Question 948208
 {{{y = ax^2+ bx + c}}}

use given points to set up the system of equations to find {{{a}}},{{{b}}},and {{{c}}}

({{{1}}}, {{{1}}}) ----> {{{a + b + c = 1}}}
({{{2}}},{{{ 2}}}) ----> {{{4a + 2b + c = 2}}}
({{{-1}}},{{{5}}}) ---> {{{a -b + c = 5}}}

Solve simultaneous equations:subtract first from second one

{{{4a + 2b + c -(a + b + c)= 2-1}}}
{{{4a + 2b + c -a - b - c= 1}}}
{{{3a + b  = 1}}}...solve for {{{a}}}

=>{{{3a = 1-b}}}
=>{{{a = 1/3-b/3}}}.......1a

add first and third one:
{{{a + b + c +a -b + c = 1+5}}}
{{{2a + 2c = 6}}} ....solve for {{{a}}}
=>{{{2a  = 6 -2c}}} 
=>{{{a  = 3 -c}}}.........2a

go to first eq., substitute {{{a}}} from 1a

{{{1/3-b/3 + b + c = 1}}}
{{{2b/3 + c = 1-1/3}}}
{{{2b/3 + c = 2/3}}}
{{{2b + 3c = 2}}}
{{{2b = 2-3c}}} ....solve for {{{b}}}
{{{b = 1-3c/2}}}...............1b

go to  eq.2, substitute {{{a}}} from 2a and {{{b}}} from 1b

{{{4( 3 -c)+ 2( 1-3c /2)+ c = 2}}}
{{{12 -4c+ 2-3c + c = 2}}}
{{{14-6c = 2}}}
{{{14-2 = 6c}}}
{{{12 = 6c}}}
{{{highlight(c=2)}}}

now find {{{a}}}
=>{{{a  = 3 -c}}}
=>{{{a  = 3 -2}}}
=>{{{highlight(a  = 1)}}}

and {{{b}}}
{{{b = 1-(3*2)/2}}}
=>{{{b = 1-3}}}
=>{{{highlight(b = -2)}}}

so, your equation is:


{{{highlight(y = x^2 -2x + 2)}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,1,.12),circle(2,2,.12),circle(-1,5,.12),
locate(1,1,p(1,1)),locate(2,2,p(2,2)),locate(-1,5,p(-1,5)),
line(1,10,1,-10),
 graph( 600, 600, -10, 10, -10, 10, x^2 -2x + 2)) }}}