document.write( "Question 28016: What is the equation of a parabola that passes through the points (1,6),(-2,27), (2,11)? \n" ); document.write( "
Algebra.Com's Answer #15259 by bmauger(101)\"\" \"About 
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A quadratic equation can be written in the form
\n" ); document.write( "\"y=ax%5E2%2Bbx%2Bc\" By putting in the three points given we can get three linear equations of the three variables a, b, & c and then solve:
\n" ); document.write( "evaluating
\n" ); document.write( "\"y=ax%5E2%2Bbx%2Bc\" for (1,6) we get:
\n" ); document.write( "\"6=a%281%29%5E2%2Bb%281%29%2Bc\" or
\n" ); document.write( "Eq 1: \"6=a%2Bb%2Bc\"\r
\n" ); document.write( "\n" ); document.write( "For (-2, 27)
\n" ); document.write( "\"27=a%28-2%29%5E2%2Bb%28-2%29%2Bc\"
\n" ); document.write( "Eq 2: \"27=4a-2b%2Bc\"\r
\n" ); document.write( "\n" ); document.write( "For (2, 11)
\n" ); document.write( "\"11=a%282%29%5E2%2Bb%282%29%2Bc\"
\n" ); document.write( "Eq 3: \"11=4a%2B2b%2Bc\"\r
\n" ); document.write( "\n" ); document.write( "Now you simply solve the three simolutanious equations for a, b, and c.
\n" ); document.write( "To get you started, let's subtract equation 1 from equation 3 to get:
\n" ); document.write( "\"11=4a%2B2b%2Bc\"
\n" ); document.write( "-\"6=a%2Bb%2Bc\"
\n" ); document.write( "=\"5=3a%2Bb\"
\n" ); document.write( "and then subtract equation 1 from equation 2 to get:
\n" ); document.write( "\"27=4a-2b%2Bc\"
\n" ); document.write( "-\"6=a%2Bb%2Bc\"
\n" ); document.write( "=\"21=3a-3b\"
\n" ); document.write( "Using the two new equations we found should be enough to get you to the answer in a few more steps. When you get there put your a, b, and c in for: \"y=ax%5E2%2Bbx%2Bc\" and that's your answer (unless you're supposed to factor or find the roots, graph it, or do anything else to the equation).
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