document.write( "Question 919341: Express x^2-5x+6 in the form of (x-a)^2-b. Hence state the coordinates of the turning point of the curve y=x^2-5x+6.\r
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\n" ); document.write( "\n" ); document.write( "I really don't understand how to work out the turning point so guidance would be very helpful. Thanks.
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Algebra.Com's Answer #557643 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
the vertex form of a Parabola opening up(a>0) or down(a<0),
\n" ); document.write( "\"y=a%28x-h%29%5E2+%2Bk\". V(h, k)
\n" ); document.write( " Completing the Square to Obtain the Vertex Form:
\n" ); document.write( "y = ax^2 + bx + c = 0
\n" ); document.write( "y = x^2-5x+6 = 0 \"b%2F%28-2a%29+=+%28-5%29%2F%28-2%29+=+5%2F2\" 5/2 the x-value for the Vertex
\n" ); document.write( "y = (x - 5/2)^2 - (5/2)^2 + 6 = 0
\n" ); document.write( "y = (x - 5/2)^2 - 25/4 + 24/4= 0 \"-a%28b%2F%28-2a%29%29%5E2+\"= -25/4
\n" ); document.write( "y = (x - 5/2)^2 - 1/4 = 0
\n" ); document.write( "V(5/2, -1/4)\r
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