Solutions
Name:
Date: Friday June 3rd.
Exam 2
1 (2pts)
2 (6pts)
2
7 (10pts)
10
6
8 (16pts)
16
3 (2pts)
4 (4pts)
2
4
9 (26pts)
EC (10pts)
26
10
5 (4pts)
6 (24pts)
4
24
Total out of (94pts)
94
Directions: Put only your answers on the exam. Do all scratch work and side calculation on a
separate piece of scratch paper that is given out in class. Write neatly and legibly. If I can’t read it
you will receive a zero for that problem. Finally be sure to show all of your work (justification) for each
problem to receive full credit.
Deep breaths, trust in yourself. I know you can do this. Good Luck!
Same speech I know, but why change what works!
1
Definitions and Concepts
Directions: Complete the following definitions by inserting the word/s or mathematical text as indicated by the
labeled underlined blanks. If asked, formulate your own definition in your own words (as long as it is still correct of
course).
1. (2pts) (Average Rate of Change)
The average rate of change of a function f(x) over the interval [x1 , x2 ] is given by the expression
f(x2 ) − f(x1 )
x2 − x1
2. (6 pts total: 2pts each)
Let F(x) be a polynomial function. If f(c) = 0 or (c, 0) is a point on the graph of f, then
We define the
b.
of c as the number of times
c.
a.
is a factor of f.
appears in the factored form of f.
(a) (x-c)
(b) Multiplicity
(c) (x-c)
3. (2pts) Complex Conjugate Pairs Theorem: Given a polynomial f(x) with real coefficients, if a + bi is a complex
zero of f s.t. b ̸= 0 then so is
a-bi
4. (4 pts total: 2pts each) Rational Zeros: If f(x) = an xn + an−1 xn−1 + ... + a1 x + a0 is a polynomial function s.t.
p
ai ∈ Z, an ̸= 0, and f has a zero that is a rational number, say with p, q ∈ Z, then p must divide
and
a.
q
q must divide
b.
.
(a) a0
(b) an
5. (4pts) (Global Maximum and Minimum)
Define the Global Maximum and Minimum of a real function f(x) in terms of the range R (assuming they both
exist).
The Global Maximum of a function f(x) (should it exist) is defined as the value a in the range R of f s.t. every
other number in the range is less than a. The Global Minimum of f (should it exist) is defined as the value b
s.t. every other number in the range is greater than b.
2
Main Problems
Directions: Show all work and read the directions carefully.
6. (24 pts total)Define m(w) = 6w2 − 4w − 3
(a) (6 pts)Transform the function into vertex form (keeping it in functional form).
Solution. Using the technique of completing the square we proceed as follows
m(w) = 6w2 − 4w − 3
(
)
2
2
=6 w − w −3
3
(
)
1 1
2
2
=6 w − w+ −
−3
3
9 9
(
)2
1
6
=6 w−
− −3
3
9
(
)2
11
1
−
= 6 w−
3
3
(b) (2 pts)Determine the vertex.
(
Solution. vtx=
1 11
,−
3
3
)
(c) (4 pts)Find the Domain and Range of m.
[
Solution. Domain is R . Range is
−
11
,∞
3
)
(d) (2 pts)Find the ”global” maximum and minimum (if they exist else state they do not and why).
Solution. There is no max since the range has no upper bound i.e. → ∞. The min is −
(e) (4 pts)Determine the x and y intercepts (if they exist)
Solution. Evaluating m(0) we get 0 − 0 − 3 = −3 and thus the y-intercept is (0, −3).
Setting m(x) = 0 and solving we get
)2
(
11
1
−
=0
6 w−
3
3
(
)2
1
11
w−
=
3
18
1
w= ±
3
)
(
)
(
√
√
1
1
11
11
+
, 0 and
−
,0 .
This gives the x-int’s as
3
18
3
18
3
√
11
18
11
.
3
(f) (6 pts)Find the average rate of change over the interval [−3, 1]
Solution. Since m(−3) = 54 + 12 − 3 = 63 and m(1) = 6 − 4 − 3 = −1 we find the average rate of change to
−1 − 63
−64
be
=
= −16
1+3
4
7. (10 pts) Write a polynomial function that could represent the following graph given that the degree of the function
is 4.
