x(t)=\frac{5 + 3e^{-4t}}{5 - 3e^{-4t}}\nchoose the correct graph below.\na.\nb.\nc.\nd.

Explanation:

Step1: Analyze limit as $t\to\infty$

As $t\to\infty$, $e^{-4t}\to0$. Then $x(t)=\frac{5 + 3e^{-4t}}{5-3e^{-4t}}\to\frac{5+0}{5 - 0}=1$.

Step2: Analyze limit as $t\to-\infty$

As $t\to-\infty$, $e^{-4t}\to\infty$. We can rewrite $x(t)$ as $x(t)=\frac{5e^{4t}+3}{5e^{4t}-3}$. Dividing numerator and denominator by $e^{4t}$, as $t\to-\infty$, $x(t)\to - 1$.

Answer:

We need to look for a graph that approaches $y = 1$ as $t$ goes to positive - infinity and approaches $y=-1$ as $t$ goes to negative - infinity. Without seeing the actual details of the graphs A, B, C, D, but based on the above limit analysis, you should choose the graph that has these horizontal asymptote behaviors. If you provide the actual characteristics of the graphs (like which one has the correct horizontal asymptotes), we can give a more definite answer among A, B, C, D.