the effect is opposite)\r<BR>\n" );
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document.write( "Hang on because here's where the tricks come into play. We're going to first force the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E2+-+4x%29 ALIGN=MIDDLE  ALT=\"%28x%5E2+-+4x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > to be a perfect trinomial square. The b term in that expression would be the -4. Cut it in half, you'll get a -2. KEEP THIS NUMBER HANDY BECAUSE YOU'LL USE IT. Now take that -2 and square it. That brings you to +4. KEEP THIS NUMBER HANDY AS WELL!!! Now, the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E2+-+4x%29+ ALIGN=MIDDLE  ALT=\"%28x%5E2+-+4x%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > term turned to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+%28x%5E2+-+4x+%2B+4%29+ ALIGN=MIDDLE  ALT=\"+%28x%5E2+-+4x+%2B+4%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > which is a \"forced\" perfect trinomial square. The process of taking the b term, halving it, and squaring the result guarantees a perfect trinomial square.\r<BR>\n" );
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document.write( "So far, our working function is \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29%28x%5E2+-+4x+%2B+4%29+-+2+ ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29%28x%5E2+-+4x+%2B+4%29+-+2+\"   DISABLED_style= \"max-width:90%; height:auto;\" > but we're not done yet. \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29%28x%5E2+-+4x+%2B+4%29+-+2+-4%281%2F2%29+ ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29%28x%5E2+-+4x+%2B+4%29+-+2+-4%281%2F2%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <----Remember the trick a(b + c) - ac = ab. Since we added a \"+ 4\" inside the parentheses, we must take it out again to preserve balance. Since the +4 really is being multiplied by the 1/2, we'll have to take out 4(1/2) = 2.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29%28x%5E2+-+4x+%2B+4%29+-+4+ ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29%28x%5E2+-+4x+%2B+4%29+-+4+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <---- We're almost done.\r<BR>\n" );
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document.write( "Remember when we were to make note of the b term when we force-created a trinomial square? It was that -2. When we rewrite the equation just above with the -2, it becomes\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+%281%2F2%29%28x+-+2%29%5E2+-+4+ ALIGN=MIDDLE  ALT=\"+y+=+%281%2F2%29%28x+-+2%29%5E2+-+4+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <---- The move from (x^2 - 4x + 4) to the (x - 2)^2 is backwards FOILING. You don't need to worry because that step where you (take the b term, divide it by two, and square the result) guarantees that the whole thing will work.\r<BR>\n" );
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document.write( "Now, the quadratic equation is in a form where you can easily pull out the vertex of the parabola. That is (2, -4). (Once in this form, the vertex's x-coordinate is the number after the x variable inside the parentheses WITH the OPPOSITE sign. The y-coordinate is whatever the last constant is, and YES, preserve the sign!)\r<BR>\n" );
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document.write( " </SPAN>\n" );
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