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Vasishta Angara
8/24/2011 10:03:38 am
For #45, did anyone get the 3rd part of the piece wise function??
Krishna Bathina
8/24/2011 10:18:44 am
You know that you have the restriction of [1,3) (due to the fact that there is an open circle on three it means its not included in the domain) so then you have to just take the slope between the two points given 1,1 and 3,0 and then implement the slope into either point slope form or slope intercept form.
Andrew Fox
8/24/2011 10:33:34 am
Hey! that was my question!
Vasishta Angara
8/24/2011 10:46:08 am
Thanks Krishna.
Dan Karesh
8/24/2011 11:40:04 am
For 57, how do you find the range? Do you just plug in 0 for x since there is an asymptote there?
Bakir Baker
8/24/2011 02:00:55 pm
Hey Dan, thats what I did and I got the right answer but I'm not completely sure. Also does anyone know how to do 65 parts b,c,d?
Ryan Pape
8/24/2011 03:19:34 pm
Bakkiiiirrrr,
Ryan Pape
8/24/2011 03:22:18 pm
*answer for b
Bakir Baker
8/24/2011 03:55:51 pm
OHHHHH, that makes a lot of sense now. So basically its just a bunch of substitution! Thanks bro
Whitney Pike
8/25/2011 03:59:48 am
so i'm confused do we do lesson 1.3 tonight? or do we just study for the quiz tomorrow?
Bakir Baker
8/25/2011 09:21:22 am
Nope, all we have to do is study for the quiz! It'll be on lesson 1.1 and 1.2
Dan Karesh
8/28/2011 03:58:59 am
For the packet, how do you graph the last one? It has three variables, so what do you do with the a^2 part?
Vasishta Angara
8/28/2011 07:04:19 am
i had the same question.
Brianne Honda
8/28/2011 11:07:20 am
what are the domain and range of a tangent(x) graph and secant(x) graphs?
Kim Baker
8/28/2011 11:09:07 am
I am also having trouble with those graphs. I have all but those. So if anyone could help with that or I can help with any others.
Neel
8/28/2011 01:12:17 pm
Which question's graphs question 4 or 5?
Kelsey
8/28/2011 02:06:59 pm
I think the tangent(x) graph is a domain of  inf., inf and the range was 1,1 ..But I'm not sure. When I tried to graph the seacant(x) graph it was a straight line at y=1, So I think I got that one wrong too. Anyone else have that? And does anyone know the domain/range of the y=log2x graph?
Kelsey Cyr
8/28/2011 02:46:35 pm
Oh my gosh, just kidding, I hadn't switched to degrees mode. I got tan to be d: (inf, inf)r: (inf,inf) and sec to be d: (inf, inf) R: (inf, 1) U (1, inf)
Vasishta Angara
8/30/2011 10:21:21 am
For 1.3 number 22,
Andrew Fox
8/30/2011 12:21:52 pm
Does anybody know how to solve number 42 in section 5?
Kelsey Cyr
8/31/2011 12:07:17 pm
Did anyone understand 5 & 6 in 1.6 pg 48 ? I saw the answer on the webpage, but can't figure out how to find it?
Brianne Honda
8/31/2011 12:57:39 pm
Okay so for inverses, I know you switch x and y but i'm stuck on what you do afterwards. help!
Rinky Patel
8/31/2011 01:06:17 pm
All you do afterwards is solve for y to get y by itself!
Emily Baker
8/31/2011 01:07:08 pm
Same with me Brianne, like in #9 in 1.5 we started in class, but then we never totally finished. The last thing I have written down after changing the x and y was x6 = y(y^24) after getting the x and y on seperate sides...
Rinky
8/31/2011 01:10:35 pm
Emily, I think she told us in class that there is no inverse for number 9. If you look at the graph it doesn't pass the vertical line test. Also, i checked the answer and it says the answer is no
Dan Karesh
8/31/2011 01:10:40 pm
Can't you also graph the line and do the horizontal line test to see if it has an inverse too? Or is that something completely different...
Rinky
8/31/2011 01:11:01 pm
*horizontal line test! sorrry :)
Emily
8/31/2011 01:13:24 pm
OH YEAH! I forgot, thanks! Well that's a relief.
Dan Karesh
8/31/2011 01:13:52 pm
Yeah vertical line test is to make sure it's a function
Dan Karesh
8/31/2011 01:14:43 pm
OH! hey guys do the chapter review problems too because it just chunks everything together which helps alot
Rinky Patel
8/31/2011 01:21:10 pm
Does anyone know how to solve #25 from section 1.6?
