document.write( "Question 64855: f(x)=x^2+6x+5
\n" ); document.write( "Find the x value of the vertex (turning point)
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Algebra.Com's Answer #45541 by funmath(2933)\"\" \"About 
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f(x)=x^2+6x+5
\n" ); document.write( "Find the x value of the vertex (turning point)
\n" ); document.write( "There are 2 ways to do this let me know if this isn't the way your teacher is doing it.
\n" ); document.write( "Because this is in f(x)=ax^2+bx+c form, I would use the formula \"highlight%28x=-b%2F2a%29\" to find the x-coordinate, then find f(-b/2a) to find the y-coordinate.
\n" ); document.write( "a=1, b=6
\n" ); document.write( "\"x=-%286%29%2F%282%281%29%29\"
\n" ); document.write( "\"x=-6%2F2\"
\n" ); document.write( "x=-3
\n" ); document.write( "y=\"f%28-3%29=%28-3%29%5E2%2B6%28-3%29%2B5\"
\n" ); document.write( "\"y=9-18%2B5\"
\n" ); document.write( "y=-4
\n" ); document.write( "The vertex is (-3,-4)
\n" ); document.write( "Happy Calculating!!!
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