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I am trying to get a graph like this to pass through the points $(1,0)$ and $(0,1)$:

enter image description here

How do I transform this, and more generally, equations of the form $\dfrac{1}{cx}$ (where $c$ is constant) so that they always pass through those $2$ points?

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    $\begingroup$ What transformations are you looking for? A translation? $\endgroup$
    – DanDan面
    Dec 2, 2022 at 19:51
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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Dec 2, 2022 at 19:54
  • $\begingroup$ I think you have in mind an equation of the form $(y-k) = \frac{1}{x-h}$, but you started off by writing $1/x$ rather than an equation (that would give you a graph). Why do you need the graph to pass through those two points? Will it be used for something? Or is it just an assigned exercise? $\endgroup$
    – hardmath
    Dec 9, 2022 at 5:51

1 Answer 1

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You could translate the curve down and to the left (or up and to the right, focusing on the $x<0$ part). The new function takes the form: $$y=\frac{1}{c(x+a)}-b$$ Plugging in points $(0,1)$ and $(1,0)$ requires: $$1=\frac{1}{ca}-b$$ $$0=\frac{1}{c(1+a)}-b$$ Which solves to: $$a=b=\frac{1}{2}\left(\pm\sqrt{1+\frac{4}{c}}-1\right)$$ In particular, the case $c=1$ gives a translated curve with parameter $a=b=\frac{\pm\sqrt{5}-1}{2}$.

You can play with this in this interactive Desmos graph.

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  • $\begingroup$ thanks, that graph is really helpful! $\endgroup$ Dec 3, 2022 at 14:19

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