Review Packet Questions 6-8

I know that we were supposed to do two but since 7 and 8 are almost the same thing, why not. so...

6a. Find a relative minimum of the function O(t)= (100t^2+100)/(t^2+4t+1) on the interval [0, +∞) and describe its significance.
If you graph this on your calculator, the only relative minimum on this interval is at t=1. This means that after 1 hour, the stream has reached its lowest concentration of oxygen.

6b. Find the average rate of change of O(t) on the interval [2,4] and describe its significance.

Since the ARC is the slope of a certain interval, it can be found by taking the coordinates for O(2) & O(4) and using the slope formula:

Δy/Δx = 51.5-38.5/4-2 = 6.5

This means that from the 2nd hour after the waste has been in the stream to the 4th hour, the oxygen concentration has increased by an average of 6.5% an hour.

7. Since the graph in #7 is only translations, you only have to add or subtract from the x and y variables.

The graph is translated to the left by 3 so x--> x+3 (when x-->x+a, translated to the left by a; when x-->x-a, translated to the right by a)

It is also translated up by 4 so y-->y-4 (when y-->y+a, translated down by a; when y-->y-a, translated up by a)

since the vertex of the graph lies on the x-axis, the function equals zero

so: abs(x+3) + abs(y-4) = 0

8. This graph was only scaled, not translated so the x and y were multiplied.

The graph was scaled in the x-direction by 3 so x--> 1/3*x (when x-->a*x, scaled by 1/a)

Since the neither the highest nor lowest points of the graph lie on the x-axis but instead lies on 1 and -1, the function equals 1

so: abs(1/3*x) + abs(y) = 1

review packet solutions to #11 and #23

11) In order to put the second equation in the desired format (in terms of one variable) you pick a variable to solve for in the first equation. I chose y. I took the equation 4x^2*y=100 and reaarranged it to y=100/4x^2. I then substituted this into the second equation for y and simplified. The unsimplified version is z=40x^2+20*x*100/4x^2. When you simplify this you get z=40x^2+500/x.

23)For this problem I used the form a*b^x. a=7 (starting amount) b=1/2 and x= 4.8/3.6 (this is the amount of time you have divided by the amount it takes for one half life to occur). Therefore, the overall equation is 7*(1/2)^(4.8/3.6) and this results with 2.78g left.

Solution to #3 on Winter Break Review Packet

Find and simplify the difference quotient of f(x) = (2x+1)^2.

We know that the formula for the difference quotient is (f(x+h)-f(x))/h, so we just plug in numbers to get the answer.

(f(x+h)-f(x))/h
((2(x+h)+1)^2)-(2x+1)^2)/h
((4x^2+8h*x+4x+4h^2+4h+1)-(4x^2+4x+1))/h
(8h*x+4h^2+4h)/h
4(2x+h+1)
8x+4h+1

Solution to #10 on Winter Break Review Packet

If 3x^2+20x+k=0 has exactly one solution, then find the value of k.

The way we approach this problem is to set the discriminant equal to 0, and then solve for k.

sqrt(b^2-4*a*c)=0
sqrt(20^20-4*3*k)=0
20^20-4*3*k=0
-12k=-400
k=100/3

I'm sorry everyone, I was wrong on that problem. I realize it now. The f(ln(x)) is which i answered Casey's question yesterday, my answer was wrong. On a brighter note, good luck to everyone on the pretest, and if anyone needs help, I'm willing to help people until 11.

for Dylan:

can you please email me about the GSP?

monica_moore91@yahoo.com

pretest

I do not understand #17. Is it a arithmetic or geometic? however is it a pattern because it is addition? and how do we find the sum?
melissa

holes

How do you find the holes in certain functions?

#9 review

Could somone go over the simplification of the numerator on this one, I was confused as to how it ended up being what the answer said it was. Thanks

Oblique Asymptotes

I was wondering if anybody had a simple rule for how to find the oblique asymptote of an equation.

Pretest #17

Does anyone know how to do #17 on the pretest? I haven't been able to compute the sum or figure out if it is arithmetic or geometric.

GSP

can someone please email me the gsp sheet

coolcat791@gmail.com

Answer to Carries Question

For 11, I set up the system according to the way we did the Britney Spears question in class.

11 = a * b^10
26 = a*b^15

I then solved fo r b and got b = the fifth root of (26/11). After this I found the amount of people who heard the rumor when the time was zero.

We can re-write the equation as f(t) = a * (26/11) ^ (t/5). We know that the amount of people who heard the rumor at t = 10. So we can divide the amount of people who heard it then (11) by 26/11. This gets us the value when t = 5. We then divide the people who heard the rumor when t = 5 by 26/11. Now we have found the value of a.

