Simultaneous Inequality Solving

Question

This originally was a quadratic question with $f(x)=ax^2 + bx + c$

If

$1 ≤ a + b + c ≤ 2$ {f(1)}

$2 ≤ 4a + 2b + c ≤ 3$ {f(2)}

$3 ≤ 9a + 3b +c ≤ 4$ {f(3)}

Prove $1 ≤ 16a + 4b + c ≤ 8$ {f(4)}

I think I got the limit of why it is $1$, but as for $8$ I am not too sure how to obtain it. Is it because its a quadratic it does that?

Answer 1

Hint: Prove that $f(4) - 3f(3) + 3f(2) - f(1) = 0 $ for a quadratic equation.

Hence, the desired inequality $1 \leq f(4) \leq 8$ follows.
When does / Can equality hold?