Hi Ms Havrot, I am currently in Grade 11 Functions and will be starting Unit 2 Lesson 2.5 tomorrow in class. I just wanted to comment that your videos have helped me so much since the start of MCR3U for me. I struggled in mathematics during grade 9 & 10, I did not do any lessons, homework etc and it caused a major setback. After watching your videos I have learned so much and truly appreciate what you do and I hope you can keep up the great work. Thank you for spreading your knowledge.
So nice of you to send me this warm message. I love hearing about how I am making a difference. I wish you the best in Functions and hope that you continue to watch next year as well. 😊
14:23, if this were a test ;) would you leave the brackets or distribute the negative? Generally, would you expand everything for final answers? Or is that preference?
I would leave it in factored form. Basically, in factored form, it is easiest to see what the zeros are for the function if you are asked to graph the function.
You need to remember that you want to know what makes the denominator equal to zero and that is what you do NOT want x to be equal to. So, if you forget, you can always set each of the bracketed terms equal to zero and solve and you will find what x can NOT be equal to or else the function would be undefined at those points as you can not have a zero in the denominator.
X- 1 divided by 1 - X Remember that 1 - x = - x + 1 = -( x - 1) So now you would have (x-1) / -(x-1) which is -1 Another way you can think of it is if you let x =3 You would have 3-1/ 1-3 = 2/-2 = -1 which proves that the answer is -1 Hope that helps! 😊
Listen to my response a second time ... The other option is to see it this way: Factor out a negative one from ( 1 - x ) and you would get - (-1 + x) right? and ( -1 + x) is the same as (x - 1) just as if I said -1 + 2 = 2 - 1 .... SO that means that once you factor out a negative the two brackets are the same, so now when you divide them out the -1 is still there. Therefore they divide into each other - 1 times. Does that help?
Hi miss, I have a test on Chapter 2 tomorrow and I'm still having some trouble identifying the restrictions and domain. I'm not really sure on what to do without looking at the textbook solutions.
Well, say for example we had (x+3) in the denominator. 0 would not be a restriction because if I set x=0 the denominator would be 3 and I CAN divide by 3. However, if x = -3 then we have a problem (and a restriction) because when I set x= -3 my denominator would be 0 and I can not divide by zero. So basically if you set the denominator equal to zero and solve for x that will be your restriction. 😊
Hi Ms Havrot,
I am currently in Grade 11 Functions and will be starting Unit 2 Lesson 2.5 tomorrow in class.
I just wanted to comment that your videos have helped me so much since the start of MCR3U for me. I struggled in mathematics during grade 9 & 10, I did not do any lessons, homework etc and it caused a major setback.
After watching your videos I have learned so much and truly appreciate what you do and I hope you can keep up the great work.
Thank you for spreading your knowledge.
So nice of you to send me this warm message. I love hearing about how I am making a difference. I wish you the best in Functions and hope that you continue to watch next year as well. 😊
your videos are carrying me through highschool math, thank you so much!!
Happy to help! Hope you find everything you need here. If you have any questions feel free to ask. : )
Very clear explanation than you for going through all the examples.
Thank you Jay. If there is anything you would like me to explain please leave me a message Good luck with your studies!
14:23, if this were a test ;) would you leave the brackets or distribute the negative?
Generally, would you expand everything for final answers? Or is that preference?
I would leave it in factored form. Basically, in factored form, it is easiest to see what the zeros are for the function if you are asked to graph the function.
On 4:05 why is it not x not equal to -3 and +2? Do we switch the signs?
You need to remember that you want to know what makes the denominator equal to zero and that is what you do NOT want x to be equal to.
So, if you forget, you can always set each of the bracketed terms equal to zero and solve and you will find what x can NOT be equal to or else the function would be undefined at those points as you can not have a zero in the denominator.
Hello Ms Havrot!
I was curious on why **x** was not equal to one in the question "Simplify and State the domain" 14:45
It cannot be equal to one because if you sub one into the equation you would get a zero in the denominator 😊
Basically restrictions are what would make the denominator equal to zero
Ah, thank you so much!
hi miss,
I was just wondering why at 14:19 the numerator became negative when you divided (x-1) and (1-x) out of the equation
X- 1 divided by 1 - X
Remember that 1 - x = - x + 1
= -( x - 1)
So now you would have (x-1) / -(x-1) which is -1
Another way you can think of it is if you let x =3
You would have 3-1/ 1-3 = 2/-2 = -1 which proves that the answer is -1
Hope that helps! 😊
Can you explain what you did in question 2? When you were crossing the (x-1) and (1-x)
Listen to my response a second time ... The other option is to see it this way: Factor out a negative one from ( 1 - x ) and you would get - (-1 + x) right? and ( -1 + x) is the same as (x - 1) just as if I said -1 + 2 = 2 - 1 .... SO that means that once you factor out a negative the two brackets are the same, so now when you divide them out the -1 is still there. Therefore they divide into each other - 1 times. Does that help?
Hi miss, I have a test on Chapter 2 tomorrow and I'm still having some trouble identifying the restrictions and domain. I'm not really sure on what to do without looking at the textbook solutions.
Hmmm is there something I can do to help you?
Maybe you could email me some questions that you have tried and I’ll comment on them. Send it to barbarahavrot at gmail dot com
why isn't zero always a restriction
Well, say for example we had (x+3) in the denominator. 0 would not be a restriction because if I set x=0 the denominator would be 3 and I CAN divide by 3. However, if x = -3 then we have a problem (and a restriction) because when I set x= -3 my denominator would be 0 and I can not divide by zero. So basically if you set the denominator equal to zero and solve for x that will be your restriction. 😊
@@mshavrotscanadianuniversit6234 thanks! i kinda caught on after i asked lol