so hmmm the idea being, you have 3 points, and thus a system of equations of 3 variables
so hmm say for example for a quadratic equation, which is what a parabolic graph is, the template is more or less => y = ax² + bx + c
now, we dunno what a,b, and c, are, but we know three points, (1,-3), (2,1), and (3,11), so let us plug those fellows in
[tex]\bf y=ax^2+bx+c\\\\
-----------------------------\\\\
\begin{cases}
-3=a(1)^2+b(1)+c\qquad &(1,-3)\\
1=a(2)^2+b(2)+c\qquad &(2,1)\\
11=a(3)^2+b(3)+c\qquad &(3,11)
\end{cases}\implies
\begin{cases}
-3=a+b+c\\
1=4a+2b+c\\
11=9a+3b+c
\end{cases}[/tex]
so, that's our system of equations of three variables, and now, we'd use say hmm elimination, to say hmmm see what "a" is, so, let's eliminate the variable "c" then, by using equations 1) and 2) there
let's multiply equation 1) by -1 then
[tex]\bf \begin{array}{llll}
-3=a+b+c&\times -1\implies &3=-a-b-c\\
1=4a+2b+c&\implies &1=4a+2b+c\\
&&--------\\
&&\boxed{4=3a+b}+0
\end{array}[/tex]
now, let us do the same, and eliminate "c" again, from equations 1) and 3)
and again, let's multiply by -1 in the equation 1)
[tex]\bf \begin{array}{llll}
-3=a+b+c&\times -1\implies &\ \ 3=-a-b-c\\
11=9a+3b+c&\implies &11=9a+3b+c\\
&&--------\\
&&\boxed{14=8a+2b}+0
\end{array}[/tex]
so.. now, let's use those "resultant" equations of two variables, and say hmmm eliminate "b", by simply multiplying 4 = 3a + b times -2
[tex]\bf \begin{array}{llll}
4=3a+b&\times -2\implies &-8=-6a-2b\\
14=8a+2b&\implies &14=\quad 8a+2b\\
&&-------\\
&&\boxed{6=2a}+0
\end{array}
\\\\\\
\cfrac{6}{2}=a\implies \boxed{3=a}\\\\
-----------------------------\\\\[/tex]
[tex]\bf \textit{now let's plug "a" in the first equation}
\\\\\\
4=3a+b\implies 4=3(3)+b\implies 4=9+b\implies 4-9=b\\\\\\ \boxed{-5=b}\\\\
-----------------------------\\\\
\textit{now, let's plug "a", and "b" in the equation 1)}
\\\\\\
-3=a+b+c\implies -3=(3)+(-5)+c\implies -3=-2+c\\\\\\ \boxed{-1=c}[/tex]
thus [tex]\bf y=ax^2+bx+c\implies \boxed{y=3x^2-5x-1}[/tex]
