# how to make the graph of $1/x$ pass through $(1, 0)$ and $(0, 1)$? [closed]

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

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?

• What transformations are you looking for? A translation? Dec 2, 2022 at 19:51
• 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.
– Community Bot
Dec 2, 2022 at 19:54
• 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? Dec 9, 2022 at 5:51

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 $$a=b=\frac{\pm\sqrt{5}-1}{2}$$.