Algebra.Com's Answer #834156 by ikleyn(52803)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "Find the equation of the parabola with axis parallel with the x-axis
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document.write( "and passing through (4,2), (-4,10) and (-4, -6). Sketch the graph
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document.write( " This problem is not to make stupid calculations (as other tutor suggests).\r
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document.write( " It is to develop your observation skills and mind, as well as the sense of beauty.\r
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document.write( "Notice that the points (-4,10) and (-4,-6) have the same x-coordinate.\r\n" );
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document.write( "Hence, the horizontal axis of the parabola is the perpendicular bisector \r\n" );
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document.write( " y=  = 2\r\n" );
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document.write( "of the segment, connecting (-4,10) and (-4,-6).\r\n" );
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document.write( "Also notice that the point (4,2) is the vertex, since it lies on the axis y= 2.\r\n" );
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document.write( "So, the vertex form of this parabola is x = a*(y-2)^2 + 4. (*)\r\n" );
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document.write( "The coefficient \"a\" is unknown now.\r\n" );
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document.write( "Find it from the condition that the point (-4,10) lies on the parabola.\r\n" );
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document.write( "Substitute x=-4, y= 10 into equation (*). You will get\r\n" );
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document.write( " -4 = a*(10-2)^2 + 4\r\n" );
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document.write( " -4 - 4 = 64a, ---> 64a = -8 ---> a =  =  .\r\n" );
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document.write( "So, the equation of the parabola is x =  .\r\n" );
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document.write( "Solved.\r
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