How can I change $y=\left(\frac{1}{x}\right)$ to have it land on $2$ specific points?

Question

I feel the need to prefix this by saying I've not touched maths for over $3$ years...

Because $y=\left(\frac{1}{x}\right)$ goes from practically vertical to practically horizontal, surely this means some part of it can be positioned between $(1, 15)$ and $(300, 1)$?

How can I find the new equation of $y=\left(\frac{1}{x}\right)$ once it has been moved? Or if it's easy to do, what is it?

what? Are you wanting to translate the graph of 1/x so it crosses both of those points? — Brevan Ellefsen, Mar 13 '17 at 02:15
if you google "graph shifting" you should find what you need. you need to add or subtract from x and y in the equation. — The Count, Mar 13 '17 at 02:34
You could use $\frac{4200}{299}\frac{1}{x}+\frac{285}{299}$ which one gets from solving $\frac ax + b = y$ for the points $(1,15)$ and $(300,1)$ — Brevan Ellefsen, Mar 13 '17 at 02:35
@BrevanEllefsen Thank you for your help, but is it me or does that not go through either of the points though? — Toby Smith, Mar 13 '17 at 15:26
@TobySmith it definitely goes through both points. I graphed it in Desmos and it crosses both. I suppose I could have written down the points incorrectly, but I doubt it — Brevan Ellefsen, Mar 13 '17 at 17:06
@BrevanEllefsen Yes, it was entirely my stupidity - thank you for your help :) — Toby Smith, Mar 13 '17 at 17:34

score 1 · Answer 1 · answered Dec 05 '18 at 06:45

From Brevan Ellefsen's comment:

The graph of the equation $$y=\frac{4200}{299}\frac{1}{x}+\frac{285}{299}$$ passes through $(1,15)$ and $(300,1)$. That equation can be found by solving a system of two equations of the form $y=\frac{a}{x}+b$ for $a$ and $b$ after plugging in the coordinates of the two given points for $x$ and $y$.

How can I change $y=\left(\frac{1}{x}\right)$ to have it land on $2$ specific points?

1 Answers1