Math help (1 Viewer)

[Shadowdude]

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

Graph




so basically the first thing we want to do is find where this graph is undefined

convince yourself that these are when x = 3, and when x = -3

so we draw those dotted lines on our graph

so we have three sections of our graph. x < -3, -3 < x < 3 and x > 3.

now let's look at some points. when x = 0, y = 0, so we plot (0, 0) on our graph

Now we want to observe the behaviour near those undefined points

so let's use our calculator and we do the same thing

plot the following points:

y at x = -100
y at x = -3.00001
y at x = -2.99999
y at x = 2.99999
y at x = 3.00001
y at x = 100

And then we join all these points with a nice curvy line

And you should get something as here:

http://www.wolframalpha.com/input/?i=y+%3D+x^2+%2F+%28x^2+-+9%29

alternatively, google: "y = (x^2)/(x^2-9)"


---

and before anyone goes "m8, you said teach this in person" - she asked nicely so... yeah.

 

[Fawun]

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

Graph




so basically the first thing we want to do is find where this graph is undefined

convince yourself that these are when x = 3, and when x = -3

so we draw those dotted lines on our graph

so we have three sections of our graph. x < -3, -3 < x < 3 and x > 3.

now let's look at some points. when x = 0, y = 0, so we plot (0, 0) on our graph

Now we want to observe the behaviour near those undefined points

so let's use our calculator and we do the same thing

plot the following points:

y at x = -100
y at x = -3.00001
y at x = -2.99999
y at x = 2.99999
y at x = 3.00001
y at x = 100

And then we join all these points with a nice curvy line

And you should get something as here:

http://www.wolframalpha.com/input/?i=y+%3D+x^2+%2F+%28x^2+-+9%29

alternatively, google: "y = (x^2)/(x^2-9)"


---

and before anyone goes "m8, you said teach this in person" - she asked nicely so... yeah.

Wait. Where did you get those numbers from? and how am I suppose to graph "y at x = -100"? also when you say "y at x" does that mean that it's on the x or y axis?

 

[RealiseNothing]

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

Wait. Where did you get those numbers from? and how am I suppose to graph "y at x = -100"? also when you say "y at x" does that mean that it's on the x or y axis?

"y at x = -100"

what is y when x = -100 is what he's saying.

 

[Shadowdude]

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

Wait. Where did you get those numbers from? and how am I suppose to graph "y at x = -100"? also when you say "y at x" does that mean that it's on the x or y axis?

x at +/- 100 is the 'limit' (technically not, but for your purposes - good enough)

and the other ones are to examine the behaviour near where the function isn't defined. try x = 3 and see you can't get anything.

 

[Sy123]

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

Hi.

I need help in graphing again (Yes I know what you're thinking).

How would you sketch this:



All I have so far are the intercepts for which when you sub x=0 and y=0, my intercepts for x and y are both 0.
I also know that it's a hyperbola because it has an x as the denominator.

:s

Lets think about our curve for a second, and understand that it is one number divided by another number.
We know that For all positive x^2 (which is always positive) .

But our concept of infinity tells us, that the closer and closer we go to infinity, the less and less the constants matter. That is, the less the -9 matters. What I mean by this, is that the closer and closer we get to infinity, the closer and closer the denominator will equal the numerator.

This means that at positive infinity, we approach y=1, however we approach y=1 from the top. This is because we dont actually have a number infinity, but we approach it, so that no matter how high of a number we go But this is only true for x^2-9>0 (inequality logic). So we approach y=1 at positive infinity...

If you dont care about why asymptotes are asymptotes, then just follow this rule:



IF m=n, so that the degree of the dividing functions are the same: Like in your case:



n=m here since the highest degrees are the SAME.

This means our horizontal asymptote is

Remember a and b are our co-efficients of the HIGHEST degrees. In our above case, a=1, b=1. Hence our horizontal asymptote is y=1/1=1

IF n<m

Such as

Our horizontal asymptote is always y=0

This is asymptote logic.

