To answer the questions for the function $f(x) = 3e^x - 5$, let's analyze each question.

1. Finding the $y$-intercept:

   • The $y$-intercept occurs when $x = 0$.
   • Substitute $x = 0$ into $f(x)$: $f(0) = 3e^0 - 5 = 3 \cdot 1 - 5 = 3 - 5 = -2$
   • So, the $y$-intercept is $-2$.

2. Determining the asymptote:

   • As $x \to -\infty$, $e^x \to 0$.
   • Therefore, $f(x) = 3e^x - 5 \to -5$ as $x \to -\infty$.
   • This means the horizontal asymptote is $y = -5$, so $a = -5$.

Answers:

1. $y$-intercept: $-2$
2. Asymptote $y = a$ where $a = -5$

Would you like further explanation on this solution?

Additional Questions:

1. What is the behavior of $f(x)$ as $x \to +\infty$?
2. How does the coefficient of $e^x$ (in this case, 3) affect the graph of $f(x)$?
3. If $f(x) = 3e^x - 5$ were transformed to $f(x) = 3e^{x+1} - 5$, what would be the effect on the graph?
4. How would the asymptote change if the function were $f(x) = 3e^x + 2$ instead?
5. Can you identify the domain and range of $f(x) = 3e^x - 5$?

Tip: When finding the $y$-intercept of a function, substitute $x = 0$ into the function equation, as it represents the point where the function crosses the $y$-axis.