TMP question #6a

I know we've done these types of problems in class before but I wasn't sure if I was remembering how to solve them correctly. Since the graph of 2x^2-4x=o is a parabola would the value of k have to equal the vertex of that parabola in order for it to have only one solution?

Overview of Working With Functions on TI-89

To define a function on the calculator, go to F4 > Define, and type in "f(x)=FUNCTION," where FUNCTION represents an equation containing the variable of the function. The screen should now say something like "Define f(x)=x^2," at which point you can press enter. From here you can enter in something such as f(5), and the calculator will display the value of x^2 where x=5. This is extremely handy to use for the homework over the weekend.

Assignment 10: 3.2.44, 3.2.59, and TMP(10b)

I just spent the first part of my extremely exciting Friday night doing my math homework, and I had trouble with a couple problems...

For number 44 in 3.2, we were to answer questions about a piecewise g(x) function. I was able to answer it, and did so by graphing, but my calculator wouldn't let me type the rule -1<=x<=2 into the "y=" screen. When we are graphing things like y= 2x-3 | x<-1 or y= |x|-5 | -1<=x<=2 can we only include a one-sided inequality? Am I typing something wrong?

For number 59 in 3.2, I knew that to find the domain, I had to solve (x-9)^2 <= 9 (since what's inside the square root sign must be greater than or equal to zero). These were the steps of my solution:
(x-9)^2<=9
|x-9|<=sq.rt.(9)
x-9<=3 and x-9<= -3
x<=6 and x<=12
However, the solution the book gave is [6,12], meaning that I should have gotten x>=6. Where should I have switched the direction of the inequality sign? I am sure there's an easier way to solve this on a calculator, but I'm just curious to find my mistake.

For 10b on the TMP sheet, I am just really confused about how to approach it. In the third row of the table, I got a final result of y=(x^2/4)-5. And if I made the changes given in 10b, the final result would be y=(0.5(x-k))^2+x-4. I tried setting these two equations equal to one another on my calculator and solving for k, but I got a complicated sq.rt. answer with x involved. Did anyone find an actual numerical value for k?

Thanks guys

response to #5

Run a horizontal line through the graph, seeing where it intersects four lines. At roughly above -60 and below 130, it has 4 solutions.

SECRET BONUS DIAMOND SURPRISE!

If you press the diamond key, then the EE button (two above the on key) a screen will pop up that tells you what different buttons do when diamond is selected.

#5 practice test

On number 5 on the practice test I dont understand how to get the value of k based on the number of solutions. Also, I dont know how you would find the number of solutions based on a certain value like in problem #2 in station 1 for the around the room exercises. What am I missing?

14c on practice test

i don't understand 14c on the practice test. i got an expression for the surface area but dont know what to set it equal to or how to solve it. maybe i did something wrong. i got Surface area =4(200000/x)+ x^2 what do i do next?

problem 14

In a problem dealing with a box or cylinder with no top, does the surface area include the interior of the object? For example, in question 14 of the practice test, would the formula for surface area include the inside of the box as well as the outside? It seems like logically the material used would only include the surface of the outside, but last time i asked a question about a problem like this someone said that the interior would be included as well, so i'm sort of confused.

Tangent Lines to Parabola

Did anyone understand how to find the line tangent to the parabola on the practice test? Is there something we're supposed to infer based on prior knowledge?

Absolute Max/Min vs. Relative Max/Min

I am so confused about finding absolute and relative max/min values.
For example, in practice test no calculator section, #4a find the approximate max on [-5,5]. What is the correct answer, (5,500) or (-1,130)?
Continue #4, 4b, is there a absolute max value? and the same thing on the "I didn't want to go to the talk" sheet, #4a-4d.

Practice Test Questions

What did other people get for #5? I wasn't really sure how to do it but I said that -8<=x<=4.
For #12 how do you make x=0 at 1990 in the data matrix program?
For 14c how do you find the minimum? I got y1=x^2 +800,000/x and graphed it but had trouble finding an upper bound in order to use the minimum function in the math menu.
How do you solve #16. I know you use the distance formula and set it equal to 4 but i'm not sure how to set up the rest of the equation.

Last year's test

If anyone has done the EC practice test, can you help me with 11b? It says to find a slope for the line found in part A so that this line is tangent to the parabola y=x^2. The line I got in part A was y=9-m(x-3), but I'm a bit confused about where to go from here. I tried setting the two equations equal to each other (so that 9-m(x-3)=x^2) and solving for m, but of course that gave me a slope that was in terms of x, not a numerical value. Any ideas?

Apps Desktop on TI-89

There's a couple of different ways to display the Apps screen on your calculator, one of which uses icons and is refered to as the Apps Desktop, and the other which uses a standard drop-down menu.

To toggle the Apps Desktop on and off you can go to Menu screen, and all the way at the bottom is an option to control it.



Basically the only difference seems to be how it looks visually and how long it takes to find something. In the Apps Desktop mode you just have to find the icon of the program that you're looking for, and press enter to run it. However, all the programs show up in this screen.



When you have Apps Desktop turned off however, you are presented with the all too familiar drop-down list.



The only difference is that its organized into a list, and the Flash applications are separated from the rest, but you can use Diamond+Apps to go directly there. The reason I personally leave the desktop turned though is because i can just hit the number of the program I want to launch without scrolling around. For instance to get directly to the Data/Matrix Editor, you can just type in "6."

Setting Up Scatter Plots and Linear Regression Problems

If any of you are still confused about how to set up a scatter plot or linear regression problem on your TI-89, I typed up some directions for how to do so as a reponse to Sarah's blog, "assignment 6". Hope it might be helpful.

assignment 6

I had a lot of trouble with the problems, in assignment 6. since they are all similar how would you go about setting the problem up and actually solving it.
-sarah iqbal

2.4#21

I set up equation of y=10*(300/x) +20 and graphed looking for minimum. However, in first quadrant graph looks like it approaches x axis and i thought we were looking for minimum value. can someone point me in the right direction for a better way to approach the problem?

scatter plot

How do you make a scatter plot?

If you want to find a good window...

when viewing your graph, hit f2/ZOOM and go up until you see ZOOM FIT. This usually gives you a pretty good window.

2.4.15?

I made the cost per hundred graph, and am wondering what to do from there. Any thoughts?

2.3.31
what is the best way to set up this problem, concerning the fact that the height is at least 3?

The way I find windows

Usually to find a good viewing window for any equation, especially if the graph does not even cross the standard viewing window, is to use the trace key to find a point on the graph. After you find a point on the graph create a window centered around this point of a decent size. Atleast now you have a window with the line in it. From here you can tweak the window to best fit your use. It is not the quickest or neatest way to do it, but i find that it works just fine.

Tricks for finding windows?

Does anyone have any tricks based on looking at equations for finding good viewing windows?

#11 of 2.4

How do we do this blog thing?? I am so confused... anyway I don't know how to do number 11. I am Melissa. Thanks for helping.

Number 9

Can anyone help me set up the problem for section 2.4 number 9? I got that the surface area = x^2+4(x*[30000/x^2]), where x is the length of the base and width and 30000/x^2 is the height, because the volume has to be 30000. When you graph this there doesn't seem to be a minimum, but the back of the book has a numerical answer. I know that x has to be greater than 0, but is there another constraint for x that determines what part of the graph applies to this problem? Otherwise, I can't see how there could be a minimum value for the surface area.

A quicker way to find the zeros of any equation is to use the zero function. In the hom screen go to F2--zeros(--then type in your equation then ,x) and tadaa. It's ike two strokes quicker, so it's not incredibly useful. Also, you want to do summation sequences or multiplication sequences, then you can find those under the F3 menu. it's a litt trick
for summation do as follows-- [sigma symbol](variable expression,variable,lower limit,upperlimit)

Graphing Window Find Tricks

Some of you may already know about this but anyways, for problems such as 1,5, and 7 in 2.4, in which finding a small or large graph is required, using the Zoom Fit feture makes things alot easier. In number 5 for example the graph is hard to see unless you zoom in alot. If you push F2 and then scroll down to "ZoomFit" the calculator will show you the graph in vivd detail.

How Do I solve Problems Like problems 2.3:31 Correctly

I have been having a little trouble with some of problems like 31 from 2.3. I am not sure what is the best way to determine the equation that is to be graphed. How do I form an equation that will encorporate the prices and the possible dimensions? If anyone can help, it would be greatly appreciated.

Quiz #1-1B

For number 1b. do I have the right equation? I got 0.95=2.56 - .18/5(x-20) Cuz the answer I got for x was 64.722 So I added that to the year 1980 and I got 2044.72 so I rounded to the year 2045 and I got it partially wrong. What did I do wrong?

How to Perform a Quadratic Regression on a TI-89

In class we went over how to perform a linear regression over a set of data, so I'm going to take that a step further and show how to do a quadratic regression.

Open up Apps > Data/Matrix Editor > New, type in something under variable, and press Enter. Note: Variable names cannot match system variables (such as linreg, expreg, quadreg).



Next you have to enter in a set of data, so let's just use the sample dataset

time	population
c1	c2
1	1
2	4
3	9
4	16
5	25
6	36
7	49
8	64
9	81
10	100

An easy way to type this set of data in would be enter the 1-10 under "c1", and then if you go over and highlight "c2", type in "c1^2," and the "c2" column will be populated with the squares of the "c1" values.

Now go to F2 > Plot 1 > F1. Under "Plot Type" choose scatter, under "x" and "y" put "c1" and "c2" respectively, and press Enter.

When you're back at the Data/Matrix Editor screen, go to F5, and change "Calculation Type" to "QuadReg." Once again for "x" and "y" enter in "c1" and "c2" respectively, under "Store RegEQ to," choose "y1(x)," and press Enter.

Now we just have to set a window for this graph, so let's use 0 for x-min and y-min, 10 for x-max, and 100 for y-max.

So that's how to do a quadratic regression, but how about if we wanted to do the same for an exponential or cubic function. Below is a table listing a few common regressions, and under the Calculate (F5) screen, you can just change which one you enter.

Type	Command	Formula
Cubic Regression	CubicReg	y=ax^3+bx^2+cx+d
Exponential Regression	ExpReg	y=ab^x
Linear Regression	LinReg	y=a+bx
Logarithmic Regression	LnReg	y=a+b(ln(x))
Power Regrssion	PowerReg	y=ay=ax^b
Quadratic Regression	QuadReg	y=ax^2+bx+c

Problems of the Week & Scavenger Hunts

This is your first opportunity for Extra Credit. Rules:

1. No collaboration.
2. Research is okay, but must be correctly documented.
3. Each correct answer is worth +2 points on a test.
4. Answers to be handed in, on paper, by Friday, September 22

Ready?

1. How is “modal logic” different from ordinary logic?
2. Four positive integers are given. Take the mean of three of the integers and add it to the fourth. The four possible results are 29, 23, 21, 17. What are the original integers?
3. What is the Cantor Set?
4. Increasing the radius of a cylinder 6 units increases its volume by y cubic units. Increasing its height by 6 units would also increase its volume by y cubic units. If the original height is 2 units, what is the original radius?

My TI-89 didn't simplify...

When I was doing the homework for Thursday, I had trouble using the solve function on my TI-89 to solve one of the problems. For 48b in section 2.2, were were to determine how many real solutions the equation .2x^5-2x^3+1.8x+k=0 has when k=1. I entered solve(.2x^5-2x^3+1.8x+1=0,x) and the calculator gave me an unsimplified answer: x(x^4-10x^2+9)=-5. I ended up having to graph the equation to determine that there were 3 solutions. Did this happen to anyone else when solving this problem? Perhaps I am still just having trouble getting used to my TI-89, and I can't remember if at any point in class we went over dealing with calculators that give unsimplified solutions.

Also, in response to Carrie's blog, if part of what you are asking is whether you can program your TI-89 to always give an exact answer, I think you can do this by going to the MATH menu, scrolling down to the exact/approximate row, and choosing "exact". However, you might be referring to a way to get exact answers by typing in approximations, in which case I can't be of much help.

Ti-89 Approx v. Exact Answers

Is there a way on an 89 to enter an approximate value without doing any operations and get an exact answer?
Ex. Enter in 1.73205(approx) and get out sqrt(3)?
Or is there a way to find a zero of an equation and or graph and get out an exact answer as opposed to an approximation?

Welcome to the Period 4 Blog!

Welcome to the period 4 blog! Notice, first of all, that this is a blog about precalculus, not an excuse to blog in precalculus. Having gotten the prepositions out of the way (as it were) ...

You'll need to make a blog post or substantial comment every week, due on Fridays. (Yes, they're still due on Fridays on weeks when the last day isn't Friday, but you can always do them early.)

A posting or substantial comment can be ...

• A question or hint about a math concept or problem.
• A solution to a math question or problem.
• A "read" on how math class is going for you right now.

Each blog posting is worth up to 4 points; postings that do not discuss math explicitly (e.g. the third category above) are worth 4/4 points during the first quarter, and 3/4 points after that.

Have fun!