MAC 1105
Final Exam Review
And Practice Final Exam Solutions
• You will need to have your own graphing
calculator for the test.
• You may not share calculators or use any type
of communication device in place of a
calculator.
• The final exam is comprehensive over the
entire course (not just the second half).
• Students who do not take the final exam will
receive a zero for the final exam unless prior
arrangements have been made for a rare
incomplete.
• If you miss the final and do not email me by the
day of the exam, I will assume you wish to take
a zero on your final exam. Communication
with me via email is key!
How to Study for the Final Exam
• Complete all assigned homework. Remember that an overall
(for the entire semester) homework score of at least 80% gives
you 10 bonus points on the test.
• Review your notes and applicable worksheet questions for each
of the 25 objectives in the chart below.
• Set up a study system (note cards for example) for each of the 25
objectives in the chart below. See the column below labeled
“important ideas”.
• Complete the practice test and check your solutions in class.
• Work the suggested text problems for each of the 25 objectives
in the chart below and check your solutions. The problems are
odd numbered so answers should be readily available.
Exam Topics 1 – 5
Objective
Sections
1) Solve a quadratic
equation using the
quadratic formula.
1.4
2) Use the discriminant to
determine the number and
type of solutions of a
quadratic equation.
1.4
3) Solve equations
graphically.
1.5
Suggested
Text
Problems
Important Ideas
p. 165: 31, 33,
35
π₯=
−π ±
π 2 − 4ππ
2π
π· = π 2 − 4ππ
p. 165: 47, 49
D = 0 → one real root
D > 0 → two real roots
D < 0 → two imaginary roots
See Notes
Enter each equation.
(Y = )
Begin with the standard window.
(ZOOM 6)
Adjust the window to see the intersection
(WINDOW)
Find the intersection
(2nd TRACE (CALC) 5)
The x coordinate is the solution.
4) Solve an absolute value
inequality.
1.7
5) Given two points, find
the distance between the
two points and the
midpoint.
2.1
p. 166:
99, 101
p. 293:
9a, 11a, 13a
|u| < a means
|u| ≤ a means
|u| > a means
|u| ≥ a means
−a < u < a
−a ≤ u ≤ a
u> a or u < −a
u≥ a or u ≤ −a
π=
(π₯2 − π₯1 )2 + (π¦2 − π¦1 )2
π₯1 + π₯2 π¦1 + π¦2
π=(
,
)
2
2
Exam Topics 6 – 10
6) Test an equation for
symmetry.
2.2
p. 293:
25, 27, 29, 31,
33
(symmetry
only)
1) Substitute:
x−axis symmetry: (−y) for y
y−axis symmetry: (−x) for x
origin symmetry: (−x) for x and (−y) for y
2) Determine if the result is the original equation.
7) Find the center and
radius of a circle.
2.2
p. 189:
57, 59, 71
8) Write the equation of a
circle.
2.2
p. 293:
35, 37
9) Find the slope of a line.
2.3
p. 293:
9c, 11c, 13c
10) Write a linear equation
given two ordered pairs of
data.
2.3
p. 295:
105a
Complete the square (if necessary) to transform the equation
to
π₯ − β 2 + π¦ − π 2 = π2
ππππ‘ππ β, π , πππππ’π π
ππππ‘ππ β, π , πππππ’π π
π₯ − β 2 + π¦ − π 2 = π2
πβππππ ππ π¦
m =πβππππ ππ π₯ =
π¦2 − π¦1
π₯2 − π₯1
1) Find the slope.
2) Use the point slope formula:
y – y1 = m(x– x1)
Exam Topics 11 – 15
11) Solve problems involving
linear regression.
2.3
p. 295:
111a, 111c
Enter the data
(STAT EDIT 1)
Find the regression coefficients.
(STAT CALC 4)
Write the equation of the line.
(π¦ = ππ₯ + π)
12) Solve a system of two
equations by substitution or
elimination
5.1
p. 487:
13,15,17,19
13) Work an application of
systems of linear equations.
5.4
p. 488:
87, 91,95, 99
Set up a system of two equations in two unknowns. Solve the
system.
14) Use your calculator to
find intervals of increasing
and decreasing.
2.5
p. 236:
11
Graph the function.
Adjust the window to show all of the turning points.
The intervals of increasing/decreasing are intervals of the domain
(x).
15) Classify a function as
even, odd, or neither.
2.5
p. 237:
33, 35, 37, 41,
43, 45
1)
2)
3)
Substitution:
Solve one of the equations for one the variables (usually y).
Substitute that result into the other equation.
Solve.
Elimination:
1) Get the coefficients of one of the variables to be additive
inverses if needed (by multiplication).
2) Add the equations to eliminate that variable.
3) Solve.
4) Back substitute to find the value of the other variable.
Find −π₯ .
π −π₯ = π π₯ → ππ£ππ
π −π₯ = −π π₯ → πππ
Exam Topics 16 – 20
16) Evaluate a difference
quotient.
17) Find function values of a
piece-wise function.
18) Write the equation of a
function whose graph has been
transformed.
2.5
2.6
2.7
p. 239:
79
p. 248:
21a, 47
p. 265: 67, 69,
71, 73
π π₯ + β − π(π₯)
β
Make sure to pay attention to which “piece” of the function you should
use.