Now onto sketching graphs if you can figure out horizontal asymptote

We have our horizontal asymptote, now we need to find our vertical asymptote. Our veritcal asymptotes are when the denominator (the bottom) is equal to 0

So our vertical asymptote(s) is
So we find that we have 2 vertical asymptotes. Now lets sketch all three of these asymptotes on our imaginary graph, and plot our point of intersection with the axes (0,0)

We are nearly done, now if you noticed, our function f(x) is an even one, since f(x)=f(-x)
What this means is that our function is symmetrical about the y-axis, this is a property.

This will help us in sketching it, since we can imagine it, as if we fold the paper along the y-axis, our graphs would be touching each other exactly (symetrical).
Lets now sub in very large numbers into our function

Lets sub in our calculator:

x=100 -> You should get a number very close to 1, BUT above 1 i.e. 1.00000834382 etc. (Note, for some functions if you sub in very large numbers, the calculator will just approximate it to 1)

Now, we need to find the behaviour of our curve around the vertical asymptotes.
So we need to sub in a number very close to 3.

x=3.0001 -> y= large number, so what this tells us, is that our function approaches POSTIVE infinity AT the bigger side of x=3.

Lets sub in x=2.999 in order to find the behaviour of the curve on the otherside of it

x=2.999 -> y= very big NEGATIVE number, this tells us that the function approaches NEGATIVE infinity at the smaller side of x=3

Now, there are TWO routes you can go from here.
The smarter route:

Notice that it is an even function, so we have symetry about the y=axis, so we can just graph the first half of the graph, then replicate it on the other side.

The safety route:

Lets try the other asymptote x=-3, and test numbers on either side of them:

x=-2.999 -> y=Very big negative number, this tells us it approaches negative infinity on the bigger side of x=-3 asymptote (remember -2.99 > -3)
x=-3.001 -> y=very large postive number, tells us that approaches positive infinity on the smaller side of x=-3

Now lets test negative infinity:

x=-100 ->y= a number VERY close to 1, but bigger than one. This tells us that it approaches our horizontal asymptote.

Behaviour checking is done. Now lets figure out shape.
To be able to understand this, look at the answer graph, or go on Wolfram Alpha and look at the graph from there.
The very left and very right of the graphs are easy to graph, we know that they go from one asymptote to the other asymptote.

Now the very middle, in between the vertical asymptotes is the tricky bit.
We know that our curve needs to go from one asymptote to the other, but through (0,0). So there has to be a way for this to happen, they must connect somehow. So it will have a point in which the curve turns, and it turns into the other asymptote.

We are done graphing.

tl;dr
Here is a 'small guide to graphing'

1) Find vertical asymptotes (denominator=0)
2) Find horizontal asymptote (manually put in large values on calculator, or figure it out mentally using formula or logic)
3) Find behaviour of graphs near the asymptotes (sub in large values, close values etc.)
4) Find intercepts

5) Graph it.


This post turned out longer than I thought.

 

[theind1996]

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

Excellent work Sy123.

I am seriously admiring the effort you put into every maths problem.

I'm sure that you would make a fantastic teacher/tutor.

 

[RivalryofTroll]

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

Excellent work Sy123.

I am seriously admiring the effort you put into every maths problem.

I'm sure that you would make a fantastic teacher/tutor.

+1

Though, it makes me feel that I'm not a potential tutor at all :/

 

[twinklegal19]

Excellent work Sy123.

I am seriously admiring the effort you put into every maths problem.

I'm sure that you would make a fantastic teacher/tutor.

This

 

[nerdasdasd]

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

Excellent work Sy123.

I am seriously admiring the effort you put into every maths problem.

I'm sure that you would make a fantastic teacher/tutor.

+1

 

[Fawun]

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

Okay first of all, thank you for taking the time to help me once again! Would give a million reps if I could but here is 50^238427938742983472 hypothetical reps

This means that at positive infinity, we approach y=1, however we approach y=1

Just to clarify, when you say positive infinity, is it anything ABOVE the x axis? and where did you get the number 1 from? or did you just choose a random number?

If you dont care about why asymptotes are asymptotes, then just follow this rule:

Wow am I suppose to memorise that rule? I don't even get that rule :s and is m ALWAYS equal to n? or is it just this case?

