Year 13 Maths Help Thread

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    (Original post by kiiten)
    Ok, ill sketch it out - i usually avoid drawing graphs because i normally mess it up and get the question wrong

    So whats the difference between having a domain of R and domain of x E R ?
    You don't have a domain of  x\in \mathbf{R} , that notation means that  \mathbf{R} is the domain, the real numbers. The notation is notation commonly used when you're talking about sets.
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    (Original post by RDKGames)
    Quite simple. You are finding a certain y coordinate that the graph approaches but never reaches as you keep going along the x-axis, so you are basically going off to infinity along the x-axis in order to get the asymptote. From our equation, we can see that as we go off to infinity, the denomintor will get bigger and bigger, so the overall fraction will go to 0.
    So the asymptote will never touch the axis? - you work it the value if it does and then make the range less than that value?

    Sorry this topic has been bothering me for ages
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    (Original post by kiiten)
    So the asymptote will never touch the axis? - you work it the value if it does and then make the range less than that value?

    Sorry this topic has been bothering me for ages
    Sorry but what you've said does not make any sense whatsoever to me.

    An asymptote is a straight line (in this case) that the graph approaches, and this case it is also a line that the graph never crosses. y=2 IS that line as we found it (along with x=3 but that is a different asymptote), because f(x) \not= 2. The range is everything EXCEPT that value of the asymptote which is why it is important to find it.

    In order to work out the asympotote we need to consider what value the function goes to as we pump up x to infinity. Our initial function form does not make this value seem obvious therefore we manipulate it to get something divided by x. As x goes to infinity, those fractions go to 0 and whatever is left make up that value we need, as I have shown with my examples in the long post.
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    (Original post by RDKGames)
    Sorry but what you've said does not make any sense whatsoever to me.

    An asymptote is a straight line (in this case) that the graph approaches, and this case it is also a line that the graph never crosses. y=2 IS that line as we found it (along with x=3 but that is a different asymptote), because f(x) \not= 2. The range is everything EXCEPT that value of the asymptote which is why it is important to find it.

    In order to work out the asympotote we need to consider what value the function goes to as we pump up x to infinity. Our initial function form does not make this value seem obvious therefore we manipulate it to get something divided by x. As x goes to infinity, those fractions go to 0 and whatever is left make up that value we need, as I have shown with my examples in the long post.
    Ye you lost me here. Sorry if i wasted your time but i think i need someone to go through this with me irl - ill ask my teacher but thanks anyway
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    (Original post by kiiten)
    Ye you lost me here. Sorry if i wasted your time but i think i need someone to go through this with me irl - ill ask my teacher but thanks anyway
    Do you understand why the fractions go to 0 as x goes to infinity? If you understand this, then you should understand how we are left with the asymptote value.
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    (Original post by RDKGames)
    Do you understand why the fractions go to 0 as x goes to infinity? If you understand this, then you should understand how we are left with the asymptote value.
    not really - i thought that it would go to 0 when x goes towards 3
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    (Original post by kiiten)
    not really - i thought that it would go to 0 when x goes towards 3
    But x isn't going towards 3?? x is going towards infinity. The 3 is something we consider with the VERTICAL asympotote and we already established this previously. Now we are dealing with the horizontal asymptote so the 3 has nothing to do with it. If x was to approach 3 then the function's value would shoot to infinity because x=3 is an asymptote.

    1/1=1
    1/10=0.1
    1/100=0.01
    1/1000=0.001

    and so on. As the denominator goes to infinity, the fraction goes to 0. The numerator can be anything, the fraction will still go to 0, which is what we apply to the question. Simple observation you should be aware of.
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    (Original post by RDKGames)
    But x isn't going towards 3?? x is going towards infinity. The 3 is something we consider with the VERTICAL asympotote and we already established this previously. Now we are dealing with the horizontal asymptote so the 3 has nothing to do with it. If x was to approach 3 then the function's value would shoot to infinity because x=3 is an asymptote.

    1/1=1
    1/10=0.1
    1/100=0.01
    1/1000=0.001

    and so on. As the denominator goes to infinity, the fraction goes to 0. The numerator can be anything, the fraction will still go to 0, which is what we apply to the question. Simple observation you should be aware of.
    Ohh yea i understand that. I just need to go over considering the horizontal and vertical asymptotes separately again
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    (Original post by kiiten)
    Ohh yea i understand that. I just need to go over considering the horizontal and vertical asymptotes separately again
    Okay, just ask your teacher concerning the two then. I don't think I can explain it any simpler than that, sorry. You can also see it from the graph that for the horizontal asympotote you must go off to infinity along the x-axis in order to approach it, so it makes sense from that perspective.
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    (Original post by RDKGames)
    Okay, just ask your teacher concerning the two then. I don't think I can explain it any simpler than that, sorry. You can also see it from the graph that for the horizontal asympotote you must go off to infinity along the x-axis in order to approach it, so it makes sense from that perspective.
    I appreciate your help anyway - thanks
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    Despite bumping it numerous times, the Year 12 Maths Thread is still dead so far. What can I do to revive it?
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    (Original post by Palette)
    Despite bumping it numerous times, the Year 12 Maths Thread is still dead so far. What can I do to revive it?
    Leave it dead. Y12s clearly don't need it
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    (Original post by Palette)
    Despite bumping it numerous times, the Year 12 Maths Thread is still dead so far. What can I do to revive it?
    Skip to may, then they ask what y=mx+c is before the C1 exam


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    (Original post by physicsmaths)
    Skip to may, then they ask what y=mx+c is before the C1 exam


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    10 minutes before a uni exam for a module called ODEs and Control, someone jokingly asked 'what is an ODE and how do I control one?'. I cracked up.

    Jellyxx are you still stuck?
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    (Original post by SeanFM)
    10 minutes before a uni exam for a module called ODEs and Control, someone jokingly asked 'what is an ODE and how do I control one?'. I cracked up.

    Jellyxx are you still stuck?
    No, but thank you for inquiring!

    Edit: Probably won't be the last of me you will see. I'll come back soon with another question
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    (Original post by RDKGames)
    Leave it dead. Y12s clearly don't need it
    What explains why the Year 12 thread has died while the Year 13 thread is thriving?
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    (Original post by Palette)
    What explains why the Year 12 thread has died while the Year 13 thread is thriving?
    Year 12s became yr 13s lol


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    (Original post by Palette)
    What explains why the Year 12 thread has died while the Year 13 thread is thriving?
    Because Y13 maths is harder? Dunno.
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    (Original post by SeanFM)
    10 minutes before a uni exam for a module called ODEs and Control, someone jokingly asked 'what is an ODE and how do I control one?'. I cracked up.

    Jellyxx are you still stuck?
    Lol.
    Hahahahaha


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    Why were fp3 edexcel grade boundaries so much higher than last year and do people reckon they will remain higher or is it common for one year a papers boundaries just shoot up and back down
    Also nhow does fp3s difficulty compare to s3 m2 fp2
 
 
 
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