We are given three points: $(4, -8)$, $(3, -2)$, and $(5, -1)$, and we need to determine the quadratic equation passing through these points in standard form:

$y = ax^2 + bx + c$

To find $a$, $b$, and $c$, we will set up a system of equations by plugging in each point into the general quadratic form. Let's do this step by step.

1. For $(4, -8)$: $-8 = a(4)^2 + b(4) + c \quad \Rightarrow \quad 16a + 4b + c = -8$

2. For $(3, -2)$: $-2 = a(3)^2 + b(3) + c \quad \Rightarrow \quad 9a + 3b + c = -2$

3. For $(5, -1)$: $-1 = a(5)^2 + b(5) + c \quad \Rightarrow \quad 25a + 5b + c = -1$

Now, we have the system of equations:

16a + 4b + c &= -8 \\ 9a + 3b + c &= -2 \\ 25a + 5b + c &= -1 \\ \end{aligned}$$ We can solve this system of equations to find $$a$$, $$b$$, and $$c$$. I'll calculate that for you.