T is the tangent to the curve y= x^2 +6x -4 at (1,3) and N is the normal to the curve y= x^2 - 6x + 18 at (4,10). Find the coordinates of the point of intersection of T and N
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1 Answer

y=x^2+6x-4 . . .

dy/dx=2x+6....tangent

at x=1, tangent=8=slope

thats nuff ....up tu yu tu du the sekond half

the "normal" is 90 deg tu the kerv, & slope av normal=-1/tangent
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