Download Chapter 1 Exam Review

College Algebra
Chapter 1 Review
Created by: Lauren Atkinson
Math Coordinator, Mary Stangler Center for Academic Success
Note:
This review is composed of questions from the
chapter review at the end of chapter 1. This
review is meant to highlight basic concepts from
chapter 1. It does not cover all concepts
presented by your instructor. Refer back to your
notes, unit objectives, handouts, etc. to further
prepare for your exam.
Classify each number listed as one or more of the following:
natural number, integer, rational number, or real number.
104 3
55,1.5,
, 2 , 3, −1000
17
Natural numbers: 1,2,3,4, ⋯
Integer: ⋯ , −3, −2, −1,0,1,2,3, ⋯
𝑝
𝑞
Rational numbers: can be written as where p and q are integers; includes
repeating and terminating decimals.
Real numbers: any number that can be expressed in decimal form; includes
rational and irrational numbers.
55 = natural, integer, rational, real
1.5 = rational, real
104
= rational, real
17
23 = natural, integer, rational, real
3 = real
−1000 = integer, rational, real
Write the number in scientific notation:
0.0001001
We have moved the decimal point 4 “spots” to the right
so our scientific notation becomes the following:
1.001 × 10−4
Write the number in standard form:
−7.2 × 10−3
We have moved the decimal point 3 “spots” to the left
so our standard form becomes the following:
−.0072
Evaluate each expression. Write your answer in
scientific notation and in standard form.
a) 4 × 103 5 × 10−5
= 4000 .00005 = 0.2, 2 × 10−1
3×10−5
b)
6×10−2
.00003
=
=
.06
0.0005, 5 × 10−4
Evaluate by hand:
2
3∙3 ÷
3−5
6+2
−2
−1
=3∙9÷
= 27 ÷
= −108
8
4
Determine the minimum and maximum values
of the set of data. Calculate the mean and
median.
8.9, −1.2, −3.8,0.8,1.7,1.7
Arrange data from smallest to largest:
−3.8, −1.2, 0.8, 1.7, 1.7, 8. 9
Maximum: 8.9 Minimum: -3.8
−3.8+−1.2+0.8+1.7+1.7+8.9
Mean:
6
Median: 0.8 + 1.7 = 1.25
2
= 1.35
Express the data as a relation S.
Find the domain and range of S.
𝑆=
X
-0.6
-0.2
0.1
0.5
1.2
Y
10
20
25
30
80
−0.6,10 , −0.2,20 , 0.1,25 , 0.5,30 , (1.2,80)
domain refers to all of the x-coordinates, range refers to all of the
y-coordinates. DOMAIN always refers to x, RANGE always refers to
y.
𝑑𝑜𝑚𝑎𝑖𝑛 = −0.6, −0.2,0.1,0.5,1.2
𝑟𝑎𝑛𝑔𝑒 = {10,20,25,30,80}
Determine if the relation is a function: YES
{ 1.5,2.5 , 0,2.1 , −2.3,3.1 , 0.5, −0.8 , −1.1,0 }
A relation is a function if for every input of x, there is
exactly only one output for y. To illustrate this:
1.5
0
-2.3
0.5
-1.1
2.5
2.1
3.1
-0.8
0
Since each input (green
circle) value corresponds to
only one output (blue
circle), the relation is a
function.
Find the distance between the two points:
1.2, −4 , (0.2,6)
𝑑=
(𝑥2 − 𝑥1 )2 +(𝑦2 − 𝑦1 )2
𝑑=
(0.2 − 1.2)2 +(6 − −4 )2
𝑑 = 101
Find the midpoint of the line segment with the
given endpoints.
1 5
,
2 4
,
1 −5
,
2 2
𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 =
=
1 1 5 5
+
−
2 2 4 2
2
,
2
=
𝑥1 +𝑥2 𝑦1 +𝑦2
,
2
2
1 −5
,
2 8
Find the standard equation of a circle with
center (-5,3) and radius 9.
