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**Unformatted text preview: **on gives us
2
x4 + 1
= x2 − 1 + 2
2+1
x
x +1
The remainder is not zero so r(x) is already reduced.
3. To ﬁnd the x-intercept, we’d set r(x) = 0. Since there are no real solutions to
have no x-intercepts. Since r(0) = 1, so we get (0, 1) for the y -intercept. x4 +1
x2 +1 = 0, we 4. This step doesn’t apply to r, since its domain is all real numbers.
5. For end behavior, once again, since the degree of the numerator is greater than that of the
denominator, Theorem 4.2 doesn’t apply. We know from our attempt to reduce r(x) that we
can rewrite r(x) = x2 − 1 + x22 , and so we focus our attention on the term corresponding
+1
to the remainder, x22 It should be clear that as x → ±∞, x22 ≈ very small (+), which
+1
+1
means r(x) ≈ x2 − 1 + very small (+). So the graph y = r(x) is a little bit above the graph
of the parabola y = x2 − 1 as x → ±∞. Graphically,
y
5
4
3
2
1 x 14 But rest assured, some graphs do! 258 Rational Functions 6. There isn’t much work to do for a sign diagram for r(x), si...

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