How many vertical asymptotes does the equation $y=\frac{x-1}{x^2+6x-7}$ have?

Guest Aug 4, 2020

#1**0 **

To determine vertical asymptotes, you **must** first factor both the numerator and denominator and cancel

out those factors which are found in both the numerator and denominator.

y = (x - 1) / (x^{2} + 6x - 7) = (x - 1) / [ (x - 1)·(x + 7) ] = 1 / (x + 7)

Because there is a factor of (x + 7) remaining in the denominator, there is **one** vertical asymptote,

at x = -7.

The factor (x - 1) **does not** give a vertical asymptote, instead, it creates a "hole" in the graph at x = 1.

geno3141 Aug 4, 2020