Question 999688
You're going to have to slog through it.
Start with the general equation for a quadratic.
{{{y=ax^2+bx+c}}}
Use the points,
{{{-10=a(4)^2+b(4)+c}}}
1.{{{16a+4b+c=-10}}}
.
.
{{{5=a(11)^2+b(11)+c}}}
2.{{{121a+11b+c=5}}}
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.
.
{{{17=a(23)^2+b(23)+c}}}
3.{{{529a+23b+c=17}}}
.
.
.

The quickest way is to use Cramer's rule,
{{{A=(matrix(3,3,
16,4,1,
121,11,1,
529,23,1))}}}
{{{abs(A)=-1596}}}
.
.
.
{{{A[a]=(matrix(3,3,
-10,4,1,
5,11,1,
17,23,1))}}}
{{{abs(A[a])=96}}}
.
.
.
{{{A[b]=(matrix(3,3,
16,-10,1,
121,5,1,
529,17,1))}}}
{{{abs(A[b])-4860}}}
.
.
.
{{{A[c]=(matrix(3,3,
16,4,-10,
121,11,5,
529,23,17))}}}
{{{abs(A)=33864}}}
.
.
.
{{{a=abs(A[a])/abs(A)=-(96/1596)=-(8/133)}}}
{{{b=abs(A[b])/abs(A)=4860/1596=405/133}}}
{{{c=abs(A[c])/abs(A)=-(33864/1596)=-(2822/133)}}}
.
.
.
{{{y=-(1/133)(8x^2-405x+2822)}}}
.
.
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*[illustration v9.JPG].