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I'm looking for a function with the following characteristics:

  • Vertical asymptote at $0$ (i.e. function never touches negative $x$-values)
  • Horizontal asymptote at $7$ (i.e. function never results in $y$-values larger than $7$)
  • $x$-intercept at $0.25$ (i.e. function crosses $x$-axis at $(0.25,0)$)

I tried working with a log-function (e.g. $y=\ln(x+0.75)$). This generally helped to achieve the intercept, but I still couldn't make the asymptotes work. Any advice?

Thanks

Ѕᴀᴀᴅ
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Berbatov
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  • function crosses x-axis at (0,0.25) ??

    How can a function cross the x-axis above the x axis on the horizontal asymptote? Did you mean (0.25,0)?

     – Neo Oct 10 '18 at 15:20
  • Typo, sorry. Intercept at (0.25,0) – Berbatov Oct 10 '18 at 15:22
  • Logarithms never have upper bounds, so that will never work. Also, a horizontal asymptote does not mean that x values cannot be higher than 7. y=1/x has horizontal and vertical asymptotes, yet the value of a function can be higher than the horizontal asymptote. – Neo Oct 10 '18 at 15:46
  • It doesn't have to be a log-function. It just has a shape very close to what I'm looking for so that why I started with it. Also, there is no restriction with respect to x-values, you are correct. I just want that even with an x-value of 1,000, the y-value never ends up above the asymptote of 7 – Berbatov Oct 10 '18 at 15:58

1 Answers1

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Define the function:

$$ y = \begin{cases} undefined & x\leq 0 \\ 7(1-\frac{1}{4x}) & x > 0 \end{cases} $$

This statisfies your requirements.

A logarithmic function is growing continuously and will never work. It has no upper bound.

A few more points:

  • A horizontal asymptote does not mean that y values cannot be higher than the value of the asymptote.
  • A vertical asymptote does not mean that x values cannot be lower than the value of the asymptote.

So, according to that, the function below also satisfies your requirements: $$y = 7(1-\frac{1}{4x})$$

y = 7(1-1/(4x))

Neo
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