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

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22-11-2016
Mathematics

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

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MissPhiladelphia MissPhiladelphia · Answer 1 · 2016-11-24T16:31:05+01:00

MissPhiladelphia MissPhiladelphia

24-11-2016

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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