document.write( "Question 1106926: Obtain the equation of the curve y=ax^2+bx+c,using the following conditions
\n" ); document.write( "(i) d²y/dx²=2
\n" ); document.write( "(ii) The curve is at minimum when x=-1/2
\n" ); document.write( "(iii) The curve passes through the point (-2,- 4).
\n" ); document.write( "

Algebra.Com's Answer #721922 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
So, the first derivative is,
\n" ); document.write( "\"dy%2Fdx=2ax%2Bb\"
\n" ); document.write( "and the second is,
\n" ); document.write( "\"d2y%2Fdx2=2a\"
\n" ); document.write( "So,
\n" ); document.write( "\"2a=2\"
\n" ); document.write( "\"a=1\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"y=x%5E2%2Bbx%2Bc\"
\n" ); document.write( "Converting to vertex form,
\n" ); document.write( "\"y=%28x%5E2%2Bbx%2B%28b%2F2%29%5E2%29%2B%28c-%28b%2F2%29%5E2%29\"
\n" ); document.write( "\"y=%28x%2Bb%2F2%29%5E2%2B%28c-%28b%2F2%29%5E2%29\"
\n" ); document.write( "So the minimum occurs at,
\n" ); document.write( "\"-b%2F2=-1%2F2\"
\n" ); document.write( "\"b=1\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"y=x%5E2%2Bx%2Bc\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "Finally using the point,
\n" ); document.write( "\"-4=%28-2%29%5E2%2B%28-2%29%2Bc\"
\n" ); document.write( "\"-4=4-2%2Bc\"
\n" ); document.write( "\"c=-6\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"highlight%28y=x%5E2%2Bx-6%29\"\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );