Question 1183874


Find a quadratic model for each set of values
1. 
({{{-1}}},{{{1}}})
({{{1}}},{{{1}}}) 
({{{3}}},{{{9}}})


{{{f(x)=ax^2+bx+c}}}...........use point ({{{-1}}},{{{1}}})

{{{1=a(-1)^2+b(-1)+c}}}
{{{1=a -b +c}}}.........solve for {{{a}}}
{{{a=1+b-c}}}.........eq.1


{{{f(x)=ax^2+bx+c}}}...........use point ({{{1}}},{{{1}}})

{{{1=a(1)^2+b(1)+c}}}
{{{1=a +b +c}}}.........solve for {{{a}}}
{{{a=1-b-c}}}.........eq.2


{{{f(x)=ax^2+bx+c}}}...........use point ({{{3}}},{{{9}}})

{{{9=a(3)^2+b(3)+c}}}
{{{9=9a +3b +c}}}.........solve for{{{ a}}}
{{{9a=9-3b-c}}}
{{{a=1-b/3-c/9}}}.........eq.3


from eq.1 and eq.2 we have

{{{1+b-c=1-b-c}}}.........simplify
{{{b=-b}}}
{{{2b=0}}}
{{{b=0}}}

from eq.1 and eq.3

{{{1+b-c=1-b/3-c/9}}}........substitute{{{ b}}}
{{{1+0-c=1-0/3-c/9}}}
{{{1-c=1-c/9}}}..........multiply by {{{9}}}
{{{9-9c=9-c}}}
{{{c-9c=9-9}}}
{{{-8c=0}}}
{{{c=0}}}


then
{{{a=1-0-0}}}.........eq.2
{{{a=1}}}


since{{{ a = 1}}}, {{{b = 0}}}, {{{c = 0}}}, your equation is {{{f(x)=x^2 }}}


{{{ drawing(600, 600, -10, 10, -10, 10,
circle(-1,1,.12), locate(-1,1,p(-1,1)),
circle(1,1,.12), locate(1,1,p(1,1)),
circle(3,9,.12), locate(3,9,p(3,9)),
graph(600, 600, -10, 10, -10, 10, x^2 )) }}}