1. **State the problem:** Find the x-intercepts of the curve given by the equation $$y=\frac{16-(x-2)^2}{x^2-1}$$. 2. **Recall the definition of x-intercepts:** The x-intercepts occur where the graph crosses the x-axis, which means where $$y=0$$. 3. **Set the function equal to zero:** $$\frac{16-(x-2)^2}{x^2-1} = 0$$ 4. **Solve the numerator equal to zero:** Since a fraction is zero only when its numerator is zero (and denominator is not zero), solve: $$16-(x-2)^2 = 0$$ 5. **Expand and simplify:** $$(x-2)^2 = 16$$ $$x-2 = \pm 4$$ 6. **Find the values of x:** $$x = 2 \pm 4$$ So, $$x = 6 \quad \text{or} \quad x = -2$$ 7. **Check denominator for these x values:** Denominator is $$x^2 - 1$$. For $$x=6$$: $$6^2 - 1 = 36 - 1 = 35 \neq 0$$ For $$x=-2$$: $$(-2)^2 - 1 = 4 - 1 = 3 \neq 0$$ Both values are valid. **Final answer:** The x-intercepts are at $$x=6$$ and $$x=-2$$.