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# Transformations of reciprocal functions watch

1. 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
2. (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 from

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

Similarly, note that

multiplies the x coordinates by

and

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).
3. (Original post by Indeterminate)
If you want to sketch from

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

Similarly, note that

multiplies the x coordinates by

and

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
4. In terms of stretches, the sketch will not look any different

looks pretty much like
5. (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

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

then

and

6. (Original post by Indeterminate)
By definition, the x in the bracket of the equation

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

then

and

thanks!, the reflections are the same as well I assume?
7. (Original post by Lay-Z)
thanks!, the reflections are the same as well I assume?
Yes, with

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

and

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

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