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The normal (perpendicular) line to the curve y=sqrt(8-x^2) at (-2,2) has slope

A. -2
B. 1/2
C. -1/2
D. 1
E. -1

I got A

1 answer

y = √(8-x^2)
y' = -x/√(8-x^2)
y'(-2) = 2/2 = 1
so, the normal has slope -1

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