Math 121 – Section 4.2 Solutions
11. The domain of R(x) = 4x
x−3is all real numbers xexcept x= 3.
15. The domain of F(x) = 3x(x−1)
2x2−5x−3is all real numbers xexcept x=−1
2,3.
18. The domain of F(x) = x
x4−1is all real numbers xexcept x=±1.
24. Using the given graph of y=f(x), we have:
(a) the domain of f(x) is all real numbers xexcept x=−1; the range of f(x) is y > 0
(b) the y-intercept is y= 2; there are no x-intercepts
(c) y= 0 is the horizontal asymptote
(d) x=−1 is the vertical asymptote
(e) there is no oblique asymptote
27. Using the given graph of y=f(x), we have:
(a) the domain of f(x) is all real numbers xexcept x=±2; the range of f(x) is (−∞,0] ∪(1,∞)
(b) the y-intercept is y= 0; the x-intercept is x= 0
(c) y= 1 is the horizontal asymptote
(d) x=±2 are the vertical asymptotes
(e) there is no oblique asymptote
41. For the function R(x) = 3x
x+ 4 , the vertical asymptote is x=−4 and the horizontal asymptote is
y= 3.
44. For the function G(x) = −x2+ 1
x2−5x+ 6 , the vertical asymptotes are x= 2,3 and the horizontal asymp-
tote is y=−1.
51. First, rewrite the function G(x) = x3−1
x−x2as:
G(x) = (x−1)(x2+x+ 1)
−x(x−1)
We note that there is a hole at x= 1 since the multiplicity of 1 as a root of the numerator is the same
as that of the denominator. Therefore, x= 0 is the only vertical asymptote.
Further rewriting G(x), we have:
G(x) = −x2+x+ 1
x=−x−1−1
x
Therefore, the oblique asymptote is y=−x−1.
1