Be sure to know the ten different transformations from section 2.7:
π π₯ ± π → π’π πππ€π π π’πππ‘π (π’π ππ +)
π π₯ ± π → ππππ‘ πππβπ‘ π π’πππ‘π (ππππ‘ ππ +)
ππ(π₯) → π π‘πππ‘πβππ π£πππ‘ππππππ¦ ππ¦ π ππππ‘ππ ππ π
ππ π > 1,
1
π βππ’ππ ππ¦ ππ 0 < π < 1
π
π ππ₯ → π βππ’ππ βππππ§πππ‘ππππ¦ πππ π > 1,
π π‘πππ‘πβππ βππππ§πππ‘ππππ¦ πππ 0 < π < 1
−π π₯ → πππππππ‘ππ ππππ’π‘ π₯
π −π₯ → πππππππ‘ππ ππππ’π‘ π¦
19) Find the vertex of a
quadratic function.
3.1
p. 307: 53, 55
(vertex only)
Use the formula:
π
π
The vertex of π₯ = ππ₯ 2 + ππ₯ + π is − 2π , π − 2π
OR
Graph the function and find the max/min.
20) Find any vertical or
horizontal asymptotes for a
given rational function.
3.6
p. 373: 25, 27,
29, 31, 35, 37,
39, 41
Vertical: Simplify the function completely. The vertical line π₯ = π is a
vertical asymptote of the function if π causes the denominator to be
zero (after simplification).
Horizontal: n = degree of numerator, m = degree of denominator
π < π → π₯ − ππ₯ππ
π = π → π¦ = πππ‘ππ ππ πππππππ πππππ.
π > π → ππ π». π΄.
Exam Topics 21 – 25
21) Work a variation
application.
22) Solve an exponential
equation by writing both sides
to the same base and/or by
taking the (natural/common)
logarithm of both sides.
3.8
p. 393:
29, 33, 35, 37
1.
2.
3.
4.
4.1, 4.5
p. 473:
55, 57, 59, 63
Convert both sides to the same base and set the exponents equal to
one another (if possible).
Use the given information to set up an equation.
Solve for k → the constant of variation.
Write a new equation using all variables and the constant k.
Input the new information and compute the answer.
OR
Take the (natural/common) logarithm of both sides. Use the third
property of logarithms to “move” the variable to the outside.
Solve for the variable.
23) Work with compound
interest formulas (including
compounding continuously).
4.2
24) Write a logarithmic
expression in expanded form
or condensed form.
4.4
p. 425:
19, 21
π
π΄ = π(1 + )ππ‘
π
π΄ = ππ ππ‘
p. 453:
19, 21, 23, 25,
39, 43
ππππ ππ = ππππ π + ππππ π
π
ππππ
= ππππ π − ππππ π
π
ππππ ππ = πππππ π
25) Solve a logarithmic
equation by consolidating the
logarithms to a single
logarithmic expression.
4.3, 4.5
p. 467:
53, 55, 57, 59
1) Isolate all logarithmic expressions on one side of the equation
and use logarithmic properties to rewrite the logarithmic
expressions as a single logarithmic expression .
2) Rewrite the logarithmic expression as an exponential
expression.
3) Solve.
4) Reject values that produce the logarithm of a negative number
or the logarithm of zero.
Special Note
• It is entirely possible that the questions on
the test may have additional parts that differ
slightly from those in the practice test.
• To prepare for these additional parts, be
sure you are familiar with your notes, your
worksheets, and your homework.
• Objectives will remain unchanged.
Practice Test Solutions
1a) Solve: π₯ 2 = 5 − 8π₯
π₯ 2 + 8π₯ − 5 = 0 β¨ π = 1, π = 8, π = −5
−π ± π2 − 4ππ
π₯=
2π
−8 ± 82 − 4(1)(−5)
=
2(1)
−8 ± 64 + 20
=
2
−8 ± 84
=
2
−8 ± 4 β 21
=
2
−8 ± 2 21
=
2
= −4 ± 21
{−4 − 21, −4 + 21 }
1b) Solve: π₯ 2 + 4π₯ + 8 = 0
π₯ 2 + 4π₯ + 8 = 0 β¨ π = 1, π = 4, π = 8
−π ± π 2 − 4ππ
π₯=
2π
−4 ± 42 − 4(1)(8)
=
2(1)
−4 ± 16 − 32
=
2
−4 ± −16
=
2
−4 ± 4π
=
2
= −2 ± 2π
{−2 + 2π, −2 − 2π }
1c) Solve: π₯ 2 = 15 + 9π₯
π₯ 2 − 9π₯ − 15 = 0 β¨ π = 1, π = −9, π = −15
−π ± π2 − 4ππ
π₯=
2π
−(−9) ± (−9)2 − 4(1)(−15)
=
2(1)
9 ± 81 + 60
=
2
9 ± 141
=
2
9 − 141 9 + 141
{
,
}
2
2
2) Find the discriminant and the number and type of roots:
8π¦ 2 = −5π¦ − 3
8π¦ 2 + 5π¦ + 3 = 0 β¨ π = 8, π = 5, π = 3
π· = π2 − 4ππ
= 5 2− 4 8 3
= 25 − 96
= −71
π· ππ πππππ‘ππ£π β¨ 2 πππππππππ¦ ππππ‘π
π· = −71, two imaginary roots
3) Use the intersection – of – graphs method to approximate the
solution to nearest hundredth: −3.5 20 + 7 π₯ + 6.4(14ππ₯ +
π1 = −3.5 20 + 7 π₯ + 6.4 14ππ₯ + 0.6
π2 = −5
π΄πππ’π π‘ π‘βπ π₯ π‘π −2,2 π‘βππ πβπππ π 2ππ πππππ πΆπππ 5
πβπ π πππ’π‘πππ ππ π‘βπ π₯ − πππππππππ‘π ππ π‘βπ
πππππ‘ ππ πππ‘πππ πππ‘πππ.