The exponential equation of the problem should be:

f(t) = 1.96894 * (26/11) ^ (t/5)

Now, just find f(31). I hop this is right Carrie and everyone else, if it is not please tell me so I don't get it wrong too.

Reponse to sam's question

First, use differences of squares on X^2-1 to make log ((x-1)(x+))= 2=+log (X+1) then rewrite the log((x-1)(x+1)) to log(x+1) + log(x-1) cancel out the log(x+1) and you should be fine from there

Book Review #s 51 and 52

These problems were kind of difficult for me, so I spent a lot of time thinking about them. Just thought I'd share some ideas...

For 51, I figured that as x went up, y would go down (since it's -logbaseb(x)). At first I thought it would be a regular concave up exponential growth graph except reflected over the x-axis. So I thought it was graph VI. Then I realized x must be greater than 0, so it had to be graph V.

For 52 I had a harder time... as x gets bigger in the positive direction, y gets smaller (if you take b - a fraction - to the millionth root it's smaller than taking it to the third root). As x gets bigger in the negative direction, y gets bigger (b to the negative million is 1/(b^1000000), which 1 over a really small fraction... basically a really big number). So graph IV is the only one that seems reasonable. Also... you know that when x=0, g(x) must equal 2. 2 is the y-intercept of graph IV, so it all works out!

Domain Question

how do u figure out the domain of f(ln(x)) if u no the domain of f(x)?

Practice Test Questions

For number 11 can someone explain how you would go about setting up a system to solve the problem.
Also, for 16 I said that log11/log2=-log11/log .5 but i wasn't sure how to prove it. Any suggestions?

tips for finding asymptotes or holes

holes:

factor the numerator and see if it cancels out the denominator, if it does then the function has a hole

oblique asymptotes:

Use calculator--> F2 PropFrac and insert the function. take the equation you get and basically ignore the remainder (the fraction you get in your final equation) and that is the equation for the asymptote.

w/o calculator--> just divide the numerator by the denominator and ignore the remainder once again.

horizontal asymptotes:

big little principle, basically. these are hard to determine.

Vertical asymptotes:

when the denominator equals zero and doesnt cancel out with the numerator otherwise its a hole

what does use solver mean

NOTE: "Solve" does not equal "Use solver" if there is an algebraic solution, what does use solver mean.

HW #34

For question 30 in section 5.4, how do you solve for x? I keep getting an equation where the x's cancel out. (x/2x=t)

Also, I don't understand 75b (in the same section). How do you get ln(2)/ln(5)?

Tips for 4.5 19

Does anyone have any tips for finding problems like this on the hw?

5.3-106,107

alright this is probably a stupid question, but how exactly do u graph the second function which is like f(sub n) (x).

4.5.29

Well, you can get the horizontal asymptotes by the big-little principle, little over big =little= horizontal asymptote at zero, but I thought the vertical asymptotes were a lot more tricky.

questions

I have the same problems...I am not sure how to find the horizontal asytopes and the holes and I do not understand #62A thus not able to continue.
melissa

Problem 4.5.29

I am having similar trouble as Taylor, however my question stems from the arrangements of the parts of the graph. I am very confused as to how you are supposed to draw the lines of the graph. Sometimes the part of the lines on the graph will be in different quadrants than other graphs. I am confused about how you determine in what quadrants you should draw the lines and in what quadrant you should not.

horizontal asymptotes

I'm not really sure if I understand how to find the horizontal asymptotes correctly. For #62, in section 4.5, is the horizontal asymptote y=50 because for questions 19, 29, and 31 i did not get the correct value for the asymptote.

Response to Asymptotes v. Holes

Casey Blue, I believe you answered your own question. It is true that when you have a graph with a numerator and denominator where both the numerator and denominator have the same factor to cancel out. Of course the result in the case we refer to we get a line from the resulting equation, but to prevent the original equation from being undefined, the line has a hole where the zero of the denominator is located.
Secondly, I do hope someone can find an easier way to do 107 in 5.3 because the only way I see is guess and check. Boy is this long.

