Question 986161
Find the values of {{{a}}}, {{{b}}}, and {{{c}}} such that the equation {{{y = ax^2 + bx + c}}} has ordered pair solutions ({{{-1}}},{{{ -4}}}), ({{{2}}},{{{ -7}}}) and ({{{3}}},{{{ 0}}}):

for ({{{x}}},{{{ y}}})=({{{-1}}},{{{ -4}}})

{{{-4 = a(-1)^2 + b(-1) + c}}}

{{{-4 = a -b + c}}}.............eq.1



for ({{{x}}},{{{ y}}})= ({{{2}}},{{{ -7}}})

{{{-7 = a(2)^2 + b(2) + c}}}

{{{-7 = 4a +2b + c}}}.............eq.2


for ({{{x}}},{{{ y}}})=({{{3}}},{{{ 0}}})

{{{0 = a(3)^2 + b(3) + c}}}

{{{0= 9a +3b + c}}}.............eq.3


solve this system:

{{{-4 = a -b + c}}}.............eq.1
{{{-7 = 4a +2b + c}}}.............eq.2
{{{0= 9a +3b + c}}}.............eq.3
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start with
{{{-4 = a -b + c}}}.............eq.1
{{{-7 = 4a +2b + c}}}.............eq.2...subtract 1 from 2
-----------------------------------

{{{-7 -(-4)= 4a-a +2b-(-b) + c-c}}}
{{{-7 +4= 3a +2b+b }}}
{{{-3= 3a +3b }}}
{{{-1=a +b }}}
{{{b= -1-a  }}}.............eq.1a

go with

{{{-7 = 4a +2b + c}}}.............eq.2
{{{0= 9a +3b + c}}}.............eq.3..........subtract 2 from 3
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{{{0-(-7)= 9a-4a +3b-2b + c-c}}}
{{{0+7= 5a +b }}}
{{{b=7-5a }}}..........eq.2a

from eq.1a and 2a we have

{{{-1-a=7-5a }}}........solve for {{{a}}}
{{{5a-a=7+1 }}}
{{{4a=8 }}}
{{{a=2 }}}
{{{highlight(a=2) }}}

go to 
{{{b= -1-a }}}.............eq.1a.....substitute {{{2}}} for {{{a}}}

{{{b= -1-a }}}

{{{b= -1 -2 }}}

{{{highlight(b=-3) }}}


go to {{{0= 9a +3b + c}}}.............eq.3 ...substitute {{{2}}} for {{{a}}},{{{-3}}} for {{{b}}}

{{{0= 9(2) +3(-3) + c}}}
{{{0= 18 -9 + c}}}
{{{0= 9 + c}}}
{{{highlight(c= -9)}}}

so, your equation is: {{{y = 2x^2 -3x -9}}}



{{{drawing( 600, 600, -10, 10, -10, 10,
circle(3,0,.13),circle(-1,-4,.13),circle(2,-7,.13),
locate(3,0.5,p(3,0)),locate(-1,-4,p(-1,-4)),locate(2,-7,p(2,-7)),
 graph( 600, 600, -10, 10, -10, 10, 2x^2 -3x -9)) }}}