{0.22}
4a) Solve: 8π₯ − 8 ≥ 9
8π₯ − 8 ≤ −9 ππ 8π₯ − 8 ≥ 9
8π₯ ≤ −1 ππ 8π₯ ≥ 17
1
17
π₯ ≤ − ππ π₯ ≥
8
8
1
17
(−∞, − ]π [ , ∞)
8
8
4b) Solve: 9π₯ + 3 < 19
−19 < 9π₯ + 3 < 19
−22 < 9π₯ < 16
22
16
−
<π₯<
9
9
22 16
(− , )
9 9
5a) Find the distance between P and Q.
π 4, −4 , π −4, 2
π=
(π₯2 − π₯1 )2 + (π¦2 − π¦1 )2
=
((−4) − 4)2 + (2 − (−4))2
=
−8
2
+ 6
2
10
=
100 = 10
5b) Find the midpoint of P and Q.
π 4, −4 , π −4, 2
π₯1 + π₯2 π¦1 + π¦2
π=(
,
)
2
2
4 + −4 −4 + 2
=(
,
)
2
2
0 2
=
,−
2 2
(0, −1)
6a) Test the equation for symmetry:
π¦ = −2π₯ 3 + 5π₯
π₯: −π¦ = −2π₯ 3 + 5π₯ β¨ π¦ = 2π₯ 3 − 5π₯ ππ
π¦: π¦ = −2 −π₯
3
+ 5 −π₯ β¨ π¦ = 2π₯ 3 − 5π₯ ππ
ππππππ: −π¦ = −2 −π₯ 3 + 5 −π₯ β¨
−π¦ = 2π₯ 3 − 5π₯
β¨ π¦ = −2π₯ 3 + 5π₯ π¦ππ
ππππππ
6b) Test the equation for symmetry:
π¦ = 2π₯ 2 − 2
π₯: −π¦ = 2π₯ 2 − 2 β¨ π¦ = −2π₯ 2 − 2 ππ
π¦: π¦ = 2 −π₯
2
− 2 β¨ π¦ = 2π₯ 2 − 2 π¦ππ
ππππππ: −π¦ = 2 −π₯ 2 − 2 β¨
−π¦ = 2π₯ 2 − 2 ππ
π¦ − ππ₯ππ
7) Find the center and radius of the circle.
π₯ 2 + π¦ 2 + 8π₯ + 10π¦ = 23
π₯ 2 + 8π₯ +
+ π¦ 2 + 10π¦ +
1
8 = 4, 42 = 16
2
1
10 = 5, 52 = 25
2
= 23
π₯ 2 + 8π₯ + 16 + π¦ 2 + 10π¦ + 25 = 23 + 16 + 25
π₯+4
π₯ −β
2
+ π¦ + 5 2 = 64
2 + π¦ − π 2 = π2
−4, −5 , π = 8
8) Find the equation of the circle that satisfies the given
conditions: Cπππ‘ππ ππ‘ 0, −7 , πππππ’π 9
π₯ −β
π₯−0
2
2
+ π¦ −π
2
= π2
+ π¦— −7
2
= 92
π₯2 + π¦ + 7
2
= 81
9a) Find the slope of the line through the given pair of points.
−9, −1 , (4, 7)
π¦2 −
π=
π₯2 −
7 − (−1)
=
4 − (−9)
8
=
13
8
13
π¦1
π₯1
9b) Find the slope of the line through the given pair of points.
−7, −1 , 8, −1
π¦2 −
π=
π₯2 −
−1 − −1
=
8 − −7
0
=
15
=0
0
π¦1
π₯1
10) After three seconds, a bottle has 15 ounces of water. After 10
seconds, the bottle has 43 ounces of water. Write a linear equation
that models the amount of water in the bottle in terms of x.
3 π ππππππ β¨ 15 ππ’ππππ : 3, 15
10 π ππππππ β¨ 43 ππ’ππππ : 10, 43
43 − 15 28
π=
=
=4
10 − 3
7
π¦ − 15 = 4 π₯ − 3
π¦ − 15 = 4π₯ − 12
π¦ = 4π₯ + 3
CHECK:
π¦ = 4 3 + 3 = 12 π¦ππ , π¦ = 4 10 + 3 = 43 (π¦ππ )
π¦ = 4π₯ + 3
11) The paired data consists of the costs of advertising (in thousands
of dollars) and the number of products sold (in thousands). Find the
equation of the regression line that models the data.
Cost
9
2
3
4
2
5
9
10
Number
85
52
55
68
67
86
83
73
πππππ π‘βπ "Costs" ππ πΏ1 πππ π‘βπ "ππ’πππππ " ππ πΏ2
πππ΄π β¨ πΆπ΄πΏπΆ β¨ 4
π ≈ 2.79, π ≈ 55.8
π¦ = 2.79π₯ + 55.79
12) Solve the system of equations.
6π₯ + 9π¦ = −69
3π₯ − 2π¦ = −2
6π₯ + 9π¦ = −69
−2 3π₯ − 2π¦ = −2
6π₯ + 9π¦ = −69
−6π₯ + 4π¦ = 4
13π¦ = −65
π¦ = −5
6π₯ + 9 −5 = −69
6π₯ − 45 = −69
6π₯ = −24
π₯ = −4
{ −4, −5 }
13) Jim wants to plan a meal with 127 grams of carbohydrates and 1150 calories. If green
beans have 7 grams of carbohydrates and 30 calories per half – cup serving and if fried
shrimp have 9 grams of carbohydrates and 190 calories per three – ounce serving, how
many servings of green beans and shrimp should he use?
