Download Math 101 – Exam 2 – Concept Review

Math 101 – Exam 1 – Review
Reminder: Exam 1 will be on Friday, October 14, 2011 at 8am.
It will cover sections 1.1, 1.2 and 10.1-10.3
Room Assignments:
Room
Sections
Nesbitt 111
9, 14, 23, 24, 28
Nesbitt 125
20, 29
CAT 61
4, 5, 30
Stratton 113
12, 16, 19
Main Auditorium 6, 7, 8, 10, 11, 13, 15, 18, 21, 22, 25, 26, 27
Section 1.1 Recap
Slope and equations of Lines:
Slope
Formula for slope: m  y  y
2
1
x2  x1
1. Slope of a horizontal line is 0
2. Slope of a vertical line is undefined
3. Parallel Lines: have the same slope
4. Perpendicular lines: slopes are negative reciprocals
Equations of Lines:
1. Use slope-intercept form (y = mx + b) if you are given
The slope and the y-intercept
2. Use point-slope form ( y  y1  m  x  x1  ) if you are given
- The slope (m) and any point on the line,
- You are given two points on the line only (find the slope first, using
m
y2  y1
x2  x1
, then pick a point
and use point-slope form)
3. Equation of Horizontal line through the point (a, b): y = b
4. Equation of Vertical line through the point (a, b): x = a
Section 1.2 Recap
Definition of a Linear Function:
y =f(x) = mx +b for real numbers m and b.
Supply and Demand:
How many items a company (“producers” )is willing to supply, p=S(q) and how much buyers
(“consumers”) are willing to buy p = D(q) are both dependent upon the price at which the item is
being sold. Even though it seems that quantity demanded and supplied depends on price, when it
comes to supply and demand you have to remember that price is the independent variable and
quantity is the dependent variable.
Equilibrium point:
The point at which supply and demand are equal is called the equilibrium point; you will have
neither a surplus nor a shortage. The equilibrium price is the price at that point, and the
equilibrium quantity is the quantity at that point.
Cost Analysis:
Linear cost functions describe how much it costs to produce x items. All linear cost functions have
to types of costs, 1. The per-item cost (Variable cost) and 2. The fixed costs (the cost associated
with producing nothing)
Linear Cost functions have the form
C(x) = Variable costs *x + Fixed Costs
Marginal Cost:
Defined as the cost to produce one additional unit beyond your current production level, or more
simply put the slope of the linear cost function.
Break-Even Analysis:
The Revenue Function, R(x) = price per item * the number of items = px
The Profit Function, P(x) = R(x) - C(x)
When the profit > 0, you made money
When profit < 0, you lost money
Break- Even Point: R(x) = C(x) or P(x) = 0
The x-coordinate is the break-even quantity
Section 10.1
Definition of a Function
A function is a rule that assigns to each element from one set exactly one element from another set. This rule is
usually a mathematical expression that relates an input value to an output value.
Domain: The set of all possible input values (another word is meaningful replacements for x)
Range: the set of all possible output values (or in other words, possible values y can take on)
To find the domain of a function, you want to begin by assuming that the domain is all real numbers….UNLESS:
1. There is a denominator. Division by zero is undefined so any value of x that makes the denominator
zero cannot be used in the function.
2. There is a square root. You cannot take the square root of a negative number, so whatever expression
placed under the square root must be  0.
Evaluating Functions
“Evaluating a function” really means finding the value of the function (or of y) for a given x value. So if I want to
evaluate f x   2 x 2  6 x  1 at x = 4. I need to replace x with 4 and calculate.
The Vertical Line Test
The vertical line test is a quick and easy way for you to determine if a graph represents a function. Since each x
must be paired with only 1 y, if you cross a function with a vertical line, any vertical line will cross the graph at
most once.
Section 10.2
Transformations:
Let f be any function and h and k be positive constants.
1. The graph of y = f(x) + k is the graph of y= f(x) shifted up k units
2. The graph of y = f(x) – k is the graph of y = f(x) shifted down k units
3. The graph of y = f(x-h) is the graph of y = f(x) shifted to the right h units
4. The graph of y = f(x+h) is the graph of y = f(x) shifted to the left h units
5. The graph of y = -f(x) is the graph of y=f(x) flipped over the x-axis
6. The graph of y = f(-x) is the graph of y = f(x) flipped over the y-axis.
This allows us to graph familiar functions by just performing translations and reflections in the plane.
Definition of a Quadratic Function
The first thing you need to know about a quadratic function is that it is a function in which the highest power of
x appearing is 2. In general, a quadratic function is defined as f  x   ax2  bx  c where a, b and c are real
numbers and a  0 .
The graph of a quadratic function is a parabola. Each parabola either opens up or opens down.
y
f(x)=-x^2+3
8
y
f(x)=x^2-3
8
6
6
4
4
2
2
x
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
x
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
-2
-2
-4
-4
-6
-6
-8
-8
A parabola that opens up will have a minimum point, and a parabola that opens down will have a maximum
point. The minimum (or maximum) point on the parabola is called the vertex. Parabolas also have the property
of being symmetric across the vertical line that passes thru the vertex. That line is called the axis of symmetry.
Sometimes it is useful to change the equation to an equivalent form, called Vertex Form. f ( x)  a( x  h)2  k .
Then, using our transformations, we realize that the vertex (which is the origin for y  x 2 ) is just shifted h units
to the left and k units up to a new vertex of (h,k)
Finding the Vertex of a Quadratic Function:
Example: Find the vertex of f  x   2x2  8x  8 .
Option 1: We have to find a way to write this in the form: y  a  x  h   k
2
To do this we are going to use the process completing the square.
1. Move the -8 away from the rest of the equation. Factor a 2 out of the first two terms.

