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If z = ( y - x ) ^2
Find the targent plane at ( 1 , 2 , 1 )
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z = 2( y - x ) - 1
z = 2( x -y ) + 2
z = 2( x - y ) + 1
z = 2( y - x ) -2

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In the xy plane is a line

( y - x ) = 0and on it there is a parabola. If you imagine an axis a, perpendicular to( y - x ) = 0, on the xy plane. You'll seez = a^2. So, forz= 1we havea = 1. Differenciatez( a )fora = 1and find the targent linez = 2a - 1. Asz = ( y - x )^2, andz = a^2. We havea = ( y - x ). Therefore, the targent plane is justz = 2( y - x )-1