Probability: Deriving The Moment Generating Function Given the Definition of a Continuous Random Variable

Question

Here's a question I'm working on in my textbook.
It says:

Let $Y$ be a Normally distributed random variable with mean $μ$ and variance $σ^2$ . Derive the moment-generating function of $X = −3Y+4$

Here's what I have:
$M_x(t) = E[e^{tx}]$
$M_x(t) = E[e^{-3ty + 4t}]$
$M_x(t) = e^{4t}E[e^{-3ty}]$

I'm not sure how to use what I know about the moment generating function when it comes to the $E[e^{-3ty}]$ part. There's a -3 in the way, and I'm still new with all of this stuff, so any help would greatly be appreciated.
Thanks.

Answer 1

Hint: $$\mathbb{E}\left[e^{-3tY}\right] = \mathbb{E}\left[e^{(-3t)Y}\right] = M_{Y}(-3t)$$ where $M_{Y}$ is the moment-generating function (MGF) of $Y$.

Do you know the formula for the MGF of a normally-distributed random variable, mean $\mu$ and variance $\sigma^2$?