f  x   2 x2  4x

8
2. You are going to attempt to make the expression in parentheses a perfect square. So we are going to
put a perfect squared expression underneath. Now look at the terms in parentheses. Take the
coefficient of the second term and divide it by 2. You are going to put the square root of x^2 and the
number you just came up with in ( )
f  x   2  x 2  4 x
 2  x  2
2

8
3. Now you have to go back to the first line and adjust it based on what you just did.
f  x   2  x 2  4 x
 2  x  2

8
2
Adjust how you ask? You have to put in what needs to be there. Take the 2 in the second line and
square it and put it back into the first line, so you put in what is now missing from the parentheses.
Then you have to balance out that line.
f  x   2  x 2  4 x 4 
88
 2  x  2   16
2
You just added 2*4 = 8 in, so you have to subtract it back out. Which is where
f  x   2  x  2   16 comes from. This means the vertex is (-2,-16)
2
Option 2: There is a shortcut formula for finding the vertex. To find the vertex of the parabola
f  x   ax2  bx  c
b
x
2a
 b 
y f 
 2a 
Intercepts: To find x-intercepts, you have to set y =0; to find the y-intercept you set y = 0.
Section 10.3
I. Polynomial Functions
A polynomial function of degree n, where n is a non-negative integer is defined by f  x   an xn  an1x n1  ...
Where an ,...a0 are real numbers called coefficients with an  0 . The number a n is called the leading
coefficient.
Properties of Polynomial functions:
1. A polynomial function of degree n can have at most n -1 turning points. Conversely if the graph of a
polynomial function has n turning points must have degree at least n + 1
2. In the graph of a polynomial function of even degree, both ends either go up or go down. For a
polynomial function of odd degree one end goes up and one end goes down.
3. If the graph of x becomes large, the leading coefficient is positive. If the graph goes down as x becomes
large, the leading coefficient is negative.
4. A polynomial of degree n can have at most n zeros (x-intercepts)
II. Rational Functions
A rational function is defined by f  x  
p  x
q  x
where p(x) and q(x) are polynomial functions and q(x) not = 0
Rational functions will have graphs with one or more breaks because you cannot use any value of x which makes
the denominator = 0.
Asymptotes:
If a function gets larger and larger in magnitude without bound as x approaches the number k then the line x = k
is a vertical asymptote. If the values of y approaches a number k as x gets larger and larger the line y = k is a
horizontal asymptote. NOTE: ASYMPTOTES ARE LINES, IN PARTICULAR VERTICAL ASYMPTOTES ARE VERTICAL
LINES (X = #) AND HORIZONTAL ASYMPTOTES ARE HORIZONTAL LINES (Y = #)
To find vertical asymptotes
1. Cancel any common factors from the numerator and denominator. You will have a “hole” at these
points.
2. Set the simplified denominator =0
3. Factor to get the root(s), for example, x = k
4. The vertical line x = k is a vertical asymptote
To find horizontal asymptotes
You will have to compare the numerator and the denominator
1. If the degree of the numerator is lower than the degree of the denominator, the line y = 0 is the
horizontal asymptote
2. If the degree of the denominator is the same as the degree of the numerator, the horizontal asymptote
will be the leading coefficient of the numerator divided by the leading coefficient of the denominator
3. If the degree of the numerator is larger than the degree of the denominator there is no horizontal
asymptote.
Graphing Rational Functions:
1. Find the domain of the function
2. Check if you can factor and cancel, producing a hole in the graph
3. Find any vertical asymptotes, drawing them in your graph as dashed lines
4. Find any horizontal asymptote, drawing as a dashed line
5. Find any x intercepts (set y =0)
6. Find any y-intercepts (set x = 0)
7. Make a number line chart to determine where the function lies above and below the x axis around the
vertical asymptotes.
8. Graph, hugging all of the asymptotes.
Math 101 – Exam 2 – Review Problems
Section 1.1:
#1. Find the slope of the line through (4,5) and (-1,2)
#11. Find the slope of the line y = 8
#15. Find the equation of the line through the point (1,3) with m =-2
#21. Find the equation of the line through (2/3, ½) and (1/4, -2)
#23. Find the equation of the line through (-8,4) and (-8,6)
#29. Find the equation of the line through (-4, 6) parallel to 3x+2y=13
#31. Find the equation of the line perpendicular to x + y = 4
Graph the following equations:
#45. y = 4x + 5
#49. 3x – y = -9
#59. 3x - 5y = 0
