Important Notices, Handouts, and Assignments
Discussion of course goals, policies, and rules of the test center. See the assignment due on 1-17. A graph (JPG) worth a thousand words. Is it a function?
(Substitute) went over questions from 1.1. A quiz was given over the prerequisites needed for the course. Here's the key. See the assignment due on 1-22.
Rules for proportionality were presented, and in groups we worked 2 and 15 from Section 1.2. In teams, we worked 8 from Chapter 1 Review (page 31) and discussed how we could match the table to the graph to the verbal description. Then we computed the average rates of change over each interval to see how this quantitatively described what was just discussed. This led to the answer to the title of Section 2.2, i.e., What Makes a Function Linear?
since a linear function must have a constant average rate of change. We talked about functions which were increasing and decreasing, and how to define this using the average rate of change. Looking at the graph similar to 10 from Section 1.3, we asked which of the intervals was the average rate of change the greatest, and to justify why:
I. From a = -1 to b = 1 or II. From c = -9 to d = 0 or III. From d = 0 to e = 3.
The answer is interval I, since the slope of the segment connecting (a, f(a)) and (b, f(b)) is the largest.
See the assignment due on 1-24. A video is available on reserve at the service desk at the library.
We next began Section 2.1, and, again in teams, worked on Problem 1 from Section 2.1, using a similar analysis as in Problem 8 mentioned above. Next, we found formulas for each of the functions in Problem 1. Found a simple linear modeling problem such as Problem 9 from Section 2.1 and talked about the general formula for the family of linear functions as y = b + mx, written this way since b is the initial value and m is the average rate of change. Examined problem 15 from Section 2.2 to make sure we knew how to get a function graphed in our graphing calculator (enter the equation, set a window) and observed that we needed to change Ymax to see the initial value. Need help for your calculator model? See the Assistance with Graphing Calculators Web Site. Sometimes we are not given the initial value, such as in Problem 11 from Section 2.2, but we can still find it. One strategy shown was to find m first, then plug the values of x and y from another point into y = b + mx to find b. Looked briefly at a budget constraint example. For homework, see the assignment due on 1-29.
Consider y = Mx + P and y = Nx + Q, with N < M < 0 and P < Q. What are the relative positions of the lines (which line is above the other) as x tends toward positive infinity/ negative infinity. Do the lines cross, and if so will the intersection be for x > 0, x < 0, or x = 0. Explain why. Is there are any conditions under which the
lines will not cross. This led to a discussion of parallel lines, and from there to the conditions for perpendicular lines. Worked on choice of Problem 3, 6, or 7, and then a power company problem similar to 19 or 20. For homework, see the assignment due on 1-31.
Showed how to use the grapher to solve a problem like 19. (For a TI-83, click here. For others, see Assistance with Graphing Calculators Web Site.) Discussed Example 9 from Section 3.1. Looked at Problem 10 and 11 from Section 3.2. Investigated the following: Let S(t) and J(t) be the population (in thousands) of Smithville and Jonesville, respectively, t years after 1970. Suppose J(t) = S(t
2) + 3 . We first interpreted the units of the 2 and 3 in this expression. (units on the inside are years and the outside unit is thousands.) We determined how the population of Jonesville is related to that of Smithville: the population of Jonesville in a given year is 3000 more than the population was in Smithville two years earlier.
For homework, see the assignment due on 2-5.
The notation for inverse functions is introduced in this section. This is a very brief introduction to the concept and interpretation of an inverse function. Worked problems 16 and 17 from Section 3.2. Defined domain and range of a function. Explain that if there are no restrictions on the domain (either given or due to the context of the problem), the domain will be considered as the set of input values which yield a real value as an output. Looked at the domain and range of a function represented as a graph. For homework, see the assignment due on 2-7.
Now that we found the domain and range of a function represented as a graph, we tried problem 8 from Section 3.3, followed by problem 10. Next talked about the domain of a function like
and
how the graphing calculator can mislead. The test cycle starts next week. If you missed class today, you need to get a test ticket before taking the test. Be sure to check out the Review for Test 1 and sample quizzes (See top)
For homework, see the assignment due on 2-12.
We invested $2300 the day my child was born and watched it grow by 12% each year. Predicted the amount we would have when the child would be18, then 65. Derived the formula A = 2300(1.12)t which gives the amount A when the child is t years old. Used our grapher to find when A(t) =1 million. Discussed an exponential decay problem, then worked on problem 8. Summarized Section 4.1 by drawing graphs of an exponential function which shows exponential growth, and one which shows exponential decay. Put the graphs of the investment function and decay function on the overhead calculator and discuss intercepts, rate of change, etc.
