Quadratic transform of the uniform distribution

Question

Consider $X$ as a uniform distribution over $[-1, 3]$. Let $Y = X^2 + 4X$
(a) Find $f_Y(y)$.
(b) Verify that $f_Y(y)$ is a PDF.

Answer 1

HINT

$$ \begin{split} f_Y(y) &= \frac{d}{dy} \mathbb{P}[Y \le y] \\ &= \frac{d}{dy} \mathbb{P}\left[X^2 + 4X \le y\right] \\ &= \frac{d}{dy} \mathbb{P}\left[X^2 + 4X + 4\le y + 4\right] \\ &= \frac{d}{dy} \mathbb{P}\left[(X+2)^2 \le y + 4\right] \end{split} $$ Now you convert this to an expression of the $F_X(y)$ and when you differentiate you get an expression in $f_X(y)$, both of which are known.