1
π₯ = ππ’ππππ ππ ππ’π π πππ£ππππ ππ πππππ πππππ ,π¦ = ππ’ππππ ππ π‘βπππ ππ’πππ βπππππππ ππ π βππππ
2
7π₯ + 9π¦ = 127
30π₯ + 190π¦ = 1150
ππππππππ‘πππ
−30 7π₯ + 9π¦ = 127
7 30π₯ + 190π¦ = 1150
−210π₯ − 270π¦ = −3810
210π₯ + 1330π¦ = 8050
1060π¦ = 4240
π¦=4
7π₯ + 9 4 = 127
7π₯ + 36 = 127
7π₯ = 91
π₯ = 13
13 βπππ ππ’ππ ππ πππππ πππ 4 π‘βπππ ππ’πππ βπππππππ ππ π βππππ
14) By using a graphing calculator, find the increasing interval(s)
for the function f x = 2π₯ 3 − 3π₯ 2 − 36π₯ + 30
Choose windows large enough to allow
you to see the turning points:
π₯: −10, 10
π¦: −200, 200
−∞, −2 , (3, ∞)
15a) Determine if the given function is even, odd, or neither.
π π₯ = 4π₯ 2 − 2
π −π₯ = 4 −π₯ 2 − 2
= 4π₯ 2 − 2
πΈπ£ππ
15b) Determine if the given function is even, odd, or neither.
π π₯ = −5π₯ 5 + 9π₯ 3
π −π₯ = −5 −π₯ 5 + 9 −π₯
= −5 −π₯ 5 + 9 −π₯ 3
= 5π₯ 5 − 9π₯ 3
= −1 −5π₯ 5 + 9π₯ 3
πππ
3
15c) Determine if the given function is even, odd, or neither.
π π₯ = 9π₯ 4 − 4π₯ − 6
π −π₯ = 9 −π₯
4
− 4 −π₯ − 6
= 9π₯ 4 + 4π₯ − 6
ππππ‘βππ
15d) Determine if the given function is even, odd, or neither.
π₯
π π₯ = 3
π₯ − 7π₯
−π₯
π −π₯ =
−π₯ 3 − 7(−π₯)
−π₯
=
−π₯ 3 + 7π₯
−1 π₯
=
−1 π₯ 3 − 7π₯
π₯
= 3
π₯ − 7π₯
Even
π π₯+β −π π₯
16) Determine the difference quotient,
β
2
π π₯ = 10π₯ + 13π₯ − 11
, for the function
π π₯+β −π π₯
β
10 π₯ + β 2 + 13 π₯ + β − 11 − (10π₯ 2 + 13π₯ − 11)
=
β
10 π₯ 2 + 2π₯β + β2 + 13π₯ + 13β − 11 − 10π₯ 2 − 13π₯ + 11
=
β
10π₯ 2 + 20π₯β + 10β2 + 13π₯ + 13β − 11 − 10π₯ 2 − 13π₯ + 11
=
β
20π₯β + 10β2 + 13β
=
β
= 20π₯ + 10β + 13
20π₯ + 13 + 10β
17) In Country X, the average hourly wage in dollars from
1945 to 1995 can be modeled by
π π₯ =
0.076 π₯ − 1945 + 0.36 ππ 1945 ≤ π₯ < 1970
0.184 π₯ − 1970 + 3.07 ππ 1970 ≤ π₯ ≤ 1995
Use π to estimate the average hourly wages
in 1950, 1970, and 1990.
π 1950 = 0.076 1950 − 1945 + 0.36 = 0.74
π 1970 = 0.184 π₯ − 1970 + 3.07 = 3.07
π 1990 = 0.184 1990 − 1970 + 3.07 = 6.75
$0.74, $3.07, $6.75
18) Write an equation for the function whose
graph fits the following description.
The graph of π π₯ = |π₯| is shifted three units left,
reflected in the y-axis and shifted two units
down.
π βπππ‘ππ π‘βπππ π’πππ‘π ππππ‘: π¦ = π₯ + 3
πππππππ‘ππ ππ π‘βπ π¦ − ππ₯ππ : π¦ = −π₯ + 3
π βπππ‘ππ 2 π’πππ‘π πππ€π: π¦ = −π₯ + 3 − 2
π¦ = −π₯ + 3 − 2
19) Find the vertex of the parabola π¦ = π₯ 2 + 6π₯ + 10.
π = 1, π = 6, π = 10
π
6
6
−
=−
= − = −3
2π
2β1
2
π −3 = −3
2
+ 6 −3 + 10
= 9 − 18 + 10
=1
OR
graph the function and find the minimum.
(−3,1)
20a) Find the vertical asymptotes, if any, of
π₯+5
π π₯ =
π₯−4
π₯−4=0
β¨π₯=4
π₯=4
20b) Find the horizontal asymptotes, if any, of
−4π₯ 2 + 7
β π₯ =
4π₯ 2 − 1
πππππππ πππ πππ’ππ
4
β¨π¦= −
4
β¨ π¦ = −1
π¦ = −1
21) The volume V of a gas varies inversely as the pressure P on it. The volume of a
ππ
gas is 190 ππ3 under a pressure of 21 ππ2 . What will its volume be under a
ππ
pressure of 35 2?
ππ
Assume the temperature is constant.
