Unformatted text preview: ons are x = ±1, both of
which must be excluded from the domain. Hence, the domain of k is (−∞, −1) ∪ (1, ∞). To
build the sign diagram for k , we need the zeros of k . Setting k (x) = 0 results in √x2 −1 = 0.
We get 2x = 0 or x = 0. However, x = 0 isn’t in the domain of k , which means k has no zeros.
We construct our sign diagram on the domain of k below alongside the graph of k . It appears
that the graph of k has two vertical asymptotes, one at x = −1 and one at x = 1. The gap
in the graph between the asymptotes is because of the gap in the domain of k . Concerning
end behavior, there appear to be two horizontal asymptotes, y = 2 and y = −2. To see why
this is the case, we think of x → ±∞. The radicand of the denominator x2 − 1 ≈ x2 , and as
such, k (x) = √x2 −1 ≈ √ x2 = |2x| . As x → ∞, we have |x| = x so k (x) ≈ 2x = 2. On the other
hand, as x → −∞, |x| = −x, and as such k (x) ≈ −x = −2. Finally, it appears as thoug...
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