Math help (1 Viewer)

[Shadowdude]

Re: Math help (Because Carrot made me do this)

No one will help me 1v1

My sister doesn't remember any of her maths
My tutor encourages r0te learning and i'm too scared to ask her.

did you not read my post at all

 

[Sy123]

Re: Math help (Because Carrot made me do this)

Wasn't it large negative values? Like I know you can't have a negative in a square root but is it easier to test large negative values?

How do you know if negative infinity is outside or inside the domain?

What's testing neg/pos infinity? is that the one where you test those random numbers?

Just to confirm the whole domain thing so lets say this question:



So the domain would be x is less than or equal to 1 right? because 1-1=0 and zero is the lowest number you can have in a root right?
So how would I find the range? is it just 0? because square root 1-1=0 which is 0?

So then I would test large negative numbers? say if I test x=-100 then it would be:

which is



what do I do next? is this how you do it? :S

Ok, in reality what we do is we test both negative and positive large values, the thing is, for certain functions it is redundant to test negative large values because we know they dont work. We know that any large positive value wont work, we test a large negative value and we get a large positive y value. So for example , any value of x less than 1 will not work (so any large negative wont work) we dont test this because we dont need to, because we know we will get math error if we put it in the calculator.

Negative large values (as they approach to neg infinity), are part of the x values of the graph itself.
Now this is important, if we look at the domain x>1, this means that x can be any value BUT it must be greater than 1, is 0 part of this domain? No
Is -1 part of this domain? No, is negative infinity part of this domain? no, BUT positive infinity is, you can think of it like this: the domain for x>1 is from 1 to positive infinity, so pos infinity is counted.

Testing pos/neg infinity is testing very large values on your calculator, they arent really that random, but rather they are there so we can get an idea of how the curve looks like.

For the curve , you are right with the logic of the domain. Now for the range, it is always greater than or equal to 0. This is because, the result of a square root is always positive, so no matter how big our x goes (or how small within its domain), our y will always be greater than or equal to zero, hence our range is as such.

To test large negative values, take x=-100, we get . They will never ask you to plot this, nor do you need to, what this shows us is that when you get a very large negative value of x, we get a large (but a much smaller) value for y which is positive. This concept is best shown graphically.

So we find intercepts, starting point of square root function and the domain and range, we test a large positive or negative value. From this information, start to sketch from the starting point, through our intercepts. Now looking at the points we tested sqrt(101)=10.something. So although we have an incredibly big x value we only get a 'little' big y value, so our curve will look like such.

hence our curve goes through the 2nd quadrant (the section of positive y negative x), but its closer to the x axis than the y.

 

[enoilgam]

Re: Math help (Because Carrot made me do this)

No one will help me 1v1

My sister doesn't remember any of her maths
My tutor encourages r0te learning and i'm too scared to ask her.

Given this comment and your questions in this thread, I would question how useful your current tutor is.

 

[twinklegal19]

Re: Math help (Because Carrot made me do this)

Given this comment and your questions in this thread, I would question how useful your current tutor is.

+1000

 

[Shadowdude]

Re: Math help (Because Carrot made me do this)

inb4 carrot crucifies the tutor for roat learnin'

 

[deswa1]

Re: Math help (Because Carrot made me do this)

+1000

Are you finished all your exams now?

 

[Fawun]

Re: Math help (Because Carrot made me do this)

Ok, in reality what we do is we test both negative and positive large values, the thing is, for certain functions it is redundant to test negative large values because we know they dont work. We know that any large positive value wont work, we test a large negative value and we get a large positive y value. So for example , any value of x less than 1 will not work (so any large negative wont work) we dont test this because we dont need to, because we know we will get math error if we put it in the calculator.

Negative large values (as they approach to neg infinity), are part of the x values of the graph itself.
Now this is important, if we look at the domain x>1, this means that x can be any value BUT it must be greater than 1, is 0 part of this domain? No
Is -1 part of this domain? No, is negative infinity part of this domain? no, BUT positive infinity is, you can think of it like this: the domain for x>1 is from 1 to positive infinity, so pos infinity is counted.

Testing pos/neg infinity is testing very large values on your calculator, they arent really that random, but rather they are there so we can get an idea of how the curve looks like.