For homework, see the assignment due on 2-14.
Emphasized the difference between the verbal descriptions for exponential functions vs. linear functionsi.e., exponential functions change by a constant percent rate over equally spaced intervals and linear functions change by a constant quantity over equally spaced intervals. We summarized Section 4.1 by drawing graphs of an exponential function which shows exponential growth, and one which shows exponential decay. The formula A = 2300(1.12)t gives the amount A when the child is t years old if 2300 is invested at the year of their birth and the account grows by 12% each year. It's growth factor is 1.12 and its growth rate is 12%. If I have 5000 acres of property at year t = 0 and Bluto the Bully moves next to me and takes 25% of my property each year thereafter, I then have A = 5000(0.75)t acres in year t. We could say that its decay factor is 0.75, and its decay rate is 25% (or growth rate is 25%, where negative growth is another name for decay.)
Observe the general pattern for y = a(b)t.
It's initial value is a and growth/decay factor is b.
If b > 1, it increases (exponential growth) and if 0< b <1,
it decreases (exponential decay).
Examples:
A = 2300(1.12)t = 2300(1
+ 0.12)t. Increases by 12% each year.
A = 5000(0.75)t = 5000(1
+ 0.25)t. Decreases by 25% each year.
Next we attacked the tables on pg 118, problems 16-19. Given (in the instructions) that the data is from either an exponential or a linear function, we worked in groups to determine possible formulas for each table. The distinction was made between linear data (difference in consecutive ys is constant) versus exponential data (ratio of consecutive ys is constant). After finding formulas for each, we had fun with problem 21, making certain that we can find an exponential formula when the initial value is not given in the table, and when the input values are not all evenly spaced. Followed this with something like Chapter 4 Review, Problem 8 on page 172. A good strategy in finding the formula y = abx is to estimate values for a (the y-intercept) and b (less than 1 in problem 8 since it is decay) before getting any values. If the fractions cause panic, you can find the ratio of the outputs and use the FRAC feature on their calculator. For homework, see the assignment due on 2-19.
The Malthusian example (Section 4.3, Example 2) is a model you might have heard about in some other courses. It is an excellent example of the need for graphical or numerical techniques to determine where the linear and exponential models meet. We found intersection points using the grapher, and then placed the following four functions in a grapher: f(t) = 20(1.4)t , g(t) = 28(1.2)t , h(t) =10(0.8)t , i(t) = 15(0.6)t
We asked how many times f(t) and g(t) intersect. Some thought more than once. We show the answer after Section 4.4.. The number e is introduced in this section as a number between 2 and 3 that is often used as a base for exponential functions. The understanding and usefulness of e as a base will more fully understood after 4.8. We defined a logarithm and looked at the problems on the logarithm worksheet. We gave the mantra A logarithm is an exponent. Discussed common logarithm (base 10) and natural logarithm (base e). For homework, see the assignment due on 2-21.
We went back to Section 4.3 briefly to mention the idea of horizontal asymptotes. Used the notation, introduced in this section, for of Q = abt describing this as
.We noticed that you can not take logs of zero or negative numbers, regardless of the base. We introduced the inverse properties of logs for 10, e, and any base b. We looked at three properties of logarithms. We introduced the three properties of logarithms and rewrote some expressions, including something like
.
We solved a simple logarithmic equation such as
We reviewed the inverse properties of logs and noticed one could make each side
an exponent to the base 10 to find x. Then we solved exponential
equations such as
2x = 32 and then 2x
= 31.
Showed graphical solution (approximate) and analytical solution (exact). Logs
proved their usefulness in determine the solution to 2300(1.12)x
= 1000000, which was how old my son would be when he gets 1 million from his
breast milk fund.
For homework, see the assignment due on 2-26.
Also, the first Internet
assignment is due Monday March 3. Here are the directions.
As promised, we solved when f(t) = 20(1.4)t and g(t) = 28(1.2)t intersect and determined it was only once. We showed the equation 20(1.4)t = 28(1.2)t and (1.4/1.2)t = 7/5 were equivalent. We know the latter only has one solution from our graphical representation of the solution. With logs, we arrived at
which is consistent with the picture we obtained.
We reviewed the definition of the common log function and produced a graph of y = log x by exploring the relationship between log x and 10x . We noticed the reversal of the roles of domain and range values. The definition of vertical asymptote is given in this section. You can relate the horizontal asymptote on the exponential function to the vertical asymptote of the log function. Horizontal asymptotes are present if y approaches a finite value as x tends to infinity (either positively or negatively or both) and that vertical asymptotes have the opposite behavior, i.e., y tends to infinity ('explodes' in a positive or negative direction) as x approaches a finite number. The notation indicating "from the left" and "from the right" was also introduced:
for y = log x, we have that as x → 0+, then y → ∞.
