Functions argument help

Find the quadratic functions y=ax^2 +bx +c coefficients, if x=6 is the functions zero points and the function has the smallest value of -8 if the value of the argument is 4.

Basically what I can make out of this is: f(4)=ax^2 +bx +c=-8. The 4 in the brackets is the argument I believe, although I have no idea what that tells me. The zero points are (0;6)
That's about it, I know I have to find a, b and c, but how I don't know.
Results should be: a=2, b=-16. c=24

Could someone also explain to me what an argument exactly is, as simple as possible?

Thanks in advance!

Argument is X. The statement "the function has the smallest value of -8 if the value of the argument is 4" means that when X = 4, the function is -8. It also implies that the minimum of the curve is at (4,-8). Since x=6 is one of the zeroes, and the minimum (or maximum) of a parabola should be the midpoint of the two zeroes, the implication is that x=2 is the other zero.

So from the two zeroes of x= 2 and 6, you get:
4a+2b+c = 0
36a+6b+c= 0

Solving that set of linear equations gives you:
32a+4b = 0
Which reduces to
8a+b=0

You are also given y=-8 when x=4. So:
16a+4b+c= -8

Let's use 36a+6b+c=0 to solve:
36a+6b+c=0
16a+4b+c=-8
Subtracting the 2nd equation from the first gets you:
20a+2b=8
Or
10a+b=4

We also have 8a+b=0 So:
10a+b=4
8a+b=0
Implies:
2a=4 or a=2

Plugging back a=2 into the previous equalities gets you:
8a+b=0 ---> 16+b=0 or b=-16
And
4a+2b+c= 0 ---> 4(2) + 2(-16) + c= 0 ----> -24+c=0 -----> c=24

The wording is very odd.
The argument is the x-value for which y=8, which should have been stated.
You don't give in the question that x=0 is a zero of the function.
The f(4) refers to f(x)=ax^2+bx+c and f(4) means sub x=4 giving 16a+4b+c=-8
If f(6)=0 then 36a+6b+c=0