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    I don't understand Q5 on the Jan 2011 Edexcel C1 Paper

    *mark scheme*
    http://www.edexcel.com/migrationdocu...s_20110309.pdf


    If anyone would be so helpful and explain part b in non examiner terms and just talk through the Q and what to do
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    (Original post by Bevin)
    I don't understand Q5 on the Jan 2011 Edexcel C1 Paper

    *mark scheme*
    http://www.edexcel.com/migrationdocu...s_20110309.pdf


    If anyone would be so helpful and explain part b in non examiner terms and just talk through the Q and what to do
    It would help if you gave the question rather t5han just the mark scheme.
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    Show the Question..
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    (Original post by calster102)
    Show the Question..
    (Original post by brianeverit)
    It would help if you gave the question rather t5han just the mark scheme.

    http://www.edexcel.com/migrationdocu...e_20110110.pdf
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    Please help :/ This is the last bit of maths hmw I have for this week
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    Why not just get it wrong and have your teacher talk it through with you so that you understand it? You wont be able to look through the mark scheme in the exam.
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    (Original post by Bevin)
    I don't understand Q5 on the Jan 2011 Edexcel C1 Paper

    *mark scheme*
    http://www.edexcel.com/migrationdocu...s_20110309.pdf


    If anyone would be so helpful and explain part b in non examiner terms and just talk through the Q and what to do
    

'(b) Find the coordinates of the points where the curve with equation $y = f(x-1)$ crosses the coordinate axes.'

\vspace{0.2in}



The question is asking what the coordinates are when the curve crosses the $x$ and $y$-axis. When a function crosses the $x$-axis the $y$-value at that point must be 0, and when a function crosses the $y$-axis the $x$-value at that point must be 0.

\vspace{0.2in}



Assuming you got the answer right in part $(a)$, you could use that to find the point the curve crosses the $x$-axis. The original graph crosses at the origin (0,0), and $f(x-1)$ translated it so it crosses at (1,0).

\vspace{0.2in}



To find where it crosses the $y$-axis you put $f(x-1)=0$, so first you will have to change $f(x) = \frac{x}{x-2}$ (given in the question) into $f(x-1)$. To do this, every time you see an '$x$' in $f(x)$ you need to replace it with '$x-1$'. 

\vspace{0.2in}



so $y = \frac{x-1}{(x-1)-2}$

= $\frac{x-1}{x-3}$

\vspace{0.2in}



And when $x$ is 0 you're left with $\frac{-1}{-3}$ which is the same as $\frac{1}{3}$.
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    (Original post by Monaa123)
    

And when $x$ is 0 you're left with $\frac{-1}{-3}$ which is the same as $\frac{1}{3}$.
    

So your final answer will be (1,0) and (0,$\frac{1}{3}$). Hope that helps!
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    (Original post by Monaa123)
    

'(b) Find the coordinates of the points where the curve with equation $y = f(x-1)$ crosses the coordinate axes.'

\vspace{0.2in}



The question is asking what the coordinates are when the curve crosses the $x$ and $y$-axis. When a function crosses the $x$-axis the $y$-value at that point must be 0, and when a function crosses the $y$-axis the $x$-value at that point must be 0.

\vspace{0.2in}



Assuming you got the answer right in part $(a)$, you could use that to find the point the curve crosses the $x$-axis. The original graph crosses at the origin (0,0), and $f(x-1)$ translated it so it crosses at (1,0).

\vspace{0.2in}



To find where it crosses the $y$-axis you put $f(x-1)=0$, so first you will have to change $f(x) = \frac{x}{x-2}$ (given in the question) into $f(x-1)$. To do this, every time you see an '$x$' in $f(x)$ you need to replace it with '$x-1$'. 

\vspace{0.2in}



so $y = \frac{x-1}{(x-1)-2}$

= $\frac{x-1}{x-3}$

\vspace{0.2in}



And when $x$ is 0 you're left with $\frac{-1}{-3}$ which is the same as $\frac{1}{3}$.

    That makes sense!
    Thank you very much
    I'll give rep to both posts very soon, can't do two in a short space of time
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    (Original post by Bevin)
    That makes sense!
    Thank you very much
    I'll give rep to both posts very soon, can't do two in a short space of time
    No problem
    I don't know what rep is but thanks!
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    (Original post by Monaa123)
    No problem
    I don't know what rep is but thanks!
    Just the thumbs up next to the post
    It's kind of like a "like" on facebook
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    (Original post by Bevin)
    Just the thumbs up next to the post
    It's kind of like a "like" on facebook
    Got it, thanks!
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    (Original post by pryngles)
    Why not just get it wrong and have your teacher talk it through with you so that you understand it? You wont be able to look through the mark scheme in the exam.
    I know I won't, I already did this as a practise mock and was marking it myself and I did get it wrong.
    I was just asking someone to talk this through with me

    Also my teacher isn't very good and we've reported her so the student room is my only hope of passing AS maths
 
 
 
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