The pH of a substance was given as
It measures chemical acidity. Chemists sometimes say that the p in pH stands for the negative of the logarithm. The notation
represents the hydrogen ion concentration, which is measurably so very small that recording
this value is cumbersome. For example, for
pure water
is 0.0000001 or 10-7 and for milk of magnesia it is
10-10. (Its units are moles per liter.) Instead of writing this, chemists just look at the magnitude
of the exponent and report 7 and 10 respectively. Ask for the pH if
= 0.001. Then if
= 0.007. Go backwards and give
of a beer if its pH = 4.2. For homework, see the assignment due on
2-28.
The following gives different populations as functions of time, where t is in years.
|
A. The population doubles in size every 2 years. |
 |
Match the description of the population with the formula. Then construct your own description for the formula that is not used. Also determine which population is increasing the fastest.
Answers:
A. Doubles
in size every 2 years. P5 .
B. Doubles every 6 months (which is
½ of a year). P1
C. Grows by 3% every 6 years
P3
D. Doubles
in size every 2 months (which is 1/6 of a year) P2
For the one that is not used, P4
=8000(2)^(t/6), the description would be The population doubles in size every 6
years.
We noticed that the 8000 in the formula gave the initial population. We
talked about doubling time and half life. Suppose we want to write formulas for
which relate the size of a population to its time t, where t is in
years. (Note that this is very important to specify how the
independent variable is measured; otherwise you would have a different formula.)
What formula describes a population which initially is at 250 and triples every
7 years? Ans: P = 250(3)^(t/7).
What about something that starts at 250 and in 5 years reduces to half of
its value? Ans: P = 250(1/2)^(t/5).
Same as before, but the half-life is every 3 months?
Ans: P = 250(1/2)^(4t). Note 3 months = 1/4 year. By
the way, back to our original populations in the above box, which of P1, P2, P3, P4, or P5
is growing the fastest? determined which of P1, P2, P3, P4, or P5
is growing the fastest. The winner was P2,
which doubles
in size every 2 months. If we write each of the populations in the form 8000* b^t, we'd see that b would be
26
for P2, which is the highest. We talked about annual growth
rate, and found the annual growth rate of P4 by writing it as
8000* b^t and determining b = 2^(1/6). Notice b gives the
growth factor 1.122, so its growth rate is 12.2% each year. If a population
initially of 8000 decreases by half every 6 years, it's population is P =
8000(0.5)t/6.
By what percent does it decrease each year? We want the decay factor b
such that 8000(0.5)t/6
= 8000(b)t,
so rewrite:
Since 8000(0.5)t/6
= 8000((0.5)1/6)t
,
then 8000((0.5)1/6)t
= 8000(b)t
This means b
= (0.5)1/6
or about 0.891.
(Be
careful. On a calculator you need the ^ key and parentheses.The answer is
NOT b = 0.833.)
So if P = 8000(0.5)t/6
= 8000(0.891)t
, it decreases by 10.9 % each year.
The first Internet assignment is extended to noon Wednesday March 5. Here are the directions.
Cycle for Test 2 begins Friday March 7.
Review for Test 2 (4.1-4.6)
Chapter 4 Review 1, 2 through 21, 26, 27, 30, 31 (annual growth rate
only), 32i, iii, 34, 35, 36 (Here are the
answers to the evens)
For homework, see the assignment due on 3-05.
We began Section 4.8, and explored compound interest for different compounding periods. Then we explored continuous compounding, which answered the long desired question What is e? We worked with the formula A = Pert. We found the annual growth rate of an account which paid 12% APR that was compounded 1 times per year, then 4 times per year, then monthly, weekly, daily, etc. and looked at A = Pe0.12t = P(1.127496852)t to see a growth rate of 12.7496852% per year if we compounded continuously. This led to a discussion of a continuous annual growth rate. See the assignment due on 3-7.
After practicing a few problems from Section 4.8, we worked on problem 2 from Section 5.1 involving the heating schedule of a building. The function H = f(t) is given as a graph (it doesn't have an easy equation). The new function p(t) = f(t)
2 describes lowering the temperature
2 degrees, and results in a vertical shift of f(t) down 2.
This is an "outside" change, so a change to the output. The function q(t)
= f(t
2) describes altering the timing of the schedule
by 2 hours, and results in a horizontal shift of f(t) to
the right 2. (Note: Not to the left, but to the right.)
