Question 172691
If {{{f}}} is a function, then {{{f(a)}}} is the value of the function evaluated at {{{a}}}.  You are given the values of the function evaluated at 0, 1, and 2.  So:

(1)  {{{f(0)=5}}} means that {{{a(0)^2+b(0)+c=5}}}, simplified {{{c=5}}}


(2)  {{{f(1)=10}}} means that {{{a(1)^2+b(1)+c=10}}}, simplified {{{a+b+c=10}}}


(3)  {{{f(2)=19}}} means that {{{a(2)^2+b(2)+c=19}}}, simplified {{{4a+2b+c=19}}}


Since from (1) we have {{{c=5}}}, that can be substituted into (2) and (3), thus:


(4)  {{{a+b+5=10}}} yielding {{{a+b=5}}} and,


(5)  {{{4a+2b+5=19}}} yielding {{{4a+2b=14}}} yielding {{{2a+b=7}}}


Multiply (5) by {{{-1}}}:


(6)  {{{-2a-b=-7}}}


Add (6) to (4) term-by-term:


(7)  {{{-a+0b=-2}}} or {{{a=2}}}


Substitute {{{a=2}}} into (4):


(8)  {{{2+b=5}}} or {{{b=3}}}


Now you can say that {{{f(x)=2x^2+3x+5}}} by substituting the values determined for {{{a}}}, {{{b}}}, and {{{c}}} when the system of linear equations was solved.


Check:

{{{f(0)=2(0)^2+3(0)+5=0+0+5=5}}}  Check


{{{f(1)=2(1)^2+3(1)+5=2+3+5=10}}}  Check


{{{f(2)=2(2)^2+3(2)+5=8+6+5=19}}}  Check