For the curve , you are right with the logic of the domain. Now for the range, it is always greater than or equal to 0. This is because, the result of a square root is always positive, so no matter how big our x goes (or how small within its domain), our y will always be greater than or equal to zero, hence our range is as such.

To test large negative values, take x=-100, we get . They will never ask you to plot this, nor do you need to, what this shows us is that when you get a very large negative value of x, we get a large (but a much smaller) value for y which is positive. This concept is best shown graphically.

So we find intercepts, starting point of square root function and the domain and range, we test a large positive or negative value. From this information, start to sketch from the starting point, through our intercepts. Now looking at the points we tested sqrt(101)=10.something. So although we have an incredibly big x value we only get a 'little' big y value, so our curve will look like such.

hence our curve goes through the 2nd quadrant (the section of positive y negative x), but its closer to the x axis than the y.

Ohhhhh I get it.

I finally get this!

So how would you add ordinates for this question?



Because Deswa tried to explain it to me yesterday but I don't get it Do you have to add ordinates or is there an alternative way to solve this?

Given this comment and your questions in this thread, I would question how useful your current tutor is.

Tbh, yesterday my dad said "I like this tutor. He's very good" just because he tests me lolwat

 

[Fawun]

Re: Math help (Because Carrot made me do this)

did you not read my post at all

Imagine that awkwardness.

 

[kazemagic]

Re: Math help (Because Carrot made me do this)

Ohhhhh I get it.

I finally get this!

So how would you add ordinates for this question?



Because Deswa tried to explain it to me yesterday but I don't get it Do you have to add ordinates or is there an alternative way to solve this?

Tbh, yesterday my dad said "I like this tutor. He's very good" just because he tests me lolwat

wow ur new tutor is brutal, btw is that question given to u by ur tutor? or was it something u thought up? below is the graph for ur equation
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427e19qp5ibkf3

 

[Fawun]

Re: Math help (Because Carrot made me do this)

wow ur new tutor is brutal, btw is that question given to u by ur tutor? or was it something u thought up? below is the graph for ur equation
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427e19qp5ibkf3

Yes it is. My tutor just writes the questions on the board and before I even have the chance to copy it down, she's already writing the solutions to it. 5 minutes before the end of the lesson, she tried to teach us the square root graphs and says "copy it down and go home and complete the exercise for homework". The question is part of my homework.

 

[kazemagic]

Re: Math help (Because Carrot made me do this)

Yes it is. My tutor just writes the questions on the board and before I even have the chance to copy it down, she's already writing the solutions to it. 5 minutes before the end of the lesson, she tried to teach us the square root graphs and says "copy it down and go home and complete the exercise for homework". The question is part of my homework.

O... now im curious in how to draw that above graph lol

 

[deswa1]

Re: Math help (Because Carrot made me do this)

O... now im curious in how to draw that above graph lol

Its not hard- add ordinates. Fawun, I'll try and find a website that explains it because I honestly don't know how to explain it online. Absolute worst case if no one can help you get it, I might be able to do one arvo at a library or something if you want it.

 

[Sy123]

Re: Math help (Because Carrot made me do this)

Ohhhhh I get it.

I finally get this!

So how would you add ordinates for this question?



Because Deswa tried to explain it to me yesterday but I don't get it Do you have to add ordinates or is there an alternative way to solve this?

Tbh, yesterday my dad said "I like this tutor. He's very good" just because he tests me lolwat

Addition of Ordinates (and stuff like that) is a 4U topic I think, so I dont think you will ever see it again but they might test you to try and filter out the smarter students from the rest.

Here it goes:



Ok to simplify our explanation we are going to express y as this:



Now to add ordinates, take x=1. Our final function at x=1 which is seen as y, will be the addition of f(x)+g(x) when x=1, that is: f(1)+g(1)
Take x=2, our final function will be the addition of f(2)+g(2) where we sub in x=2 into both functions and add the result to get the resultant y-value (of our final function)

Now as the more practical example take x=2:


That is our resultant y-value.
We havent started how to add ordinates graphically though, so now we begin:

Take y=f(x)+g(x) again, we have to graph y such that it is the addition of these two functions f(x) and g(x), so what we can do is we graph BOTH f(x) and g(x).