The circle with center (h,k) and radius r has the
equation: (𝑥 − ℎ)2 +(𝑦 − 𝑘)2 = 𝑟 2
= (𝑥 + 5)2 +(𝑦 − 3)2 = 81
Graph 𝑦 = 𝑓(𝑥) by first plotting points to
determine the shape of the graph.
𝑓 𝑥 = 3𝑥
𝑓 𝑥 = 2𝑥 − 3
𝑓 𝑥 =
1 2
𝑥
2
−1
𝑓 𝑥 = 3−𝑥
Evaluate 𝑓(𝑥) at the indicated values of x. Find the
domain of f.
𝑓 𝑥 = 4 − 5𝑥 𝑓𝑜𝑟 𝑥 = −5,6
𝑓 −5 = 4 − 5 −5 = 4 + 25 = 29
𝑓 6 = 4 − 5 6 = 4 − 30 = −26
𝑑𝑜𝑚𝑎𝑖𝑛 = 𝑎𝑙𝑙 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
𝑓 𝑥 = 𝑥 3 − 3𝑥 𝑓𝑜𝑟 𝑥 = −10, 𝑎 + 1
𝑓 −10 = −10 3 − 3 −10 = −970
𝑓 𝑎 + 1 = 𝑎 + 1 3 − 3 𝑎 + 1 = 𝑎3 + 3𝑎2 − 2
𝑑𝑜𝑚𝑎𝑖𝑛 = 𝑎𝑙𝑙 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
Evaluate 𝑓(𝑥) at the indicated values of x. Find
the domain of f.
𝑓 𝑥 = 𝑥 + 3 𝑓𝑜𝑟 𝑥 = 1, 𝑎 − 3
𝑓 1 = 1+3= 4=2
𝑓 𝑎−3 =
𝑎−3 +3= 𝑎
𝑑𝑜𝑚𝑎𝑖𝑛 =
𝑎𝑙𝑙 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 − 3
Write 5 ≤ 𝑥 < 10 in interval notation.
[5,10)
Determine if S represents a function:
𝑆 = −1,3 , 0,2 , −1,7 , 3, −3
This is similar to the table/relation question
from before. Does every output of x only have
one output of y?
NO, since -1 (input) corresponds to both
3 and 7, this is not a function.
State the slope of the graph f:
𝑓 𝑥 =
1
𝑥
3
−
2
3
Slope is always the 𝑚 in 𝑓 𝑥 = 𝑚𝑥 + 𝑏;
so in our equation, the 𝑚 =
1
.
3
Find the slope of the line passing through each
pair of points.
1, −4 , (2,10)
𝑚=
𝑦2 −𝑦1
𝑥2 −𝑥1
−1 2
,
3 3
𝑚=
,
=
10−(−4)
2−1
=
14
1
= 14
−1 −5
,
3 6
−5 2
−
6 3
−1 −1
−
3
3
=
3
−
2
0
= 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
Decide whether the function f is constant, linear
or nonlinear.
𝑓 𝑥 = 2𝑥 2 − 3𝑥 − 8
𝑓 𝑥 =6
NONLINEAR, because the highest
power on the x variable is not 1.
This function has a power of 2 on x.
CONSTANT (and linear),
because there is not an x
variable in the equation.
𝑓 𝑥 = 3𝑥 + 24
LINEAR, because the highest
power on the variable is 1.
Determine where the graph of 𝑓 𝑥 = 𝑥 − 3 is
increasing and where it is decreasing.
Graph this function first:
From the graph, we can see where it is
increasing (positive slope) and where it is
decreasing (negative slope):
Increasing: 3, ∞ Decreasing: (−∞, 3]
Find the average rate of change of
𝑓 𝑥 = 𝑥 2 − 𝑥 + 1 from 𝑥1 = 1 to 𝑥2 = 3.
Average rate of change, means to find the slope:
𝑓 1 = (1)2 − 1 + 1 = 1
𝑓 3 = (3)2 − 3 + 1 = 7
Giving two ordered pairs: 1,1 , 3,7
Then compute the slope:
7−1
3−1
Average rate of change is 3.
6
2
= =3
The rest of the chapter review covers the
application problems. There will be application
problems on the exam so study accordingly.