That’s so nice to hear. 😊 make sure to check out the playlists as I have posted all of the grade 12 advanced functions and calculus and vectors course material. Please like, share and subscribe to increase the channel’s visibility 😊
If you go to this website mshavrot.pbworks.com/w/page/21867096/Ms%20Havrot's%20Exciting%20Math%20Classes you will find the notes and at the end of the notes I write what homework questions I assigned. Thanks for watching : )
hello how would you solve the oblique asymptote when for example x^3/x^2+1, I did x^3/x^2 + x^3/1 and I got x + x^3 but apparently the equation of the o.a. is y=x
You need to check your division. I am assuming it was x^3 /(x^2 +1) which is not what you did. You cannot split it up like that. It would be like me saying that 9/(2 + 1) = 9/2 + 9/1Try it again using long division and dividing it properly. Let me know if you find your mistake. : )
Hi,Ms. Havrot love the videos and have learned a lot. I had one question about graphing so how do you estimate the scale to use? Because when I basically finish my whole graph and its in small intervals. When the question asks me for a test point to determine local/max points i do that last. And what ends up happening is my y value is way too big for my scale to fit it so I have to basically regraph everything which isn't efficient for a test or something. Another question I had was for the test points let's say I have 1 vertical asymptote and 1 Oblique parabola asymptote, and f(x) ends up being a parabola in between the two how would I predict the test point to use to best find the local max and min. An example i was doing that had this exact question was x³+3/x-1. And is it wrong to choose a test point and that test point ends up not being the actual local max or min. Sorry for the amount of questions its just these 3 have really been confusing me. I'm good with the graphing and everything else. Thanks
I understand your concerns and the best answer I can give you is to do all your points before deciding on a scale. As for determining the best test points you are generally limited by the vertical asymptote and the oblique. Not until you do curve analysis in calculus will you be able to be more accurate in finding the exact max and min values which is a big part of calculus. In advanced functions you are not expected to be exact although you should be able to test a few points to see if you are close. Also, the question that you mentioned would not have an oblique asymptote unless the degree in the numerator was meant to be 2 and not 3. Remember that for an OA the degree is one greater than the denominator. Hope that helps!
@@mshavrotscanadianuniversit6234 That helps a lot thanks for the detailed response. Amazing teacher honestly these videos are years old and you are still commenting and helping out. You are much help to the community. Thanks!
For 5:04, example 2, how is it that expanding got them x^3 -3x -2, wouldn't the remainder be +2? I even typed the equation into google & it said expanding it would make it x^3 -3x +2. Am I missing something?
Ah yes … now I see what you are asking. Your expansion answer is correct but in the final step the question has been written to combine the last two terms after you divide by x^2. So, if you just look at the last two terms they are - 3X/x^2 -(-2/x^2) which would give you the +2x/x*2
What happens if you’re dividing by more than just x though? Like (x^2 + 5) / (x -4) Now you can’t just divide each term in the numerator by x, so what do you do?
This has helped me so much! Words can not describe how thankful I am for you!! Keep posting these math videos they are so helpful
That’s so nice to hear. 😊 make sure to check out the playlists as I have posted all of the grade 12 advanced functions and calculus and vectors course material. Please like, share and subscribe to increase the channel’s visibility 😊
This chapter is harder than previous ones. Followed your channel (so helpful) watched a few times, pass through with no issues. BIG THANKS :)
Thank You Ms Havrot!
You are so welcome! Hope you are finding the lessons helpful. Please like, subscribe and share to increase the channel's visibility. : )
do you have any textbook homework questions along with the lessons or is it just the lessons
If you go to this website mshavrot.pbworks.com/w/page/21867096/Ms%20Havrot's%20Exciting%20Math%20Classes
you will find the notes and at the end of the notes I write what homework questions I assigned.
Thanks for watching : )
hello how would you solve the oblique asymptote when for example x^3/x^2+1, I did x^3/x^2 + x^3/1 and I got x + x^3 but apparently the equation of the o.a. is y=x
You need to check your division. I am assuming it was x^3 /(x^2 +1) which is not what you did. You cannot split it up like that. It would be like me saying that 9/(2 + 1) = 9/2 + 9/1Try it again using long division and dividing it properly. Let me know if you find your mistake. : )
Hi,Ms. Havrot love the videos and have learned a lot. I had one question about graphing so how do you estimate the scale to use?
Because when I basically finish my whole graph and its in small intervals. When the question asks me for a test point to determine local/max points i do that last. And what ends up happening is my y value is way too big for my scale to fit it so I have to basically regraph everything which isn't efficient for a test or something.
Another question I had was for the test points let's say I have 1 vertical asymptote and 1 Oblique parabola asymptote, and f(x) ends up being a parabola in between the two how would I predict the test point to use to best find the local max and min. An example i was doing that had this exact question was x³+3/x-1. And is it wrong to choose a test point and that test point ends up not being the actual local max or min. Sorry for the amount of questions its just these 3 have really been confusing me. I'm good with the graphing and everything else.
Thanks
I understand your concerns and the best answer I can give you is to do all your points before deciding on a scale. As for determining the best test points you are generally limited by the vertical asymptote and the oblique. Not until you do curve analysis in calculus will you be able to be more accurate in finding the exact max and min values which is a big part of calculus. In advanced functions you are not expected to be exact although you should be able to test a few points to see if you are close. Also, the question that you mentioned would not have an oblique asymptote unless the degree in the numerator was meant to be 2 and not 3. Remember that for an OA the degree is one greater than the denominator. Hope that helps!
@@mshavrotscanadianuniversit6234 That helps a lot thanks for the detailed response. Amazing teacher honestly these videos are years old and you are still commenting and helping out. You are much help to the community.
Thanks!
For 5:04, example 2, how is it that expanding got them x^3 -3x -2, wouldn't the remainder be +2? I even typed the equation into google & it said expanding it would make it x^3 -3x +2. Am I missing something?
Ah yes … now I see what you are asking.
Your expansion answer is correct but in the final step the question has been written to combine the last two terms after you divide by x^2. So, if you just look at the last two terms they are - 3X/x^2 -(-2/x^2) which would give you the +2x/x*2
What happens if you’re dividing by more than just x though?
Like (x^2 + 5) / (x -4)
Now you can’t just divide each term in the numerator by x, so what do you do?
You do long division.
How would i solve if there was vertical asymptote for example (x^2 + 2)/(x -6)
You would have both a vertical asymptote when x = 6 and also an oblique asymptote.
@@mshavrotscanadianuniversit6234 ty
i love you
That’s sweet of you to say ❤️
thank you Ms Havrot!
You’re most welcome 😊
yo rishie, did u go to westh ill