#71. The number of unmarried, opposite se couples in the U.S. who are living together has been rising
at a roughly linear rate in recent years. The number of cohabitating adults was 1.59 million in 1980 and
5.08 million in 2004.
a.
Write an equation representing the number of cohabitation adults (in millions), y, in terms of the
number of years after 1980, x.
b. Use your result in part a to predict the number of cohabitating adults in the year 2010
Section 1.2
For the following exercises, let f(x) = 7 – 5x and g(x) = 2x – 3 Find
#1. f(2)
#5. g(1.5)
#19. Write a linear cost function for each situation (#19 and #25). Identify all variables used: A Lake
Tahoe resort charges a snowboard rental fee of $10 plus $2.25 per hour
#25. Assume that each situation can be expressed as a linear cost function. Find the cost function:
Marginal Cost: $75; 50 items cost 4300 to produce.
#27. Suppose the demand and price for a certain model of youth wristwatch are related by
p  Dq  16 1.25q where p is the price (in dollars) and q is the demand (in hundreds). Find the price
at each level of demand.
a. 0 watches
b. 400 watches
c. 800 watches
Find the demand for the watch at each price
d. $8 e. $10
f. $12
g. Graph p  16  1.25q
Suppose the price and supply of the watch are related by p  S q   0.75q where p is the price(in
dollars) and q is the supply (in hundreds) of watches. Find the supply at each price
h. $0
i. $10
j. $20
k. Graph p = 0.75q on the same set of axes for part g.
l. Find the equilibrium quantity and the equilibrium price.
31. Joanne Ha sells silk-screened T-shirts at community festivals and craft fairs. Her marginal cost to
produce on T-shirt is $3.50. Her total cost to produce 60 T-shirts is $300 and she sells them for $9 each.
a. Find the linear cost function for Joanne’s T-shirt production.
b. How many T-shirts must she produce and sell in order to break even.
c. How many T-shirts must she produce and sell to make a profit of $500?
#37. To produce x-units of a religious medal costs C(x) = 12x+39. The revenue, R(x) = 25x. Both C(x)
and R(x) are in dollars.
a. Find the break-even quantity
b. Find the profit from 250 units
c. Find the number of units that must be produced for a profit of $130.
Section 10.1
1. For each of the following, determine the domain.
a) p ( x)  2 x  3
b)
c)
f ( x)  121  x 2
3
g ( x)  2
x  81
2. Give the domain and range of the following function:
3. Let f ( x) 
a)
b)
c)
4x  3
Compute
x5
f ( 4)
 1
f  
 2
2
f 
m
d) All values of x for which f(x)=1
4. Determine whether the following graph represents a function:
a)
b)
5. Let f x   6 x 2  5 x  8 . Compute
f x  h   f ( x)
h
Section 10.2
6. Describe how the following graph can be obtained by transforming f ( x)  x 2 . Then sketch the graph.
a)
b)
f x   ( x  5) 2  3
f x   ( x  3) 2
c) f x   x 2  3
7. Graph the following parabolas. Give the vertex, x and y intercepts, and axis of symmetry.
a) f x   x 2  4 x  3
f x   2 x 2  20 x  18
1
8
1
c) f  x   x 2  x 
3
3
3
8. Let C ( x)  4 x  3 be the cost of producing x widgets, and let Rx    x 2  8 x be the revenue.
b)
a) Graph C(x) and R(x) on the same axes.
b) Find the minimum break even quantity
c) Find the maximum revenue
d) Find the maximum profit
9. If an object is thrown upward with an initial velocity of 128 ft/sec, then its height after t seconds is given
by h(t )  128t  32t 2 .
a) Find the maximum height obtained by the object.
b) Find the number of seconds it takes to hit the ground.
Section 10.3
10. For each of the following, select an appropriate graph:
a) f x    x 3  2 x 2  x  11
b)
f x    x 4  2 x 3  10 x  15
11. The following is the graph of a polynomial. Determine:
a) Whether the degree is odd or even.
b) Whether the leading term is positive or negative.
c) Fill in the blank: Since there are ______ turning points, the degree must be at least ______.
12. For each of the following rational functions, find:
i.
The domain.
ii.
Any holes
iii.
The equations of any vertical and horizontal asymptotes
iv.
Any x and y intercepts
v.
Provide a sketch.
a)
b)
c)
x4
x3
4
f ( x) 
x4
x 2  5x  4
f ( x) 
x4
f ( x) 
In addition to reviewing your lecture notes, your quizzes, and the assigned
problems on the course syllabus, it is RECOMMENDED that you do the
following problems in preparation for the exam:
Chapter 1 Review Exercises – p.50: 3, 7, 14, 15, 19, 20, 22, 23, 25, 26, 28, 37,
38, 42, 49
Chapter 10 Review Exercises – p.642: 8(a-f), 11, 12, 13, 14, 16, 89(a-c), 91
Section 10.1 – p.568: 27, 39, 45, 55, 57
Section 10.2 – p. 582: 29, 47
Section 10.3 – p. 595: 33, 39, 47(a-c)