x/abs(x^2 - 4)
Would there be a vertical asymptote at only x = -2?
abs(x^2 - 4) = abs[(x - 2) (x + 2)]
For (x - 2), I keep thinking the absolute value bars would make the - 2 into a + 2. So I keep doing this:
abs(x - 2) = 5
abs(x) + abs(-2) = 5
abs(x) = 5 - 2
Why would the abs bars not doing anything to (x - 2)?