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    Here is a graph of y=\dfrac {3}{x}
    https://www.google.co.uk/search?q=gr..._AUIBigA&dpr=1

    Find the Co-ordinates where y=\dfrac{3}{x+2} touches any of the co-ordinate axis. x \not = -2

    how would i know the graph moves to the left using the y=f(x+2) rule which moves the graph to the left by 2 units?
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    (Original post by thefatone)
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    You have the f(x) = \frac{3}{x}. Then f(x+2) = \frac{3}{x+2} shifts the graph two units to the left, so there will be one intersection with the y-axis.

    This is given when f(0+2) = \frac{3}{0 + 2} = f(2) = \cdots. The shift is horizontal and hence there is still no x-axis intersections.
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    (Original post by Zacken)
    You have the f(x) = \frac{3}{x}. Then f(x+2) = \frac{3}{x+2} shifts the graph two units to the left, so there will be one intersection with the y-axis.

    This is given when f(0+2) = \frac{3}{0 + 2} = f(2) = \cdots. The shift is horizontal and hence there is still no x-axis intersections.
    so it doesn't matter where the x is in this one? so if this function has a 2x then i half the x -value for example?
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    (Original post by thefatone)
    so it doesn't matter where the x is in this one? so if this function has a 2x then i half the x -value for example?
    I don't understand what you're saying.
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    (Original post by Zacken)
    I don't understand what you're saying.
    so if the function instead ofy=\dfrac{3}{x+2} was  y=\dfrac{3}{2x} would the x-values be halved? i don't think this function  y=\dfrac{3}{2x} would touch any axis right?
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    (Original post by thefatone)
    so if the function instead ofy=\dfrac{3}{x+2} was  y=\dfrac{3}{2x} would the x-values be halved? i don't think this function  y=\dfrac{3}{2x} would touch any axis right?
    It would represent and horizontal stretch scale factor \frac{1}{2} or a vertical stetch scale factor \frac{1}{2} (the vertical/horizontal thingies being the same is due to the fact that 1/x is its own inverse function. Cool, right? Anyways, doesn't matter). There still wouldn't be any axis intersections, yes. It is not a shift, just a stretch.
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    (Original post by Zacken)
    It would represent and horizontal stretch scale factor \frac{1}{2} or a vertical stetch scale factor \frac{1}{2} (the vertical/horizontal thingies being the same is due to the fact that 1/x is its own inverse function. Cool, right? Anyways, doesn't matter). There still wouldn't be any axis intersections, yes. It is not a shift, just a stretch.
    ah right i see thanks for that ^-^
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    (Original post by thefatone)
    ah right i see thanks for that ^-^
    You're welcome.
 
 
 
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