If we look at the diagram, what Ive done is I have analysed the graphs in ways such that I add any two y values of these graphs I get our final grey y-value, adding the ordinates means we take the y-values of these two graphs that we will add, and then add them together to get our resultant line. If you can notice, when one of the functions has its x-intercept, such as in the graph f(x) has an x-intercept (that is when y=0) my final graph intersects with g(x).

This is because at the x-intercepts of f(x): y=f(x)+g(x), y=0+g(x), y=g(x) hence the graphs intersect at f(x), this is vice versa when g(x)=0, where y=f(x)
Also: when f(x)=-g(x) y=0
This means that when the two functions have the same value BUT DIFFERENT SIGN then we have an x-intercept for our y.
So all in all, addition of ordinates is what it is, addition of ordinates, we simply add the differnet y-values to get our resultant y-value

In your example, we graph Then we do the process of addition of ordinates.

EDIT: I deserve a billion reps for the amount of time I put into this

 

[deswa1]

Re: Math help (Because Carrot made me do this)

^That is fantastic. Solid effort to type all that up.

 

[Fawun]

Re: Math help (Because Carrot made me do this)

Addition of Ordinates (and stuff like that) is a 4U topic I think, so I dont think you will ever see it again but they might test you to try and filter out the smarter students from the rest.

Here it goes:



Ok to simplify our explanation we are going to express y as this:



Now to add ordinates, take x=1. Our final function at x=1 which is seen as y, will be the addition of f(x)+g(x) when x=1, that is: f(1)+g(1)
Take x=2, our final function will be the addition of f(2)+g(2) where we sub in x=2 into both functions and add the result to get the resultant y-value (of our final function)

Now as the more practical example take x=2:


That is our resultant y-value.
We havent started how to add ordinates graphically though, so now we begin:

Take y=f(x)+g(x) again, we have to graph y such that it is the addition of these two functions f(x) and g(x), so what we can do is we graph BOTH f(x) and g(x).
View attachment 26276
If we look at the diagram, what Ive done is I have analysed the graphs in ways such that I add any two y values of these graphs I get our final grey y-value, adding the ordinates means we take the y-values of these two graphs that we will add, and then add them together to get our resultant line. If you can notice, when one of the functions has its x-intercept, such as in the graph f(x) has an x-intercept (that is when y=0) my final graph intersects with g(x).

This is because at the x-intercepts of f(x): y=f(x)+g(x), y=0+g(x), y=g(x) hence the graphs intersect at f(x), this is vice versa when g(x)=0, where y=f(x)
Also: when f(x)=-g(x) y=0
This means that when the two functions have the same value BUT DIFFERENT SIGN then we have an x-intercept for our y.
So all in all, addition of ordinates is what it is, addition of ordinates, we simply add the differnet y-values to get our resultant y-value

In your example, we graph Then we do the process of addition of ordinates.

EDIT: I deserve a billion reps for the amount of time I put into this

Oh my gosh yes 10^1000000000 reps. Thank you so much for doing this!

1. For the final function part, why did you do x=1 and x=2?
2. For f(x): y=f(x)+g(x), y=0+g(x), y=g(x), why did you put f(x) as 0? isn't it just x only? not the whole f(x)? :s
3. I don't understand any of the bolded parts but that's okay.
4. So if it's just the addition of the y values, can't you just find the y intercept of the two equations by subbing x=0 and just add them?

Once agian, thank you!

 

[twinklegal19]

Re: Math help (Because Carrot made me do this)

Are you finished all your exams now?

almost, only french speaking left, otherwise the rest of this week is free


Also Sy123, thanks for your detailed explanations and graphs Greatly appreciated all the effort you put into this, times like this we need rep back lol

 

[Sy123]

Re: Math help (Because Carrot made me do this)

Oh my gosh yes 10^1000000000 reps. Thank you so much for doing this!

1. For the final function part, why did you do x=1 and x=2?
2. For f(x): y=f(x)+g(x), y=0+g(x), y=g(x), why did you put f(x) as 0? isn't it just x only? not the whole f(x)? :s
3. I don't understand any of the bolded parts but that's okay.
4. So if it's just the addition of the y values, can't you just find the y intercept of the two equations by subbing x=0 and just add them?