Okay I understand the asymptote part (yay!)

Lets now sub in very large numbers into our function

May I ask why we can't sub in small numbers?

So we need to sub in a number very close to 3.

Why does it have to be 3? Why can't it be large numbers like before?

x=3.0001 -> y= large number, so what this tells us, is that our function approaches POSTIVE infinity AT the bigger side of x=3.

What's the bigger side? the left or right of the asymptote of 3 or?

x=2.999 -> y= very big NEGATIVE number, this tells us that the function approaches NEGATIVE infinity at the smaller side of x=3

What is this negative infinity? like what do you mean it approaches negative infinity? like I get that you can have positive and negative infinity but like how do you know like where it is to approach it if you understand what I mean. Like is negative infinity the line that just goes on and on on the x or the y axis or?

Now, there are TWO routes you can go from here.

Which route would you take? The smarter or the safer?


The safety route:

Lets try the other asymptote x=-3, and test numbers on either side of them:

x=-2.999 -> y=Very big negative number, this tells us it approaches negative infinity on the bigger side of x=-3 asymptote (remember -2.99 > -3)
x=-3.001 -> y=very large postive number, tells us that approaches positive infinity on the smaller side of x=-3

Once again, what's the 'smaller' and the 'bigger' side? and how do you know?

Now lets test negative infinity:

x=-100 ->y= a number VERY close to 1, but bigger than one. This tells us that it approaches our horizontal asymptote.

If our horizontal asymptote is 1, and when x=-100, y= a number close to 1 BUT BIGGER than 1, doesn't it go through or past the asymptote?

To be able to understand this, look at the answer graph, or go on Wolfram Alpha and look at the graph from there.

What happens when were in like an exam and we obviously don't have the answer graph etc?

We know that our curve needs to go from one asymptote to the other, but through (0,0).

Wait where did you get (0,0) from? or was that because of our intercepts where both x and y are equal to 0?

One last question, once you graph it, do you need to label anything on the graphs besides the asymptotes?

Once again, thanks!

EDIT: Wow it took me 30 minutes to read, try to comprehend and break this down lol

 

[Sy123]

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

I didn't really choose y=1. Its just the fact that, the higher you go, the closer our function approaches y=1. Try it yourself, in your calculator, sub in x=5, x=10, x=50, x=100. You will see that the higher we go, the closer we get to y=1. The reason this is, was explained. Its because the top part is pretty much the same as the bottom, except the bottom has a -9. At infinity, our constants like that -9, dont matter, and we see why when you sub in x=5, 10, 50 and 100. (remember these are just examples to help you understand the asymptote concept, you DO NOT need to do this)


NEVER EVER, memorise what I wrote, its just a way of being able to quickly find the horizontal asympote. N is only equal to m in that case, NOT always, for example:



In the above example function, n=m (they equal 5). So our horizontal asymptote is y=4/2 (the co-efficients a and b)
So when you are graphing things like this, you can just quickly find the horizontal asymptote without having to sub in large values. (We can find the horizontal asymptote by just looking at it due to our 'rule'. If n < m, such as






If you can see, n < m, that means our asymptote is y=0. (The reason this is, is simply because the denominator will always be much greater than the numerator, but dont worry about why right now)


When we sub in large numbers, what we are doing is, is that we are trying to find how the curve is, on the very ends of the graph. Take y=x^2 for instance, when x is very large, so is y. This means that our graph will go out to both infinites. In this case however, when x is very large, our y approaches 1 from the top side of it. (it is the top side since y is always greater than 1 as x approaches infinity, when you put it into your calculator, you never ever will get y=0.9999, but always y=1.0001...). The other reason to sub in large numbers is to manually find asymptotes. Because that is what the asymptote is defined as.



Here, we are going to try and find the behaviour of the curve around x=3. Why x=3? Because it is a vertical asymptote, and we know that since it is an asymptote, we are going to approach it somehow and we need to know how it is going to approach x=3. Remember, every time we sub in values we are actually trying to find the behaviour of the curve at that point, how the curve will look like.