Asymptotes VS. Holes

Does anyone remember/know the exact rule for when a graph will have a vertical asymptote and when it will have a hole in it instead? Because I would think that any value for x that would make the denominator of an expression 0 would create a hole, but apparently sometimes it's an asymptote instead. I think in class we did one problem like this on Friday, but I can't seem to find understandable notes about it. Is it only a hole when you can cancel out a factor from both the numerator and the denominator? Like 4.5.41... The function is f(x)=(2x^2-x-6)/(x^3+x^2-6x). You can factor (x-2) out of the numerator and the denominator, and the book said this gives the graph a hole at x=2. But when a factor doesn't entirely disappear by canceling out, there seems to be an asymptote instead. Hm... i'm bad at explaining. If anyone understands the jist of what i'm talking about and has any ideas about it, let me know.

c5 supplements?

did we ever get the second half of the supplements b/c i can not find it at all!

section 5.3 #107

How are we supposed to find a value for n without guess and check? Because the answer is 30 and that would mean we'd have to type in y= x - (x^2/2)+ (x^3/3)- (x^4/4)... - (x^30/30) and thats a lot to put into the calculator.

C5 16-25

Does anyone have the Chapter 5 Supplemental Problems?
Please send to silliewillyzg@yahoo.com

graphs and asymtotes

For question 107 in section 5.3 what features should you look for in order to figure out where the graph is a good estimate of the function?
Also, in general for graphing equations after finding the various asymtotes are there any general tips for determining the shape?

Not a typo

4.5 on the homework sheet is not a typographical error. If you notice we don't even get to a 5.5 at all this chapter. The last assignment says 5.4. I think it is supposed to be review because some of this stuff is similar to what we did last chapter.

Calculating Limits on TI89 and Finding "e"

There's a function for calculating limits in the catalog in your calculator if you didn't already know. The syntax is of the form "limit(equation, variable, number)," where "equation" is an equation in a variable, variable is the variable you're limiting, and number is the number to which you want the variable to go up to. So if you typed in "limit(x/(x+1),x,1)," the calulator would give you back the answer "1/2," as it should. You can also use infinity as a number, which is found above the catalog key and selected using the diamond key.
I've heard a few people so far this year wondering exactly what the magic variable "e" represented, and it actually has to do with limits. To calculate "e," type in "limit((1+1/x)^x,x,infinity)." The calculator will simply return "e" to you.

Shapes outside of the bounds

I was wondering if anyone know whether the figure had to be within 16 ft aall the way up, or whether the base just had to fit those guidelines.

Shapes outside of the bounds

I was wondering if anyone know whether the figure had to be within 16 ft aall the way up, or whether the base just had to fit those guidelines.

Help With Project

How are people going about troubleshooting their designs for the paper to come up with the best one?

#11 and #45

can someone explainto me how to do number 11 in supp sheet and #45 in 5.2? thanks
melissa

#9 Supp

Could anyone make any progress on #9 without a calc? I couldn't figure it out.

ARC

To find the ARC of two points, find the slope. Subtract the frist y-value from the second y-value, and divide by first x minus the second x

IRC and ARC

I know we learned this a little while ago but to find the IRC the tangent line is drawn and then is it the slope that gives the answer? If so then how do u find the ARC?

NUmber 7 Supplementary Sheet

Since this is probably a useful something or other to know, could somebody explain how you would go about figuring what the values are for A&B?

Can someone post the Supp Sheet

It isnt up yet I dont believe so if someone could scan chapter 5 supplementary probs in that would be greattt. Or email them to me at pucangclan@gmail.com thanks again

limits and whatnot

Yeah, i want to second the concern for the limit problems. Secondly, i am having some issues w/ 45 in the book and # 7 on the worksheet.
And Kevin, you're a good man for inding that formula.
On a lighter note, I like these geometric sequences!

ch.5 supplementary probs. 11 & 15

for question #11, can anyone explain to me what values we are supposed to plug in? (would it be any value of x greater than 2?)
also, what is the graph supposed to show us?

for #15, im not sure if i have the correct formula for C(x) because my value for C(-10) is really small (like .8787). My equation was C(x)=208.6*(1+.78)^x, can anyone help?

-thanks

5.2 #45

For number 45, I got that 2a+c=630 by plugging (0,630) into the equation, but i don't know where to go from there. Mr Karafiol's hint on the assignment sheet says to start with a small number for k, but how can that help if the exact values of a and c are not known?

Lost with Geometric Sequence

Can someone please explain to me how to do geometric sequence when ur starting with something like:

a sub 2 = 40 and a sub 4 = 160


How do u set it up if you don't have a sub 0?

5.2 and supplement homework questions

I'm stuck on number 45 in section 5.2. I read the example that went along with the question but didn't understand how it helps. ANy suggestions on where to begin?

For question ten on the supplement I don't understand how to account for the differences between b and c. I think for one of them you raise two to the original domain and range values to get the answer but I'm not sure which part that applies to.

For numbewr 11 I'm not really sure what to be looking for on the graphs. WHat feature on a graph dictates the limit?