Question

Let the parabola $y=a x^2+b x+c$ has vertex at $M(4,2)$ and $a \in[1,3]$. If the difference between the extreme values of abc is equal to $N$, then find the digit at unit place of $N$.

Answer 1

$2=16 a +4 b + c$
$\text { and } \frac{-b}{2 a}=4 $
$\Rightarrow b=-8 a \text {, hence, } c=16 a+2$
$\because 1 \leq a \leq 3 $
$\therefore 18 \leq c \leq 50$
$\text { when } a =1, b =-8 \text {, when } a =3, b =-24$
$\therefore N _{\max }=18 \times 1 \times-8$
$N _{\min }=50 \times 3 \times-24 $
$\therefore N _{\text {diff }}=3456$