π
π=
π
π
190 =
21
π = 3990
3990
3990
π=
β¨π=
= 114
π
35
114 ππ3
22a) Solve the equation 8π₯−2 = 644π₯
8
π₯−2
= 82
4π₯
π₯ − 2 = 2 4π₯
π₯ − 2 = 8π₯
−7π₯ = 2
2
π₯= −
7
2
{− }
7
22b) Solve. Please round your answer in part b to
three decimal places. 2π₯ = 52π₯+1
ln 2π₯ = ln 52π₯+1
π₯ ln 2 = 2π₯ + 1 ln 5
π₯ ln 2 = 2π₯ ln 5 + ln 5
π₯ ln 2 − 2π₯ ln 5 = ln 5
π₯ (ln 2 − 2 ln 5) = ln 5
ln 5
π₯=
≈ −0.637
ln 2 − 2 ln 5
{−0.637}
23a) Determine the final value of the given amount:
$820 at 8% compounded continuously for 2 years
π΄ = ππ ππ‘
π΄ = 820π 0.08
= 820π .16
≈ 962.28
$962.28
2
23b) Randy invested his inheritance in an account that paid 6.2%
interest, compounded continuously. After ten years, he found that
he now had $62,978.62. What was the original amount of his
inheritance?
62978.62 = ππ .062β10
62978.62
π=
π .62
≈ 33879
$33,879
23c) Find the time required for an initial investment of $10,000 to grow to
$18,000 at an annual rate of 6% if the interest is compounded quarterly.
π
π΄=π 1+
π
ππ‘
0.06
18000 = 10000 1 +
4
1.8 = 1.0154π‘
ln 1.8 = ln 1.0154π‘
ln 1.8 = 4π‘ ln 1.015
ln 1.8
π‘=
≈ 9.87
4 ln 1.015
9.87 π¦ππππ
4π‘
24a) Write the expression in expanded form.
π₯4π§
log 8
π¦
π₯4π§
log 8
π¦
= log π₯ 4 π§ − log π¦ 8
= log π₯ 4 + log π§ − 8 log π¦
= 4 log π₯ + log π§ − 8 log π¦
4 log π₯ + log π§ − 8 log π¦
24b) Write the expression in condensed form.
1
ππππ π₯ + 5ππππ π¦ − 2ππππ π₯
2
1
ππππ π₯ 2
+ ππππ π¦ 5 − ππππ π₯ 2
= ππππ π₯ 1/2 π¦ 5 − ππππ π₯ 2
= ππππ
= ππππ
ππππ
1
π₯ 2π¦5
π₯2
π¦5
π₯ 3/2
π¦5
3
π₯2
ππ
ππππ
π¦5
π₯3
3
25a) Solve. log 27 1 − π₯ =
1
273
3
1
3
3
= 1−π₯
3
27 = 1 − π₯
27 = 1 − π₯
1 − π₯ = 27
−π₯ = 26
π₯ = −26
Note the lack of domain issues. When you substitute −26
in for π₯, the result is the cube root of 27 which is 3 and
certainly greater than zero.
{−26}
25b) Solve.
2
log(π₯ +5π₯
+ 16) = 1
101 = π₯ 2 + 5π₯ + 16
10 = π₯ 2 + 5π₯ + 16
π₯ 2 + 5π₯ + 16 = 10
π₯ 2 + 5π₯ + 6 = 0
π₯+3 π₯+2 =0
π₯ = −3, −2
Note there are no domain issues. The first step indicates
that the solutions will make the expression inside the
logarithm equal to 10, which is certainly greater than zero.
{−3, −2}
25c) Solve.
log π₯ − 9 = 1 − log π₯
log π₯ − 9 + log π₯ = 1
log π₯ − 9 π₯ = 1
101 = π₯ − 9 π₯
10 = π₯ 2 − 9π₯
π₯ 2 − 9π₯ − 10 = 0
π₯ − 10 π₯ + 1 = 0
π₯ = 10, π₯ = −1
There are domain issues.
π₯ ≠ −1 ππ log −1 − 9 ππππ πππ‘ ππ₯ππ π‘
{10}
Extra Practice
1) Solve.
5π₯ 2 − 6π₯ = 4π₯ 2 + 6π₯ − 3
1) Solve.
5π₯ 2 − 6π₯ = 4π₯ 2 + 6π₯ − 3
π₯ 2 − 12π₯ + 3 = 0
β¨ π = 1, π = −12, π = 3
π₯=
=
=
−π ±
−(−12) ±
12 ±
π2 − 4ππ
2π
(−12)2 − 4(1)(3)
2(1)
144 − 12 12 ± 132
=
2
2
12 ± 4 β 33 12 ± 2 33
=
=
2
2
= 6 ± 33
{6 − 33, 6 + 33 }
2) Find the discriminant and the number and type of roots:
16x 2 − 88x + 121 = 0
2) Find the discriminant and the number and type of roots:
16x 2 − 88x + 121 = 0
π 2 − 4ππ = −88
2
− 4 16 121
= 7744 − 7744
=0
1 ππππ ππππ‘
3) Use the intersection – of – graphs method to approximate the
solution to nearest hundredth:
4
−ππ₯ + 5 = 9 + 0.4π₯
3) Use the intersection – of – graphs method to approximate the
solution to nearest hundredth:
4
−ππ₯ + 5 = 9 + 0.4π₯
π1 = −ππ₯ + 5
4
π2 = 9 + 0.4π₯
{0.92}
4a) Solve.
3π₯ − 3 ≥ 15
4a) Solve.
