[tex]\bf (4,8)\qquad (-3,23)\qquad (-8,8)
\\\\
-------------------------------\\\\
y=ax^2+bx+c\implies
\begin{cases}
8=a(4)^2+b(4)+c\\
23=a(-3)^2+b(-3)+c\\
8=a(-8)^2+b(-8)+c
\end{cases}
\\\\\\
thus
\\\\\\
y=ax^2+bx+c\implies
\begin{cases}
8=16a+4b+c\\
23=9a-3b+c\\
8=64a-8b+c
\end{cases}
[/tex]
so, we have a system of equations of 3 variables... so... let's do some elimination, let's eliminate "c"
first by using the 1st and 2nd equations
and then by using the 1st and 3rd equations
by simply multiplying the 1st equation in both cases by -1, so we end up with -c atop, which will cancel out the +c at the bottom
[tex]\bf -8=-16a-4b-c\\
23=\quad\ 9a\ -3b+c\\
----------\\
\boxed{15=\quad 7a\quad -7b}+0\\\\
-----------\\\\
-8=-16a-4b-c\\
8=\quad\ 64a-8b+c\\
----------\\
\boxed{0=\quad 48a-12b}+0[/tex]
now, notice the 2nd resultant equation with two varibles.. ended up equals to 0... so... .might as well let's do some substitution there, to get "b" then
[tex]\bf 0=48a-12b\implies 12b=48a\implies b=\cfrac{48a}{12}\implies \boxed{b=4a}
\\\\\\
\textit{let's use that in the first resultant equation}
\\\\\\
15=7a-7b\implies 15=7a-7(4a)\implies 15=-21a\implies \cfrac{15}{-21}=a
\\\\\\
\boxed{-\cfrac{5}{7}=a}\qquad thus\qquad b=4a\implies b=4\left( -\cfrac{5}{7} \right)\implies \boxed{b=-\cfrac{20}{7}}[/tex]
now, let's use those in the 1st of the original equations
[tex]\bf 8=16\left( -\cfrac{5}{7} \right)+4\left( -\cfrac{20}{7} \right)+c\implies 8=-\cfrac{80}{7}-\cfrac{80}{7}+c
\\\\\\
56=-80-80+7c\implies 56+160=7c\implies \boxed{\cfrac{216}{7}=c}\\\\
-------------------------------\\\\
y=-\cfrac{5}{7}x^2-\cfrac{20}{7}x+\cfrac{216}{7}[/tex]
