Question 675648
{{{f(x)=ax^2+bx+c}}} 

for  f(1)=6 

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

{{{6=a+b+c}}}..............1

 f(-2)=18

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

{{{18=4a-2b+c}}}..............2

 and f(2)=6

{{{6=a*2^2+b*2+c}}}

{{{6=4a+2b+c}}}.................3

solve this system:

{{{6=a+b+c}}}..............1
{{{18=4a-2b+c}}}..............2
{{{6=4a+2b+c}}}.................3
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{{{6=a+b+c}}}..............1...solve for {{{a}}}

{{{a=6-b-c}}}........substitute in 2


{{{18=4(6-b-c)-2b+c}}}..............2

{{{18=24-4b-4c-2b+c}}}

{{{18=24-6b-3c}}}....both sides divide by {{{3}}}

{{{6=8-2b-c}}}.....solve for {{{c}}}

{{{c=8-2b-6}}}

{{{c=2-2b}}}..........substitute {{{c}}}in {{{a=6-b-c}}} 

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

{{{a=6-b-2+2b}}}

{{{a=4+b}}}


substitute {{{c}}} and {{{a}}} in 3


{{{6=4a+2b+c}}}.................3

{{{6=4(4+b)+cross(2b)+2-cross(2b)}}}.....solve for {{{b}}}

{{{6=16+4b+2}}}

{{{6=18+4b}}}

{{{6-18=4b}}}

{{{-12=4b}}}

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

now find {{{a}}}

{{{a=4+b}}}

{{{a=4-3}}}

{{{highlight(a=1)}}}

and {{{c}}}

{{{c=2-2b}}}

{{{c=2-2(-3)}}}

{{{c=2+6}}}

{{{highlight(c=8)}}}


so. your function is:

{{{f(x)=x^2-3x+8}}}


 {{{ graph( 600, 600, -6, 10, -5, 15, x^2-3x+8) }}}


let's see if all given points lie on this parabola


points are:f(1)=6 , f(-2)=18 , and f(2)=6, or (1,6),(2,6),(-2,18)

{{{drawing(600,600,-10,10,-5,20,grid(1),circle(1,6,0.2),circle(2,6,0.2),circle(-2,18,0.2),graph(600,600,-10,10,-5,20,x^2-3x+8))}}}