Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to create a sliding scale system where if variable $x$ is low, then variable $y$ will be on the lower end of a $1$ to $10$ sliding scale. If variable $x$ is high (above perhaps a threshold?), then variable $y$ will be on the higher end of the sliding scale.

I can't for the life of me figure out the logic behind that. I've tried to do something like:

$$y = \frac{10}{x}$$

But in that case, when $x$ is $100$, $y$ will be low. And when $x$ is $1$, $y$ will be high. I need the exact opposite to happen. I imagine this is a common formula but I cant figure it out. Any ideas?

share|improve this question
    
@Amzoti that question had a maximum for the input. –  Thomas Andrews Dec 12 '12 at 14:07
    
@ TA: thanks, too early in the morning! –  Amzoti Dec 12 '12 at 14:08

2 Answers 2

up vote 2 down vote accepted

If I'm not mistaken, any function of the form $f(x) = \frac{ax+b}{cx+d}$ should work if $\frac{a}{c} = 10$, $\frac{b}{d}=1$, and $\frac{d}{c} \gt 0$. This will get you a hyperbola with $f(0) = 1$, $\lim\limits_{x \to +\infty} f(x) = 10$, and the vertical asymptote in the negative $x$'s, so it approaches $10$ from below.

For example, $f(x) = \frac{10x+5}{x+5}$ would work fine.

share|improve this answer
    
This solution also seems to work well for these purposes. I'm glad you added the example because I haven't taken a math class in over 10 years (I know, blasphemy) and have no idea what a vertical asymptote is. Thanks for your solution! –  cillosis Dec 12 '12 at 15:43

Would something like $y=10-\frac9{1+x}$ suit you needs?

share|improve this answer
    
Yeah, that seems to work great! –  cillosis Dec 12 '12 at 15:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.