Suppose a curve is defined by the equation (x + y)2 = 1. what is the equation of the line tangent to the curve at (3, −2)?

Don't you mean (x + y)^2 = 1?

Take the derivative with respect to x of both sides of this equation:

2(x + y(dy/dx) = 0.  Then y(dy/dx) = -x, and (dy/dx) = -x/y.

Thus, the slope of the tangent line to this curve at (3, -2) is m = -3 / (-2), or 3/2.

The eqn of the T. L.   is then   y-(-2) = (3/2)(x-3) (answer)