document.write( "Question 136530: I am trying to help my son with a problem. He is doing parabolas. This is the one the teacher did in class and I don't understand what she did. (((y=x^2+8x+6)))....y-6+16 = x^2+8x+16....y+10=(x+4)^2....vertex (-4,-10) Domain is all real numbers, range is y is greater than or equal to -10, minimum is -10. Could you pleas explain her procedure here. Thank you! \n" ); document.write( "

Algebra.Com's Answer #100008 by scott8148(6628) $\"\"$ $\"About$

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a) subtracted the constant term (+6)\r
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\n" ); document.write( "\n" ); document.write( "b) completed the square by adding the square of half of the coefficient of the x term
\n" ); document.write( "__ (8/2)^2=16\r
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\n" ); document.write( "\n" ); document.write( "c) the equation is now in the \"vertex\" form __ y-k=(x-h)^2
\n" ); document.write( "__ where the vertex is (h,k)\r
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\n" ); document.write( "\n" ); document.write( "the domain is what values x can be
\n" ); document.write( "__ in this case, any real number value for x gives a value for y\r
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\n" ); document.write( "\n" ); document.write( "the range is what values y can be
\n" ); document.write( "__ since the x-h is squared, it can never be less than zero
\n" ); document.write( "__ so y+10 can never be less than zero __ so y can't be less than -10\r
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\n" ); document.write( "\n" ); document.write( "the y value of the vertex is the minimum value for the function (and the lower bound of the range) \n" ); document.write( "

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