Solution. f(x) =
−1
2
(x − 2) (x + 3) (x − 1)
18
4
8. (16 pts total) Define f(x) = 3x5 − 16x4 + 14x3 + 54x2 − 109x + 30.
(a) (10 pts) Find the zeros of f given that 3x3 − 4x2 − 17x + 6 is a factor.
Solution. Since 3x3 − 4x2 − 17x + 6 is a factor we can use long division to factor f as follows.
x2 − 4x + 5
)
3x3 − 4x2 − 17x + 6
3x5 − 16x4 + 14x3 + 54x2 − 109x + 30
− 3x5 + 4x4 + 17x3 − 6x2
− 12x4 + 31x3 + 48x2 − 109x
12x4 − 16x3 − 68x2 + 24x
15x3 − 20x2 − 85x + 30
− 15x3 + 20x2 + 85x − 30
0
This tell us that f(x) = (3x − 4x − 17x + 6)(x − 4x + 5) Using the rational root theorem we can choose
3
2
2
potential zeros to check using synthetic division. For instance
3
−2
3
−4
− 17
6
−6
20
−6
− 10
3
0
We now have f factored as f(x) = (x + 2)(3x2 − 10x + 3)(x2 − 4x + 5). At this point we can use the quadratic
1
formula to factor the remaining two quadratics to get f(x) = 3(x + 2)(x − )(x − 3)(x − (2 − i))(x − (2 + i)).
3
1
Thus the zeros of f are {−2, , 3, 2 + i, 2 − i}
3
(b) (6 pts) State the multiplicities of the zeros you find, and determine end behavior.
Solution. They all have multiplicity 1. As x → ∞ f(x) → ∞ and as x → −∞ f(x) → −∞.
5
9. (26 pts total) Find the zeros(2pts), state there mulitiplicities(2pts), the x, y intercepts (4pts), holes(2pts), vertical
asymptotes(2pts), slant/horizontal asymptote(4pts), end behavior(2pts), domain(2pts), and sketch a graph(6pts)(be
sure to label all the properties x/y int, v-asyms, holes etc.) for the rational function defined by
f(x) =
Solution. We can factor f as follows
3x3 − 4x2 − 17x + 6
6x2 − 5x + 1
1
3(x + 2)(x − )(x − 3)
3
f(x) =
(2x − 1)(3x − 1)
Thus the zeros are −2 and 3 with multiplicity 1 each.
The x-int are thus (−2, 0) and (3, 0). The y-int is (0, 6).
1
The hole is occurs when x = .
3
1
The vertical asymptote is x = .
2
Using long division we see that
6x2 − 5x + 1
)
1
2x
−
1
4
and thus f(x) tends toward
1
1
x − as x → ±∞
2
4
3x3 − 4x2 − 17x + 6
− 3x3 + 52 x2 − 12 x
− 32 x2 − 35
2 x +6
3 2
5
1
x
−
2
4x + 4
25
− 75
4 x+ 4
The End behavior is as x → ∞ f(x) → ∞ and as x → −∞ f(x) → −∞
1 1
The domain is R \ { , }.
2 3
6
E.C. (10 pts total: 2pts each)
• Descartes Rule of Signs: For f(x) = an xn + an−1 xn−1 + ... + a1 x + a0 .
(a) The number of
i.
real zeros is either the same as the number of
the expression f(x) or less that by an
iii.
number.
ii.
(b) The number of
iv.
real zeros is either the same as the number of
in the expression f(−x) or less that by an
vi.
number.
changes in
v.
changes
• Bounds on Zeros: When testing a potential zero c of a polynomial f(x) if
(a) c > 0 and the entire bottom line of coefficients when using synthetic division is
an
viii.
on the real zeros of f(x).
(b) c < 0 and the entire bottom line of coefficients when using synthetic division
both ±) then c is an
x.
on the real zeros of f(x).
i. positive
ii. sign
iii. even
iv. negative
v. sign
vi. even
vii. non-negative
viii. upper bound
ix. alternates in sign
x. lower bound
7
vii.
ix.
then c is
(0 taking on