NOTE that: The upper boundary (for positive values of $t$) is defined in terms of infinity, that is at $t=∞$. (There is no lower boundary for negative values of $t$.)
NOTE. To any re-reading this, I was tired when I wrote it and mixed up the equations; now the post should make much more sense!
Note that the function does not cross $y$ at the top! This upper bound is at $1.618^{3}$. As you can see, it rounds off as it approaches the upper bound!
See this Graph for further insight: https://www.desmos.com/calculator/qqqfuyrhyp
I'm looking for the parametric equations for the curve shown: I already know $y(t)$, it's $y(t)=\frac{1-φ^{-1+t}}{1-φ}+φ^{-1}$; here $φ$ is $\frac{-1+\sqrt5}{2}$. So, all I need is to find $x(t)$, which will also be defined in terms of $φ$.
Thank you all!
The given parametric equation for $y(t)$ is incompatible with the graph of the curve. According to the graph, there should exist a real number $T$ such that $$ y(T)={1-\varphi^T\over1-\varphi}=\varphi^{-3}. $$ But that implies: $$ 1-\varphi^T=\varphi^{-3}(1-\varphi)=\varphi^{-1}, \quad\text{that is:}\quad \varphi^T=1-\varphi^{-1}, $$ and that is impossible, because $1-\varphi^{-1}=-\varphi<0$.
The same conclusion can be reached studying function $y(t)$, which is monotonically increasing and has upper bound $$ \lim\limits_{t\to+\infty}y(t)={1\over1-\varphi}=\varphi^{-2}<\varphi^{-3}. $$
EDIT.
With the modified definition $y(t)={1-\varphi^{t-1}\over1-\varphi}+\varphi^{-1}$ one can repeat the above computation for $T$ to find: $$ \varphi^{T-1}=1+\varphi-\varphi^{-1}=0, $$ which is possible only in the limit $T\to+\infty$.