1. **Problem statement:** Given the function $$f_r(x) = \frac{x^2 - 4}{|x| - 1}$$ defined on its domain, determine which of the points A(0,2), B(-2,0), C(1,0), and D(0,4) belong to the curve (C1). 2. **Domain determination:** The function involves $$|x| - 1$$ in the denominator, so the domain excludes values where $$|x| - 1 = 0$$, i.e., $$|x| = 1$$, so $$x = \pm 1$$ are excluded. 3. **Evaluate $$f_r(x)$$ at each point's x-coordinate:** - For A(0,2): $$f_r(0) = \frac{0^2 - 4}{|0| - 1} = \frac{-4}{0 - 1} = \frac{-4}{-1} = 4$$ The y-value is 2, but $$f_r(0) = 4$$, so A is not on (C1). - For B(-2,0): $$f_r(-2) = \frac{(-2)^2 - 4}{|-2| - 1} = \frac{4 - 4}{2 - 1} = \frac{0}{1} = 0$$ The y-value is 0, matches $$f_r(-2)$$, so B is on (C1). - For C(1,0): $$f_r(1) = \frac{1^2 - 4}{|1| - 1} = \frac{1 - 4}{1 - 1} = \frac{-3}{0}$$ undefined, so C is not on (C1). - For D(0,4): $$f_r(0) = 4$$ as above, but y-value is 4, matches $$f_r(0)$$, so D is on (C1). 4. **Conclusion:** Points B(-2,0) and D(0,4) belong to the curve (C1). \[\boxed{\text{Points on } (C1): B(-2,0), D(0,4)}\]