Bigger always means right of the asymptote, same thing for 3 and -3. -2.999 is bigger than -3.001. 3.001 is bigger than 2.999. BUT BUT BUT, Im saying bigger in a way that you understand what it means. Bigger can be subsituted by larger. The larger the value the more right it is



Negative infinity is a very very large 'number' (pains me to say that, because its not really a number) BUT it is a number that is very negative. So like if we approach negative infinity, we are going from
x=-10, x=-100, x=-1000 . The higher negative we go, the closer we get to negative infinity, and like positive infinity, we need to test the behaviour of the curve



I like taking the speedy route, so the smarter one. Once you master graphing, you should be doing the smarter route and looking for these little tricks. But for now, safer is the best way


As clarified above, bigger is the larger value, its always on the right. So 3.001 is bigger than 2.999, 2.999 is smaller than 3.001 etc.


It is possible for graphs to pass horizontal asymptote (but it can never ever, ever ever pass through a vertical asymptote). When I say a number close to one and bigger than 1, I mean stuff like 1.0001, 1.000001. Its close to one, but it never IS 1. Never think of graphs physically, in mathematics, there are things called infinity. Our graph will never pass through 1 because IF y=1



So as you can see, our graph can never be at 1, because if it is, then we get results like above. You might want to clarify what you are saying here as well, since I dont think I quite got what you are asking here



Since this is over text and the internet, Im trying to explain how to graph is, and WHY it is through the text, and I imagine you having the graph with you while reading this in order for you to understand what Im saying. By exam time, theoretically you should of mastered this


0,0 is the point you found, the intercepts.


Label asymptote, intercepts, the x and y-axis and put arrows on the ends of your lines, and write y=f(x) next to your curve for good measure =)



Answers to your questions in order, Im sorry the layout is like this, I tried replying under your question in bold but the text editor kept bugging out and was bolding the wrong text and making some text missing

 

[Fawun]

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

Okay after reading this whole post again, I have some questions

We are nearly done, now if you noticed, our function f(x) is an even one, since f(x)=f(-x)
What this means is that our function is symmetrical about the y-axis, this is a property.

How is it an even function?





Unless i'm doing something wrong since f(x)=f(-x)? Why is it symmetrical about the y-axis and not the x-axis?

x=3.0001 -> y= large number, so what this tells us, is that our function approaches POSTIVE infinity AT the bigger side of x=3.

So when you say approaches positive infinity, you mean this right?

x=2.999 -> y= very big NEGATIVE number, this tells us that the function approaches NEGATIVE infinity at the smaller side of x=3

So it's like this? Then how come we don't sketch/draw it?

When we sub in large numbers, what we are doing is, is that we are trying to find how the curve is, on the very ends of the graph. Take y=x^2 for instance, when x is very large, so is y. This means that our graph will go out to both infinites.

Both infinities? As in both negative and positive infinities? Wouldn't it be a cubic then since it goes both up and down?

The other reason to sub in large numbers is to manually find asymptotes. Because that is what the asymptote is defined as.

Can't we just use the asymptote logic like making the denominator = to 0 to find the vertical asymptote and the coefficients of the highest degree for the horizontal asymptote instead of subbing in large numbers?

So how would you know what the graph looks like? I know by testing numbers but how does that indicate if it's a parabola, hyperbola etc?

Like how does knowing that the function approaches positive infinity on the bigger side of 3, how the function approaches negative infinity on the smaller side of 3 etc, graph this:

I understand how the parabola touches (0,0) since (0,0) are the intercepts but I don't understand how the hyperbolas came into this. Is it because of the x's in the denominator of the function that we know it's a hyperbola? Why is the parabola concave down when the x^2 in the equation is positive?

One last question, how does limits come into this?

 

[Sy123]

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

Okay after reading this whole post again, I have some questions



How is it an even function?





Unless i'm doing something wrong since f(x)=f(-x)? Why is it symmetrical about the y-axis and not the x-axis?

You have made an algebraic error:



It is symetrical about the y-axis because it is even.
The reason why even functions ARE symmetrical about the y-axis, is something you dont REALLY need to know, but it is good to know and understand the concept behind it.

