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Transformations of reciprocal functions Watch

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    Functions in the form 1/x I cant find it on the internet,
    I know that the vertical asymptote is where the denominator is 0 but i'm not sure about stretches etc, I know all the normal transformations etc f(ax),af(x),f(x)+a,f(x)-a,f(x-a),f(x+a), i'm doing c3 questions involving sketching eg, f(x)=5/2x-4 need to be able to do these for working out the range
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    (Original post by Lay-Z)
    Functions in the form 1/x I cant find it on the internet,
    I know that the vertical asymptote is where the denominator is 0 but i'm not sure about stretches etc, I know all the normal transformations etc f(ax),af(x),f(x)+a,f(x)-a,f(x-a),f(x+a), i'm doing c3 questions involving sketching eg, f(x)=5/2x-4 need to be able to do these for working out the range
    If you want to sketch af(x) from

    f(x)

    a good way would be to take any point(s) on f(x) and multiply the y coordinate by a. You'll then understand what it looks like

    Similarly, note that

    f(ax)

    multiplies the x coordinates by \frac{1}{a}

    and f(x+b)

    would move the function -b in the x direction (here, it's useful to take an asymptote as a point of reference and see where it goes to).
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    (Original post by Indeterminate)
    If you want to sketch af(x) from

    f(x)

    a good way would be to take any point(s) on f(x) and multiply the y coordinate by a. You'll then understand what it looks like

    Similarly, note that

    f(ax)

    multiplies the x coordinates by \frac{1}{a}

    and f(x+b)

    would move the function -b in the x direction (here, it's useful to take an asymptote as a point of reference and see where it goes to).
    Hey, I think you misunderstood my question, I was refering to for example how 1/x relates to 5/(x-1) or 5/(2x+3)

    At the moment I assume the translations stay the same its just how the shifts would be shown
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    In terms of stretches, the sketch will not look any different

    y = \frac{2}{x} looks pretty much like y = \frac{1}{x}
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    (Original post by Lay-Z)
    Hey, I think you misunderstood my question, I was refering to for example how 1/x relates to 5/(x-1) or 5/(2x+3)

    At the moment I assume the translations stay the same its just how the shifts would be shown
    By definition, the x in the bracket of the equation

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

    corresponds to the x denominator (think of what you'd do if I told you to evaluate f(3) or something)

    Besides that, the multiplication by scalars (numbers) have the obvious effect on the function.

    Thus, if

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

    then

    \dfrac{5}{x-1} = 5f(x-1)

    and

    \dfrac{5}{2x-3} = 5f(2x-3)
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    (Original post by Indeterminate)
    By definition, the x in the bracket of the equation

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

    corresponds to the x denominator (think of what you'd do if I told you to evaluate f(3) or something)

    Besides that, the multiplication by scalars (numbers) have the obvious effect on the function.

    Thus, if

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

    \dfrac{5}{x-1} = 5f(x-1)

    and

    \dfrac{5}{2x-3} = 5f(2x-3)
    thanks!, the reflections are the same as well I assume?
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    (Original post by Lay-Z)
    thanks!, the reflections are the same as well I assume?
    Yes, with

    -f(x)

    being a reflection in the x axis (multiplying the y coordinates by minus 1)

    and

    f(-x)

    being a reflection in the y-axis (multiplying the x coordinates by minus 1).
 
 
 
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