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Given x^2 + y^2 = 13 dy a. Find dy/dx b. Evaluate dy/dx at (3,-2). c. Find the equation of the tangent at (3,-2)

Given x^2 + y^2 = 13 dy a. Find dy/dx b. Evaluate dy/dx at (3,-2). c. Find the equation of the tangent at (3,-2)
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Given x^2 + y^2 = 13 dy

a. Find dy/dx
b. Evaluate dy/dx at (3,-2).
c. Find the equation of the tangent at (3,-2)

asked
User Blgrnboy
by
8.9k points

1 Answer

3 votes

Answer:

The equation of the tangent line at the point (3, -2) is y = (3/2)x - (13/2)

answered
User Slovenec
by
8.1k points
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