1. **State the problem:** Find the vertical asymptotes (VA) of the function $$f(x)=\frac{x^2+2x-3}{x^2-1}$$. 2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. 3. **Find the zeros of the denominator:** Solve $$x^2-1=0$$. 4. Factor the denominator: $$x^2-1=(x-1)(x+1)$$. 5. Set each factor equal to zero: $$x-1=0 \Rightarrow x=1$$ and $$x+1=0 \Rightarrow x=-1$$. 6. **Check the numerator at these points:** The numerator is $$x^2+2x-3$$. 7. Factor the numerator: $$x^2+2x-3=(x+3)(x-1)$$. 8. Evaluate numerator at $$x=1$$: $$1^2+2(1)-3=1+2-3=0$$. 9. Evaluate numerator at $$x=-1$$: $$(-1)^2+2(-1)-3=1-2-3=-4 \neq 0$$. 10. Since numerator and denominator both zero at $$x=1$$, this is a removable discontinuity, not a vertical asymptote. 11. At $$x=-1$$, denominator zero but numerator nonzero, so vertical asymptote at $$x=-1$$. **Final answer:** The function has a vertical asymptote at $$x=-1$$ only.