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    I can't figure out what to do here:
    Given that f(x)=1/x , x is not equal to 0
    (a) Sketch the graph of y = f (x) + 3 and state the equations of the asymptotes
    (b) Find the coordinates of the point where y = f (x) + 3 crosses a coordinate axis
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    (Original post by Alpha°)
    I can't figure out what to do here:
    Given that f(x)=1/x , x is not equal to 0
    (a) Sketch the graph of y = f (x) + 3 and state the equations of the asymptotes
    (b) Find the coordinates of the point where y = f (x) + 3 crosses a coordinate axis
    I hope someone knows how to do this because i would also like to know how this is done.
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    (Original post by Alpha°)
    I can't figure out what to do here:
    Given that f(x)=1/x , x is not equal to 0
    (a) Sketch the graph of y = f (x) + 3 and state the equations of the asymptotes
    (b) Find the coordinates of the point where y = f (x) + 3 crosses a coordinate axis
    Watch the videos under graph transformations and asymptotes

    http://www.examsolutions.net/a-level.../c1-tutorials/

    then revisit this question
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    Basically it's just 1/x moved up by 3. This means the asymptote at x=0 doesn't change and the asymptote at y=0 moves up by 3 to y=3.

    Then for B from the sketch you will see it crosses the x axis once so solve for x in 0=(1/x) + 3 since y is 0 when it crosses x axis
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    (Original post by Alpha°)
    I can't figure out what to do here:
    Given that f(x)=1/x , x is not equal to 0
    (a) Sketch the graph of y = f (x) + 3 and state the equations of the asymptotes
    (b) Find the coordinates of the point where y = f (x) + 3 crosses a coordinate axis
    (Original post by Vinni0001)
    I hope someone knows how to do this because i would also like to know how this is done.
    Find out what the graph of 1/x looks like. (NOTE: you can either substitute values into it to calculate coordinates, or you can cheat and use the internet or a graphical calculator.) Transform is by the column vector
    0
    3

    NOTE: there should be an asymptote
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    (Original post by metrize)
    Basically it's just 1/x moved up by 3. This means the asymptote at x=0 doesn't change and the asymptote at y=0 moves up by 3 to y=3.

    Then for B from the sketch you will see it crosses the x axis once so solve for x in 0=(1/x) + 3 since y is 0 when it crosses x axis
    thanks
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    (Original post by 04MR17)
    Find out what the graph of 1/x looks like. (NOTE: you can either substitute values into it to calculate coordinates, or you can cheat and use the internet or a graphical calculator.) Transform is by the column vector
    0
    3

    NOTE: there should be an asymptote
    (Original post by metrize)
    Basically it's just 1/x moved up by 3. This means the asymptote at x=0 doesn't change and the asymptote at y=0 moves up by 3 to y=3.

    Then for B from the sketch you will see it crosses the x axis once so solve for x in 0=(1/x) + 3 since y is 0 when it crosses x axis
    (Original post by Naruke)
    Watch the videos under graph transformations and asymptotes

    http://www.examsolutions.net/a-level.../c1-tutorials/

    then revisit this question
    (Original post by Alpha°)
    thanks
    Yh, thanks guys i understand now :banana:
 
 
 
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