Take any function, we will make this function symmetrical about the y-axis, what this means is that, at a certain x-value, say x=a. The y-value at x=a is f(a). Now if its symetical about the x-axis. When we go 'a' units to the right and we get f(a), when we go 'a' units to the left we should ALSO get f(a).
So at x=a AND x=-a, we get the same y=f(a).

Look at that graph you posted. Look at x=4. We get something around 2 from what we see as our y-value. But if you look at x=-4, you ALSO get something around y=2 (whatever this value may be they are the SAME for x=4 and x=-4)

Hence the function is the SAME for negative x and postitive x. Hence f(x)=f(-x).

So when you say approaches positive infinity, you mean this right?

Absolutely

So it's like this? Then how come we don't sketch/draw it?

This is also correct, that is the DIRECTION that we get, however we know that it is below the x-axis because when we tested these x-values we get NEGATIVE y-values. Your direction is correct. But I will get to this point at the end, of how we are going to assume things and assume certain behaviours.

Both infinities? As in both negative and positive infinities? Wouldn't it be a cubic then since it goes both up and down?

When I say both infinities, in that context Im talking about the x-values of the infinities. When I say, go out to infinity, I mean that the x goes out to infinity, the y keeps approaching 1, closer and closer (for both sides). You might want to reclarify what you mean by this, I may have misunderstood what you are trying to say

Can't we just use the asymptote logic like making the denominator = to 0 to find the vertical asymptote and the coefficients of the highest degree for the horizontal asymptote instead of subbing in large numbers?

When we are subbing in large numbers with the intention of finding an asymptote. We are always looking for horizontal asymptotes, vertical asymptotes cannot be found by subbing in values, but by making the denominator equal to 0. You can definitely use the co-efficients of the highest degrees to quickly find them, Im just saying that you can manually find them to check if you are right, or as an alternative way.

So how would you know what the graph looks like? I know by testing numbers but how does that indicate if it's a parabola, hyperbola etc?

Like how does knowing that the function approaches positive infinity on the bigger side of 3, how the function approaches negative infinity on the smaller side of 3 etc, graph this:

I understand how the parabola touches (0,0) since (0,0) are the intercepts but I don't understand how the hyperbolas came into this. Is it because of the x's in the denominator of the function that we know it's a hyperbola? Why is the parabola concave down when the x^2 in the equation is positive?

Okay this is what I am going to explain, what I was referring to you saying about that negative infinity near 2.99 thing.
Lets take a look at everything that we have found out about this graph so far, I will show you how I approach the final step of graphing, which is to predict how the curve will look like:

First we put all of the information we know mentally on the paper after drawing asymptotes and intercepts.

The Green dot is our intercept, we know the curve passes through there, we have our asymptotes in red dotted lines, and we have blue lines to indicate the direction of the curve, which we got by substiuting all those big/close values.

Assumption Number 1: Nothing crazy really happens when we dont know calculus. Without knowing how to differentiate and find turning points etc. We cant really know if anything crazy is going to happen like if it has a weird hump in the curve or something.

Given this assumption, look at the very right of the graph, we have two arrows pointing that way, there is no intercepts to go through, there isnt anything significant there except that it has two asymptotes the horizontal y=1, and the vertical x=3. Lets draw a curve that satisfies our directions. Done, simple. Same thing for the very left of the graph.

The tricky bit is in the middle. How do we know that the graph turns like that? Well look at it this way, we have two arrows going down on our graph and we have 1 point to pass through which is (0,0). Somehow the curve must turn so that x=2.9999... and x=-2.9999.... go towards negative y-infinity. So we can just sketch our graph knowing this, knowing that there shouldnt be anything weird happening, that we just go from one asymptote to the other turning at x=0.
How do we know its turning point is (0,0)? Because its an even function, because its an even function it must be symetrical, if the curve turns at a place other than x=0, then it must replicate this turn on the other side. But we assume that nothing crazy like this happens.

How do we know that it is concave down? Well for the above reason, that we must go to negative infinity AT BOTH x=2.999 and x=-2.999...
Why is it negative?