Ecuador 2011
8/31/2011 01:21:33 pm
Also, for 1116 for 1.6, the amplitude is from bottom to top right? Not just axis to top/bottom?
Dan Karesh
8/31/2011 01:36:47 pm
I think you just do tangent inverse of 2.5 = X
Dan Karesh
8/31/2011 01:41:34 pm
Oh wait just kidding... plug in tan x into y1 and 2.5 into y2 and graph them. then see where intersect... it should be the first two points to the right of the origin
Rinky Patel
8/31/2011 02:05:07 pm
OHH rightt! that makes sense, thanks!
sam jones
8/31/2011 02:09:40 pm
anyone remember how to get the function of a graph in your Y=? i remember that you need to go to stat, select the 8th one, and plug in l1,comma, l2, ,but how do you get the Y1?
Dan Karesh
8/31/2011 02:10:56 pm
Sam, you mean for like regression models?
Amanda Walsh
8/31/2011 02:13:36 pm
go to vars and then, yvars and then function and then Y1 then press equals and it will be there.
Vasishta Angara
9/1/2011 01:03:47 pm
Do we just send an email to Mrs. Johnson
Neel Thakkar
9/2/2011 01:44:58 pm
Neel Thakkar
9/2/2011 01:45:25 pm
Send Email..
heli bhatt
9/2/2011 02:17:48 pm
does anybody know how to solve #27 from section 2.1?
Krishna Bathina
9/4/2011 12:24:51 am
Yeah, in fact the problem you're looking at is one similar to our summer homework but that's ok. First what you have to do is factor out an X from the denominator to make it equal to X(2X1). Then there is a trig rule that states as the lim approaches 0 of sin(X)/X it has to equal one. So using that rule you now have as the lim approaches 0 of 1*(2X1). So now that you have something manageable just plug in your zero into this new limit and you should get the answer being 1
Krishna Bathina
9/4/2011 12:26:10 am
sorry its actually supposed to be 1/(2X1) but either way you get the same answer
heli bhatt
9/5/2011 03:19:16 am
OH YEAH! I forgot, thanks!
Bakir Baker
9/6/2011 06:27:53 am
Does anyone know how to do #11? I'm confused b/c they switched the x's to y's...
Bakir Baker
9/6/2011 06:29:32 am
Btw I know the answer is 0 by plugging in 3, but how would you support it graphically
Vasishta Angara
9/6/2011 08:23:32 am
Bakirr.
Andrew Fox
9/6/2011 09:26:18 am
does anybody know how to get 12 for the answer to #25?
Honsu Kim
9/7/2011 06:44:19 am
For problems like number 33 (pg 62) what does the "int" mean? I tried looking in the answers and it doesn't really explain much.
Ryan Pape
9/7/2011 09:27:02 am
It's a special type of function. Go to y= on your calculator, hit math, num, then int ( or 5 on your keypad. Then type in x, then ). You should be graphing int(x). It's like stepping stairs with holes on each side of each step.
Vasishta Angara
9/7/2011 12:23:43 pm
Wait soo for number 33, i just have to use the calculator?????
Brianne Honda
9/7/2011 01:40:14 pm
Yeah for all the matching ones I just plugged in the equations
Kim Baker
9/7/2011 02:37:47 pm
Can someone describe what the sandwhich theorem is?
Kim Baker
9/10/2011 10:10:03 am
in #47 of this weekends homework there is an a,b,c and d. for letter c. it says that in the piecewise function O is negative infinity and i thought it would be 1... does anyone know why?
Vasishta Angara
9/10/2011 10:29:57 am
I thought the limit was undefined, because the the left graph never touches the y axis.
Vasishta Angara
9/10/2011 10:35:41 am
Actually, for # 47 the book said the answer was  infinity. I think because the graph keeps going down to  infinity.
robert walker
9/11/2011 01:35:34 pm
did anyone else get DNE for #9 and 3 for #12 on the 2.2 homework??? i feel like its wrong
robert walker
9/11/2011 01:38:38 pm
nvm i know what i did wrong
Kelsey Cyr
9/11/2011 02:57:53 pm
Can anyone explain number 39 for 2.2? I'm really confused as to what the book is asking.
Kelsey Cyr
9/11/2011 03:11:56 pm
Oh, nvm, I figured it out.