Once agian, thank you!

1. It can be any value you like I was using them as examples, so if you want to find the y value of the final function at x=1000
Then we do f(1000)+g(1000), and we see what we get.

2. When f(x)=0, this is the x-intercept its the same thing as saying y=0. Dont get mixed up:
x=0 -> y-intercept
y=0 -> x-intercept

Therefore f(x)=0 is an x-intercept, while f(0) is the y-intercept. So when y=f(x)+g(x), when we have one x-intercept, our final graph intersects with one of the other. So it intersects g(x) when f(x)=0 and vice versa

3. The bold is a bit of a rote-y approach to this, but I think its fine if you dont get it, its just explaining the logic behind the addition

4. Yes you can indeed get the y-intercept of the function by subbing in zero and adding, but. That only gives us one value and only 1 coordinate to work with
(0, f(0)+g(0) )

So we need to do further analysis to graph it. You can graph the two little functions add ordinates graphically to get the final graph

But honestly you need a better tutor.
Also Im kind of enjoying this, since Im getting better at understanding this when Im trying to put it into words for others, (I heard a quote saying you learn the most when you teach idk)

 

[Shadowdude]

Re: Math help (Because Carrot made me do this)

Addition of Ordinates (and stuff like that) is a 4U topic I think, so I dont think you will ever see it again but they might test you to try and filter out the smarter students from the rest.

Here it goes:



Ok to simplify our explanation we are going to express y as this:



Now to add ordinates, take x=1. Our final function at x=1 which is seen as y, will be the addition of f(x)+g(x) when x=1, that is: f(1)+g(1)
Take x=2, our final function will be the addition of f(2)+g(2) where we sub in x=2 into both functions and add the result to get the resultant y-value (of our final function)

Now as the more practical example take x=2:


That is our resultant y-value.
We havent started how to add ordinates graphically though, so now we begin:

Take y=f(x)+g(x) again, we have to graph y such that it is the addition of these two functions f(x) and g(x), so what we can do is we graph BOTH f(x) and g(x).
View attachment 26276
If we look at the diagram, what Ive done is I have analysed the graphs in ways such that I add any two y values of these graphs I get our final grey y-value, adding the ordinates means we take the y-values of these two graphs that we will add, and then add them together to get our resultant line. If you can notice, when one of the functions has its x-intercept, such as in the graph f(x) has an x-intercept (that is when y=0) my final graph intersects with g(x).

This is because at the x-intercepts of f(x): y=f(x)+g(x), y=0+g(x), y=g(x) hence the graphs intersect at f(x), this is vice versa when g(x)=0, where y=f(x)
Also: when f(x)=-g(x) y=0
This means that when the two functions have the same value BUT DIFFERENT SIGN then we have an x-intercept for our y.
So all in all, addition of ordinates is what it is, addition of ordinates, we simply add the differnet y-values to get our resultant y-value

In your example, we graph Then we do the process of addition of ordinates.

EDIT: I deserve a billion reps for the amount of time I put into this

what

how so?

It's like proving the Fundamental Theorem of Algebra the way Gauss did it in his doctoral thesis, nowadays - when you can use Complex Analysis methods and it pops out in like five lines instead of fifty thousand.


just sayin'

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

what

how so?

It's like proving the Fundamental Theorem of Algebra the way Gauss did it in his doctoral thesis, nowadays - when you can use Complex Analysis methods and it pops out in like five lines instead of fifty thousand.


just sayin'

What on Earth are you on about?

 

[Sy123]

Re: Math help (Because Carrot made me do this)

what

how so?

It's like proving the Fundamental Theorem of Algebra the way Gauss did it in his doctoral thesis, nowadays - when you can use Complex Analysis methods and it pops out in like five lines instead of fifty thousand.


just sayin'

Well it was a joke about the rep, if rep was still back, my motivation for posting would be because Im bored, I want to benefit my own understanding by teaching someone else, and I want to help. Rep is as it always should be, a positive side-effect of being what I presume to be helpful.

Looking at wikipedia Fundemental Theroem Of Algebra says:


Where k is a complex number, kx has at least one complex root?