In that domain the denominator is negative. I dont know really how to explain this, but its because 3^2=9, so anything lower than x=3 must be negative because we have x=2, we square it, this result should always be less than 9, so when we minus it from 9 we get negative. Because negative^2 is positive, anything 'smaller' than x=-3, such x=-4, will make the denominator positive.

Hence we know that the 'parabola' is negative because in that part of x. The denominator is negative (AND the numerator is always positive), positive divide negative=negative.

One last question, how does limits come into this?

If you know what limits are it makes explaining this a whole lot easier, because limits have EVERYTHING to do with this

 

[Fawun]

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

Take any function, we will make this function symmetrical about the y-axis, what this means is that, at a certain x-value, say x=a. The y-value at x=a is f(a). Now if its symetical about the x-axis. When we go 'a' units to the right and we get f(a), when we go 'a' units to the left we should ALSO get f(a).
So at x=a AND x=-a, we get the same y=f(a).

So what you're saying is, is that despite whether it's negative or positive x, we will ALWAYS have the same y-value only when it's symmetrical (or an even function) yes?

Oh my gosh thank you for the diagram and the explanation! It helped a lot =)

If you know what limits are it makes explaining this a whole lot easier, because limits have EVERYTHING to do with this

Yes I have just started learning limits. The basic limits such as questions like this:



I can do and I can also do this:



That's it for limits that I have learned so far. How does limits play a role in this?

 

[Sy123]

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

So what you're saying is, is that despite whether it's negative or positive x, we will ALWAYS have the same y-value only when it's symmetrical (or an even function) yes?

Oh my gosh thank you for the diagram and the explanation! It helped a lot =)

When I say negative/positive x, I mean numbers of the same magnitude but different sign. Like x=1, x=-1 will have same y-value. x=4.5 x=-4.5, same y-value etc.
IF this is true for all values of x, then the function is symetrical and from it we can derive that f(x)=f(-x). Since what that is say is.

f(x)=f(-x)
-> Substitute any value of x in, you will get the same value for it if you substitute the negative/positive version of that number.

Yes I have just started learning limits. The basic limits such as questions like this:



I can do and I can also do this:



That's it for limits that I have learned so far. How does limits play a role in this?

Yes but do you know what the limit actually means?

It plays a role because when I 'sub in very large values'
Im actually doing this:



When you learn all of what limits actually mean and are, graphing becomes a lot easier to understand.

 

[Fawun]

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

Yes but do you know what the limit actually means?

It plays a role because when I 'sub in very large values'
Im actually doing this:



When you learn all of what limits actually mean and are, graphing becomes a lot easier to understand.

Nope. My tutor just writes it on the board and tells us to copy it down without really explaining what it means. I'm trying not to r0te learn all of this stuff hence why i'm asking so many questions. I just remember how to do it and as of now, I have no idea what limits are, and what they have to do with graphing. I just know how to do the work. I'm TRYING to stop r0te learning and TRYING to start understanding the concepts etc.

 

[Sy123]

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

Nope. My tutor just writes it on the board and tells us to copy it down without really explaining what it means. I'm trying not to r0te learn all of this stuff hence why i'm asking so many questions. I just remember how to do it and as of now, I have no idea what limits are, and what they have to do with graphing. I just know how to do the work. I'm TRYING to stop r0te learning and TRYING to start understanding the concepts etc.

Ill have a good explanation on what limits at night maybe, for now I need to study for English yearly tomorrow. Maybe ask another person on Bored of Studies in what limits are in the meantime? (because it can be a hard thing to grasp at first)

 

[Fawun]

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

Ill have a good explanation on what limits at night maybe, for now I need to study for English yearly tomorrow. Maybe ask another person on Bored of Studies in what limits are in the meantime? (because it can be a hard thing to grasp at first)

Okay good luck on your yearly

 

[asianese]

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

With divide top and bottom by x^2 to get

Clearly the 9/x^2 term disappears and you are left with 1.

 

[Fawun]

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

With divide top and bottom by x^2 to get

Clearly the 9/x^2 term disappears and you are left with 1.

Wait is that to find the asymptote?