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

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). 

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 

You can download any video except for the illegal ones