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**Unformatted text preview: **) (very small (+))
2 (very small (+))
very small (+) We conclude that as x → −2+ , f (x) → ∞.
• The behavior of y = f (x) as x → 2: Consider x → 2− . We imagine substituting
x = 1.999999. Approximating f (x) as we did above, we get
f (x) ≈ 6
3
3
=
≈
≈ very big (−)
(very small (−)) (4)
2 (very small (−))
very small (−) We conclude that as x → 2− , f (x) → −∞. Similarly, as x → 2+ , we imagine substituting
3
x = 2.000001, we get f (x) ≈ very small (+) ≈ very big (+). So as x → 2+ , f (x) → ∞.
Graphically, we have that near x = −2 and x = 2 the graph of y = f (x) looks like6
y −3 −1 1 3 x 5. Next, we determine the end behavior of the graph of y = f (x). Since the degree of the
numerator is 1, and the degree of the denominator is 2, Theorem 4.2 tells us that y = 0
is the horizontal asymptote. As with the vertical asymptotes, we can glean more detailed
3x
information using ‘number sense’. For the discussion below, we use the formula f (x) = x2 −4 .
• The be...

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