This is counterintuitive. It is an "inside" change, so a change to the
input. If q(t) = f(t
2) , observe that q(2) = f(0). Also
q(6) = f(4), q(10) = f(8), q(18) = f(16),
q(22) = f(20), and q(26) = f(24). With this we can
draw the graph to see that q is a shift of f 2 units to the right.
The schedule according to q is designed to heat the building 2 hours
later than the schedule according to f. We finished with problem 7. See the assignment due on 3-19.
We looked at a tabular problem like problem #1 and completed the tables for g(p) = f(-p) and h(p) = -f(p) (without finding the formulas for g and h first). We used the calculator to sketch the reflections. For example, have them enter Y1 = f(x) and then entered Y2 = Y1(
x) and
Y3=
Y1
[Use the VARS key to find the
Y-VARS menu.] Then moved into reflecting a function for which there is no
formula given, simply a graph. Combined reflections with shifts. Worked parts of #2. Finally, we discussed odd and
even functions. Drew an even function and demonstrated with an arbitrary point
on the graph that the first graph contains the points (a,b)
and (-a,b) while the odd function contains the points (a,b)
and (-a,-b). Pointed out
that not all functions are either even or odd--many functions are neither.
See the assignment due on
3-21.
Began by asking how f(x) = 5 ½x 3½ is related to the graph of y = ½x½. Then drew the graph of f on the board. Introduced a new function g(x) = 2 f(x), and ask the students what the notation 2f(x) indicates. [The old outputs of f are doubled.] A similar analysis was for h(x) = (-1/2)f(x). Write h as h(x) = (1/2) (-f(x)) to show that the transformations can be thought of as happening one at a time--first a reflection and then a stretch (or shrink) by a factor of 1/2. Selected some parts of problems #5 (graphing without a formula) and #6 (graphing with a formula) for in-class group work. See the assignment due on 3-26.
New group assignments for Friday. See the list of where you are.
With the tools from Chapter 5 at our disposal, it is natural to discuss the family of quadratic functions. It assumed you have seen the technique of completing the square (See pages 507-8 of Appendix E)
How are the graphs of y = x2 +1, y = (x3)2, and y = (x3)2 +1 related to the basic graph of y = x2? We graphed the functions by hand and relate each shift to the new location of the vertex and axis of symmetry. Similarly we graphed some functions of the form y = ax2 , making sure to include some cases with a < 0. From here we moved to the general case of the form of y = a(xh)2 + k and discussed the graphical interpretations of h and k [i.e., vertex at (h,k), axis of symmetry x = h] and the impact of the parameter a [i.e., the sign determines where the parabola opens up or down, magnitude whether the graph is narrower, wider, or same shape as the graph of y = x2. We reversed the above procedure by sketching a parabola, g(x) , whose vertex is at, say,(100, 25), is concave down and is the same shape as y = 5x2 .
Answer: g(x) = 5(x + 100)2 +25.
Alas, not all parabolas arrive in vertex form. Take h(x) = 2x2 8x 8 in Problem 5 on page 221 and point out it is in standard form y = ax2 + bx + c. Ask a volunteer for the values of a, b, and c. Sketch h(x) on their calculators and work 5b. Then come up with a formula for h(x). Answer: h(x) = 2(x+ 2)2 .An observation to highlight is that a = 2 for both forms. How can you check if you are correct? Enter both formulas in the calculator and check your tables and graphs, or multiply out h. See the assignment due on 3-28.
New group assignments today.
We sketched y = 3(x + 5)2 + 18 and determined the y-intercept and the x-intercepts (zeros). We could expand this and pull out the Quadratic Formula, which is time-consuming, or we could replace y with 0 for
y = 3(x + 5)2 + 18:
0 = 3(x + 5)2 + 18
(x + 5)2 = 6
Using the graph, notice the recipe for finding the zeros is to start on the x-axis at x = 5 on the axis of symmetry and walk
and
units away.Discussed the factored form of an equation. Drew a concave up parabola through 1 and 3 and asked for a formula. One was y = (x +1)(x 3), although there are others. We found formulas of other parabolas through those zeros (one that is narrower, wider, concave down, etc.) Loaded a grapher with these formulas (set a large enough window value for Ymin and Ymax) to related the graphs with their equations. This is a nice review of 5.3. Finally we asked to find a formula for the parabola whose y-intercept is, say 15.
Answer: y = 5(x +1)(x 3)
We found formulas for the parabolas shown, paying attention to when to use vertex form and when to use factored form. (If given the zeros, use factored form, if given the vertex, use vertex form).
Answer:
y = 6(x + 2)(x
6)
Answer:
y = 10(x
1)2 + 119
See the assignment due on 4-02.
