Find plausible equation for a curve

Question

Can someone give me an equation that could generate the graph below?

An equation where as X decreases, Y approaches zero but will never reach zero.

Answer 1

I disagree with the solution given by @Jamminermit. The given curve has visibly not an exponential equation : its right branch tends to be a line, i.e., the curve possesses a second asymptote, that an exponential hasn't.

A solution (hopefuly the simplest) consist in a branch of hyperbola with equation :

$$y=\tfrac12\left(\tfrac34x-2+\sqrt{1.76+(\tfrac34x-2)^2}\right)\tag{1}$$

Explanation of (1) : I have multiplied the LHS of the "estimated" two asymptotes' equations, i.e., $y=0$ and $y-\tfrac34x+2=0$ and computed the value of this product $y(y-\tfrac34x+2)$ in the midpoint $(x,y)=(0,0.2)$, giving implicit equation :

$$y(y-\tfrac34x+2)=0.44.$$

Then I have solved this quadratic in $y$, $x$ being considered as a parameter giving formula (1). If I had considered the other solution with a "minus" in front of the square root, we would have obtained the equation of the other branch (in red on the figure).

Remark : it would have been slightly simpler to express $x$ as a function of $y$ :

$$x=\frac43\left(y+2-\dfrac{0.44}{y}\right).$$

The picture below displays a very good agreement with yours.

Answer 2

As mentioned in my comment above, you are looking at an exponential graph. These have the form $y=a^x$, where $a>0$.

A graph, where $a>1$, for example, $y=2^x$, will look similar to the image you linked.

A graph where $0<a<1$, for example $y=(1/2)^x$, will look like the image you linked, but reflected in the y-axis.

A special version of this graph is $y=e^x$ which looks similar to the others, but has special properties within calculus.

Finally, it is important to note, that all exponential graphs of the form $y=a^x$ will pass through the point $(0,1)$

I hope this answers your question, have fun.

Edit: On closer inspection, it seems like the graph you posted above does $not$ follow an exponential pattern as first thought, instead the other answers referring to this being a hyperbola seem to be correct (I am not too sure as I am not familiar enough with the topic). I therefore encourage the original poster to reconsider my answer as the accepted, and instead look to the answer referring to hyperbola. Nevertheless, I will leave this answer up here to hopefully give more information.