3π₯ − 3 ≥ 15
3π₯ − 3 ≤ −15 ππ 3π₯ − 3 ≥ 15
3π₯ ≤ −12 ππ 3π₯ ≥ 18
π₯ ≤ −4 ππ π₯ ≥ 6
(−∞, −4]π [6, ∞)
4b)Solve.
3π₯ + 4 ≤ 19
4b) Solve.
3π₯ + 4 ≤ 19
−19 ≤ 3π₯ + 4 ≤ 19
−23 ≤ 3π₯ ≤ 15
23
15
−
≤π₯≤
3
3
23
[− , 5]
3
5a) Find the distance between P and Q.
π −4, 1 , π −7, −9
5a) Find the distance between P and Q.
π −4, 1 , π −7, −9
π=
=
(π₯2 − π₯1 )2 + (π¦2 − π¦1 )2
( −7 − −4 )2 + ((−9) − (1))2
=
−3
=
2
+ −10
109
2
5b) Find the midpoint of P and Q.
π −4, 1 , π −7, −9
5b) Find the midpoint of P and Q.
π −4, 1 , π −7, −9
π₯1 + π₯2 π¦1 + π¦2
π=(
,
)
2
2
−4 + −7 1 + −9
=(
,
)
2
2
−11 −8
=
,
2
2
11
(− , −4)
2
6) Test the equation for symmetry:
π¦ = 2π₯ 2 − |π₯|
6) Test the equation for symmetry:
π¦ = 2π₯ 2 − |π₯|
π₯: −π¦ = 2π₯ 2 − |π₯| ππ
π¦: π¦ = 2 −π₯
2
− |π₯| β¨ π¦ = 2π₯ 2 − |π₯| π¦ππ
ππππππ: −π¦ = 2 −π₯
2
− π₯ β¨ −π¦ = 2π₯ 2 − |π₯|
no
π¦ − ππ₯ππ
7) Find the center and radius of the circle.
π₯ 2 + π¦ 2 − 2π₯ − 2π¦ − 4 = 0
7) Find the center and radius of the circle.
π₯ 2 + π¦ 2 − 2π₯ − 2π¦ − 4 = 0
π₯ 2 − 2π₯ +
+ π¦ 2 − 2π¦ +
=4
1
−2 = −1, (−1)2 = 1
2
π₯ 2 − 2π₯ + 1 + π¦ 2 − 2π¦ + 1 = 4 + 1 + 1
π₯−1 2+ π¦−1 2 =6
π₯ − β 2 + π¦ − π 2 = π2
1, 1 , π = 6
8) Find the equation of the circle that satisfies the given
conditions: Cπππ‘ππ ππ‘ −2, −3 , πππππ’π 7
8) Find the equation of the circle that satisfies the given
conditions: Cπππ‘ππ ππ‘ −2, −3 , πππππ’π 7
π₯ −β
π₯ − (−2
2
2
+ π¦ −π
2
+ π¦— (−3)
(π₯ + 2)2 + π¦ + 3
2
= π2
2
= 7
=7
2
9) Find the slope of the line through the given pair of points.
3, −2 , (2, −3)
9) Find the slope of the line through the given pair of points.
3, −2 , (2, −3)
π¦2 −
π=
π₯2 −
π¦1
π₯1
−3 − (−2)
=
2−3
−1
=
−1
=1
10) A manufacturer produces 50 TVs at a cost of $17,500 and 75 TVs at a cost of
$21,250. Write a linear equation that models the cost in terms of the number of TVs
10) A manufacturer produces 50 TVs at a cost of $17,500 and 75 TVs at a cost of
$21,250. Write a linear equation that models the cost in terms of the number of TVs
50 πππ β¨ $17,500: 50, 17500
75 πππ β¨ $21,250: 75, 21250
21250 − 17500 3750
π=
=
= 150
75 − 50
25
π¦ − 17500 = 150 π₯ − 50
π¦ − 17500 = 150π₯ − 7500
π¦ = 150π₯ + 10000
CHECK:
π¦ = 150 50 + 10000 = 17500 π¦ππ ,
π¦ = 150 75 + 10000 = 21250 (π¦ππ )
π¦ = 150π₯ + 10000
11) The paired data consists of the number of newspapers (in thousands)
that sell at a given price (in nickels) per copy for a college newspaper.
Find the equation of the regression line that models the data.
Newspapers
1
2
3
4
5
Price
11
8
6
4
3
11) The paired data consists of the number of newspapers (in thousands)
that sell at a given price (in nickels) per copy for a college newspaper.
Find the equation of the regression line that models the data.
Newspapers
1
2
3
4
5
Price
11
8
6
4
3
πππππ π‘βπ "Newspapers" ππ πΏ1
πππ π‘βπ "πππππ" ππ πΏ2
πππ΄π β¨ πΆπ΄πΏπΆ β¨ 4
π = −2, π = 12.4
π¦ = −2π₯ + 12.4
12) Solve.
5π₯ − π¦ = 5
3π₯ + 2π¦ = −10
12) Solve.
5π₯ − π¦ = 5
3π₯ + 2π¦ = −10
2(5π₯ − π¦ = 5)
3π₯ + 2π¦ = −10
10π₯ − 2π¦ = 10
3π₯ + 2π¦ = −10
13π₯ = 0
π₯=0
5 0 −π¦ =5
−π¦ = 5
π¦ = −5
{ 0, −5 }
13) Last year Mrs. Garcia invested $50,000. She put part of the money in a real estate
venture that paid 7.5% and the rest in a small – business venture that paid 12%. The
combined income from the two investments was $5190. How much did she invest in each?
