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Help with graphs please? (Maths C1) Watch

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    How do I go about drawing this graph: y = (x+1) over (x+2)?
    I have seen what the graph looks like when drawn, but I didn't quite understand what you have to do.

    Any help would be greatly appreciated!
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    Do you know how to sketch the curve y=1/x ?
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    (Original post by B_9710)
    Do you know how to sketch the curve y=1/x ?
    Yes, but I don't understand why you have to translate the graph by -2 on the x-axis and 1 on the y axis.
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    (Original post by honour)
    Yes, but I don't understand why you have to translate the graph by -2 on the x-axis and 1 on the y axis.
    The curve  \displaystyle y=\frac{x+1}{x+2}\equiv 1-\frac{1}{x+2} . So you have to think about the transformations that map  y=1/x to  y=1-\frac{1}{x+2} .
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    For question with x on top and bottom you should use polynomial division to get something along the lines of 'a + b/(x+c)'.
    Another way I saw on here was 'adding zero', so you replace the top part by x+1 +2 -2 and then you can split the fraction into (x+2)/(x+2) and something else, giving you 1 +... which you would know how to graph, either method works but the first one is more reliable when it comes to more complicated functions.
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    (Original post by honour)
    How do I go about drawing this graph: y = (x+1) over (x+2)?
    I have seen what the graph looks like when drawn, but I didn't quite understand what you have to do.

    Any help would be greatly appreciated!
    Not sure if you are still needing help with this but...

    Like what was said above, You need to put...  y=\frac{x+1}{x+2}... into a form that you can understand. To do this you will need to know how to do "long" algebraic division in which you will divide x+1 by x+2.

    Doing so will mean the equation...
     y=\frac{x+1}{x+2}... will be re-arranged to look like....  y= 1 - \frac{1}{x+2}

    Now the equation is easier to look for any transformations that have occurred.

    So, if we let...
    f(x) = \frac{1}{x} ...which is a basic reciprocal graph that you should know the layout of.

    Now looking at the equation that we re-arranged we can easily see that  -f(x+2)+1 = 1 - \frac{1}{x+2}

    so from -f(x+2)+1 we know that the following transformations have taken place:
    1. The graph has been reflected in the x-axis
    2. A transformation of -2 (to the left) parallel to the x-axis
    3. A transformation of +1 (up) parallel to the y-axis
    I hope this clears things up
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    (Original post by Philip-flop)
    Not sure if you are still needing help with this but...

    Like what was said above, You need to put...  y=\frac{x+1}{x+2}... into a form that you can understand. To do this you will need to know how to do "long" algebraic division in which you will divide x+1 by x+2.

    Doing so will mean the equation...
     y=\frac{x+1}{x+2}... will be re-arranged to look like....  y= 1 - \frac{1}{x+2}

    Now the equation is easier to look for any transformations that have occurred.

    So, if we let...
    f(x) = \frac{1}{x} ...which is a basic reciprocal graph that you should know the layout of.

    Now looking at the equation that we re-arranged we can easily see that  -f(x+2)+1 = 1 - \frac{1}{x+2}

    so from -f(x+2)+1 we know that the following transformations have taken place:
    1. The graph has been reflected in the x-axis
    2. A transformation of -2 (to the left) parallel to the x-axis
    3. A transformation of +1 (up) parallel to the y-axis
    I hope this clears things up
    nice
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    (Original post by B_9710)
    The curve  \displaystyle y=\frac{x+1}{x+2}\equiv 1-\frac{1}{x+2} . So you have to think about the transformations that map  y=1/x to  y=1-\frac{1}{x+2} .
    Thank you dude! :^_^:
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    (Original post by Philip-flop)
    Not sure if you are still needing help with this but...

    Like what was said above, You need to put...  y=\frac{x+1}{x+2}... into a form that you can understand. To do this you will need to know how to do "long" algebraic division in which you will divide x+1 by x+2.

    Doing so will mean the equation...
     y=\frac{x+1}{x+2}... will be re-arranged to look like....  y= 1 - \frac{1}{x+2}

    Now the equation is easier to look for any transformations that have occurred.

    So, if we let...
    f(x) = \frac{1}{x} ...which is a basic reciprocal graph that you should know the layout of.

    Now looking at the equation that we re-arranged we can easily see that  -f(x+2)+1 = 1 - \frac{1}{x+2}

    so from -f(x+2)+1 we know that the following transformations have taken place:
    1. The graph has been reflected in the x-axis
    2. A transformation of -2 (to the left) parallel to the x-axis
    3. A transformation of +1 (up) parallel to the y-axis
    I hope this clears things up
    Thanks Philip, you've made this so much clearer! Can I ask how you got from
     y= 1 - \frac{1}{x+2} to
     -f(x+2)+1?
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    (Original post by honour)
    Thanks Philip, you've made this so much clearer! Can I ask how you got from
     y= 1 - \frac{1}{x+2} to
     -f(x+2)+1?
    f(x)=\frac{1}{x}

    \Rightarrow f(x+2)=\frac{1}{x+2}

    \Rightarrow -f(x+2)=-\frac{1}{x+2}

    \Rightarrow -f(x+2)+1=-\frac{1}{x+2}+1
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    (Original post by RDKGames)
    f(x)=\frac{1}{x}

    \Rightarrow f(x+2)=\frac{1}{x+2}

    \Rightarrow -f(x+2)=-\frac{1}{x+2}

    \Rightarrow -f(x+2)+1=-\frac{1}{x+2}+1
    Thank you RDKGames, you're fantastic! :^_^:
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    (Original post by honour)
    Thanks Philip, you've made this so much clearer! Can I ask how you got from
     y= 1 - \frac{1}{x+2} to
     -f(x+2)+1?
    Exactly the same process as what RDKGames has said...

    f(x)=\frac{1}{x}

    \Rightarrow f(x+2)=\frac{1}{x+2}

    \Rightarrow -f(x+2)=-\frac{1}{x+2}

    \Rightarrow -f(x+2)+1=-\frac{1}{x+2}+1
 
 
 
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