Question 1002751

{{{y=ax^2+bx+c}}} for the parabola passing through the points ({{{3}}},{{{-22}}}) ({{{1}}},{{{4}}}) ({{{2}}},{{{-4}}})

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

({{{3}}},{{{-22}}}):

{{{-22=a*3^2+b*3+c}}}

{{{-22=9a+3b+c}}}.....eq.1

 ({{{1}}},{{{4}}})

{{{4=a*1^2+b*1+c}}}

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


({{{2}}},{{{-4}}})

{{{-4=a*2^2+b*2+c}}}

{{{-4=4a+2b+c}}}.....eq.3

so, the system is:

{{{-22=9a+3b+c}}}.....eq.1
{{{4=a+b+c}}}.....eq.2
{{{-4=4a+2b+c}}}.....eq.3
----------------------------
start with

{{{-22=9a+3b+c}}}.....eq.1
{{{4=a+b+c}}}.....eq.2
-----------------------------subtract eq.1 from eq.2

{{{4-(-22)=a+b+c-9a-3b-c}}}
{{{4+22=-8a-2b}}}
{{{26=-8a-2b}}}.....both sides divide by {{{2}}}
{{{13=-4a-b}}}.....solve for {{{b}}}
{{{b=-4a-13}}}...............1a

do same with eq.2 and eq.3

{{{4=a+b+c}}}.....eq.2
{{{-4=4a+2b+c}}}.....eq.3
----------------------------subtract eq.2 from eq.3
{{{-4-4=4a+2b+c-a-b-c}}}
{{{-8=3a+b}}}
{{{b=-8-3a}}}........2a

from 1a and 2a we have:

{{{-4a-13=-8-3a}}}........solve for {{{a}}}

{{{8-13=4a-3a}}}
{{{highlight(a=-5)}}}

go to {{{b=-8-3a}}}........2a, substitute {{{-5}}} for {{{a}}}

{{{b=-8-3(-5)}}}
{{{b=-8+15}}}
{{{highlight(b=7)}}}

go to {{{4=a+b+c}}}.....eq.2, substitute {{{-5}}} for {{{a}}} and {{{7}}} for {{{b}}}

{{{4=-5+7+c}}}
{{{4=2+c}}}
{{{c=4-2}}}
{{{highlight(c=2)}}}

so, your equation is: {{{y=-5x^2+7x+2}}}

see it and given points on the graph:
({{{3}}},{{{-22}}}) ({{{1}}},{{{4}}}) ({{{2}}},{{{-4}}})

{{{drawing( 600, 600, -10, 10, -30, 10,
circle(3,-22,.12),circle(1,4,.12),circle(2,-4,.12),
locate(3,-22,p(3,-22)),locate(1,4,p(1,4)),locate(2,-4,p(2,-4)),
 graph( 600, 600, -10, 10, -30, 10, -5x^2+7x+2)) }}}