13) Last year Mrs. Garcia invested $50,000. She put part of the money in a real estate
venture that paid 7.5% and the rest in a small – business venture that paid 12%. The
combined income from the two investments was $5190. How much did she invest in each?
π₯ = πππππ¦ ππ ππππ ππ π‘ππ‘π, π¦ = πππππ¦ ππ π ππππ ππ’π ππππ π
π₯ + π¦ = 50000
0.075π₯ + 0.12π₯ = 5190 → 75π₯ + 120π¦ = 5190000
ππππππππ‘πππ
−75 π₯ + π¦ = 50000
75π₯ + 120π¦ = 5190000
−75π₯ − 75π¦ = −3750000
75π₯ + 120π¦ = 5190000
45π¦ = 1440000
π¦ = 32000
π₯ + 32000 = 50000 → π₯ = 18000
$18,000 in real estate, $32,000 in small business
14) By using a graphing calculator, find the increasing interval(s)
1
for the function f x = π₯ 3 − 4π₯
3
14) By using a graphing calculator, find the increasing interval(s)
1
for the function f x = π₯ 3 − 4π₯
3
Choose windows large enough to allow
you to see the turning points:
π₯: −10, 10
π¦: −10, 10
−∞, −2 , (2, ∞)
15) Determine if the given function is even, odd, or neither.
π₯2 + 2
π π₯ = 4
π₯ +1
15) Determine if the given function is even, odd, or neither.
π₯2 + 2
π π₯ = 4
π₯ +1
(−π₯)2 +2
π −π₯ =
(−π₯)4 +1
π₯2 + 2
= 4
π₯ +1
πΈπ£ππ
π π₯+β −π π₯
16) Determine the difference quotient,
β
2
π π₯ = 3π₯ − 2π₯ + 5
, for the function
π π₯+β −π π₯
16) Determine the difference quotient,
β
2
π π₯ = 3π₯ − 2π₯ + 5
, for the function
π π₯+β −π π₯
β
2
3 π₯ + β − 2 π₯ + β + 5 − (3π₯ 2 − 2π₯ + 5)
=
β
3 π₯ 2 + 2π₯β + β2 − 2π₯ − 2β + 5 − 3π₯ 2 + 2π₯ − 5
=
β
3π₯ 2 + 6π₯β + 3β2 − 2π₯ − 2β + 5 − 3π₯ 2 + 2π₯ − 5
=
β
6π₯β + 3β2 − 2β
=
β
= 6π₯ + 3β − 2
17) Suppose a state’s income tax code states that the tax
liability T on x dollars of taxable income as follows
T π₯ =
0.04π₯
800 + 0.06π₯
ππ 0 ≤ π₯ < 20,000
ππ π₯ ≥ 20,000
Use π to find the tax liability on each taxable income:
$12,000, $20,000, $50,000
17) Suppose a state’s income tax code states that the tax
liability T on x dollars of taxable income as follows
T π₯ =
0.04π₯
800 + 0.06π₯
ππ 0 ≤ π₯ < 20,000
ππ π₯ ≥ 20,000
Use π to find the tax liability on each taxable income:
$12,000, $20,000, $50,000
π 12000 = 0.04(12000) = $480
π 20000 = 800 + 0.06 20000 = $2000
π 50000 = 800 + 0.06 50000 = $3800
$480, $2000, $3800
18) Write an equation for the function whose
graph fits the following description.
The graph of π π₯ = π₯ 2 is shifted five units right,
stretched vertically by a factor of three, reflected in
the y-axis and shifted four units up.
18) Write an equation for the function whose
graph fits the following description.
The graph of π π₯ = π₯ 2 is shifted five units right, stretched vertically by a
factor of three, reflected in the y-axis and shifted four units up.
π βπππ‘ππ πππ£π π’πππ‘π πππβπ‘: π¦ = π₯ − 5
2
π π‘πππ‘πβππ π£πππ‘ππππππ¦ ππ¦ π ππππ‘ππ ππ π‘βπππ: π¦ = 3 π₯ − 5
πππππππ‘ππ ππ π‘βπ π¦ − ππ₯ππ : π¦ = 3 −π₯ − 5
π βπππ‘ππ 4 π’πππ‘π π’π: π¦ = 3 −π₯ − 5
π¦ = 3 −π₯ − 5
2
+4
2
+4
2
2
19) Find the vertex of the parabola π¦ = −2π₯ 2 + 10π₯ − 3.
19) Find the vertex of the parabola π¦ = −2π₯ 2 + 10π₯ − 3.
π = −2, π = 10, π = −3
π
10
10 5
−
=−
=−
= (ππ 2.5)
2π
2 β −2
−4 2
5
5
π
= −2
2
2
=−
2
+ 10
5
−3
2
25
+ 25 − 3
2
19
=
(ππ 9.5)
2
OR
graph the function and find the maximum.
