Question

for the normal (perpendicular) line to the curve y= square root of 8-x^2 at (-2,2) would the slope be 1/2? ...?

Answer 1

The correct answer to this question is -1.
The equation is
y= sqrt(8-x^2)
y' = -x/sqrt(8-x^2)
y'(-2) = 2/2 = 1
the normal would have the slope -1 for the normal (perpendicular) line to the curve y= square root of 8-x^2 at (-2,2).

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