Math 1150: Precalculus, Instructor: Dana
Ernst
Solutions to Review for Midterm Exam
#2
Here are the solutions to the review.
 (a)
(b)
(c)
(d)
(e)
 (a) All reals (b) No solution
(c)
(d)
, note that the second
interval has parentheses instead of brackets because those values are
not in the domain of the corresponding function.
 The graph is the graph of
reflected down,
stretched vertically by a factor of 2, shifted right 2, and up 1.



 See Book
 See Book
 n
 Yes, but one of the roots has to have
multiplicity 2. For example,
has 5 distinct real
roots and 1 has multiplicity 2 (for a total of 6 linear factors).
 No, complex nonreal roots always come in
conjugate pairs, and so there will always be an even number of
them. Hence there can’t be 3 of them. Note that my wording
was a bit ambiguous since I didn’t explicitly say “3 complex nonreal
roots.”

 0, 1, 4
 Many possible answers, but the most likely is
.
 The graph has xintercepts at
and
. The first of these
has multiplicity 2, and so graph bounces off xaxis at
. The other has
multiplicity 1, and so graph crosses at
.

 Only 1 is a root.

is a factor of a
polynomial if and only if h
is a zero of the polynomial.


 0, 3, 4


 Other zeros are:

 Vertical asymptote:
, Horizontal asymptote:
 I’ll do these for you in class.
 See Book