Vasishta Angara
9/12/2011 10:22:32 am
I don't know y i keep forgetting this, but can someone remind me how to find a horizontal asymptote?
Kelsey Cyr
9/12/2011 11:01:55 am
You use the powers, like you use (N(x)^n) / (D(x)^n) (same power, you compare coeff.) If the power in numerator is less than the denominator, hasy = 0, and if the power in the numerator is larger than the power in the denominator then you look at the graph for the slasy.
Vasishta Angara
9/12/2011 11:26:09 am
Oh yeaa. Thanks a lot!
Kim Baker
9/12/2011 11:38:26 am
Does anyone understand 2630s problems? It says to give a formula for the extended function that is continuous at the indicated point
Kelsey Cyr
9/12/2011 11:38:33 am
Yeah! Can anyone explain what 26, 28, & 30 are explaining, I looked in the book and still am having trouble understanding what to do?
Vasishta Angara
9/12/2011 01:25:43 pm
For number 26, all I did was simplify the graph. Then I checked the graph to see if it was continuous at that point or not.
vasishta Angara
9/12/2011 01:26:29 pm
* simplify the equation
Honsu Kim
9/13/2011 07:56:25 am
For numbers 917 on the hw i am completely lost on how to find the slope and tangent on a curved line. Did we learn this in class or did she just say do the best we can?
Vasishta Angara
9/13/2011 10:00:12 am
Do we do the first 2.4 or both??
Vasishta Angara
9/13/2011 10:19:21 am
Oh and Honsu, if you go back to page 8485 i think it explains it. Its not completely clear. but it helps a little.
Honsu Kim
9/13/2011 10:48:14 am
Just the first one. the directions were a little confusing for me. The Secant line is just like what we did in class but the points on the graph are off.
Will Doiron
9/13/2011 11:20:01 am
does anybody understand 917 on the homework? i have been looking at the book and i get how to find secant slope but how do you get the tangent slope from there? and where does the equation/coordinate Q(2+h, (2+h)^2) come from? its in example 3 on pg 84
Ryan Pape
9/13/2011 01:04:40 pm
Will,
Ryan Pape
9/13/2011 01:05:38 pm
I just can't figure out how to solve 7 and 8 because there's no equation or 13 b and 14 because I got 1, but the answer's 1, if anyone could help.
Ryan Pape
9/13/2011 01:06:38 pm
*limit as h approaches 0 for my first post haha
Will Doiron
9/13/2011 01:33:55 pm
alright thanks ryan, that helped alot. i figured out how to do 7 and 8. you have to find the slope at each point (q1, q2...) and use P as y2, x2. for example, on #7 you know the P value on the graph is (20,650) and q1 is about (10, 225) and then you just do: change in y/change in x to find slope. and repeat for each point.
Ryan Pape
9/13/2011 01:44:35 pm
ohhhhh, that makes sense. Thanks!
heli bhatt
9/14/2011 09:00:50 am
does anybody know how to solve #37 and #38 from section 2.4?
Will Doiron
9/14/2011 09:38:08 am
does anybody know how to do 21 and 31 on tonight's homework? they are related
Honsu Kim
9/14/2011 09:40:33 am
I'd really appreciate if someone could walk me through the word problems (# 23, 25, 27) I know it's rate of change but confused after that
Honsu Kim
9/14/2011 09:42:17 am
Oh and Will, i think that for number 21 you just plug in (xh) for whatever X you have. So you have 1/(x+h)1 and then 1/(x+h) all over h. When you are diving by something you are basically multiplying by it's reciprocal. so you can multiply the entire thing by 1/h
Will Doiron
9/14/2011 09:48:27 am
oh yeah i totally forgot about that.. thanks honsu
Vasishta Angara
9/14/2011 11:09:12 am
Could someone help me with #2327?
Bakir Baker
9/14/2011 02:45:37 pm
I think for those problems it wants you to interpret the word problem and use the slope of a curve formula.
Dan Karesh
9/15/2011 09:46:17 am
How do you find the horizontal asymptotes again? And for the vertical ones, its just whatever is in both the top and bottom of the equation right?
Vasishta Angara
9/15/2011 09:55:24 am
For horizontal Asymptotes, u just look at the powers. IF there are the same powers.
Dan Karesh
9/15/2011 12:41:34 pm
ooooo that makes it so much clearer Comments are closed.

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