5 19
,
ππ (2.5, 9.5)
2 2
20a) Find the vertical asymptotes, if any, of
(π₯ − 3)(2π₯ − 2)
π π₯ =
(π₯ − 3)(π₯ + 4)
20a) Find the vertical asymptotes, if any, of
(π₯ − 3)(2π₯ − 2)
π π₯ =
(π₯ − 3)(π₯ + 4)
ππππ¦ π€βπππ π₯ + 4 = 0
π₯ = −4
(πππ‘ " − 4" → π¦ππ’ ππ’π π‘ π€πππ‘π π₯ = −4
π ππππ ππ‘ ππ π ππππ)
20b) Find the horizontal asymptotes, if any, of
2π₯ − 3
β π₯ =
3π₯ + 5
20b) Find the horizontal asymptotes, if any, of
2π₯ − 3
β π₯ =
3π₯ + 5
πππππππ πππ πππ’ππ
2
β¨π¦=
3
2
2
(πππ‘ → π¦ππ’ ππ’π π‘ π€πππ‘π π¦ =
3
3
π ππππ ππ‘ ππ π ππππ)
21) The safe load that a rectangular beam can support varies
jointly as the width and square of the depth of the beam and
inversely as its length. A beam 4 inches wide, 6 inches deep,
and 25 feet long can support a safe load of 576 pounds. Find
the safe load for a beam that is of the same material but is 6
inches wide, 10 inches deep, and 20 feet long.
21) The safe load that a rectangular beam can support varies
jointly as the width and square of the depth of the beam and
inversely as its length. A beam 4 inches wide, 6 inches deep,
and 25 feet long can support a safe load of 576 pounds. Find
the safe load for a beam that is of the same material but is 6
inches wide, 10 inches deep, and 20 feet long.
ππ€π 2
π=
π
π(4)(62 )
576 =
25
14400 = 144π
π = 100
100π€π 2
100(6)(102 )
π=
β¨π=
= 3000
π
20
3000 πππ’πππ
22a) Solve.
3π₯ = 9 π₯−1 ⋅ 27
1−3π₯
22a) Solve.
3π₯ = 9 π₯−1 ⋅ 27
π₯−1
1−3π₯
1−3π₯
3π₯ = 3 2
⋅ 3 3
3π₯ = 32π₯−2 · 33−9π₯
3π₯ = 31−7π₯
π₯ = 1 − 7π₯
8π₯ = 1
1
π₯=
8
1
{ }
8
22b) Solve. Please round your answer to three decimal places.
3π₯ −2 = 43−5π₯
22b) Solve. Please round your answer to three decimal places.
3π₯ −2 = 43−5π₯
ln 3π₯−2 = ln 43−5π₯
π₯ − 2 ln 3 = 3 − 5π₯ ln 4
π₯ ln 3 − 2 ln 3 = 3 ln 4 − 5π₯ ln 4
π₯ ln 3 + 5π₯ ln 4 = 2 ln 3 + 3 ln 4
π₯(ln 3 + 5 ln 4) = 2 ln 3 + 3 ln 4
2 ln 3 + 3 ln 4
π₯=
≈ 0.792
ln 3 + 5 ln 4
{0.792}
23a) How much does Amy have to invest at 8% compounded
continuously to have a balance of $19584.76 after ten years?
23a) How much does Amy have to invest at 8% compounded
continuously to have a balance of $19584.76 after ten years?
π΄ = ππ ππ‘
19584.76 = ππ 0.08∗10
19584.76 = ππ 0.8
19584.76
π=
≈ 8800
0.8
π
$8,800
23b) How long will it take an investment of $500 to double
if the money is invested at 6% compounded monthly ?
23b) How long will it take an investment of $500 to double
if the money is invested at 6% compounded monthly ?
π
π΄=π 1+
π
ππ‘
0.06
1000 = 500 1 +
12
12π‘
2 = 1.00512π‘
ln 2 = ln 1.00512π‘
ln 2 = 12π‘ ln 1.005
ln 2
π‘=
≈ 11.58
12 ln 1.005
11.58 π¦ππππ
24a) Write the expression in expanded form.
π¦+3 2
lπ[
]
3
π₯ −2
24a) Write the expression in expanded form.
π¦+3 2
lπ[
]
3
π₯ −2
= ln π¦ + 3
2
− ln π₯ − 2
3
= 2 ln(π¦ + 3) − 3 ln(π₯ − 2)
24b) Write the expression in condensed form.
1
2 ln π₯ − ln(π₯ 2 + 1)
2
24b) Write the expression in condensed form.
1
2 ln π₯ − ln(π₯ 2 + 1)
2
1
2 ln π₯ − ln(π₯ 2 + 1)
2
= ln π₯ 2 − ln π₯ 2 + 1
1
2
= ln π₯ 2 − ln π₯ 2 + 1
= ln
π₯2
π₯2 + 1
25a) Solve.
log 2 5π₯ + 1 = −2
25a) Solve.
log 2 5π₯ + 1 = −2
2−2 = 5π₯ + 1
1
= 5π₯ + 1
22
1
= 5π₯ + 1
4
1
5π₯ + 1 =
4
5π₯ =
1
−1
4
5π₯ = −
π₯=−
3
4
3
20
1
π
πππππππ − π‘βπππ πππ ππ ππππππ ππ π π’ππ π ππππ 5π₯ + 1 π€ππ π ππ‘ π‘π .
4
1
π₯ π€πππ πππ’π π 5π₯ + 1 π‘π ππ π€βππβ ππ πππππ‘ππ π‘βππ π§πππ.
4
{−
3
}
20
25b) Solve.
log π₯ + 8 + log π₯ − 1 = 1
25b) Solve.
log π₯ + 8 + log π₯ − 1 = 1
log π₯ + 8 + log(π₯ − 1) = 1
log π₯ + 8 π₯ − 1 = 1
101 = π₯ + 8 π₯ − 1
10 = π₯ 2 + 7π₯ − 8
π₯ 2 + 7π₯ − 18 = 0
π₯+9 π₯−2 = 0
π₯ = −9 π₯ = 2
π₯ ≠ −9
{2}
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