Question 1003067
{{{f(x)=ax^2+bx+c}}}

given:

{{{f(0)=-1}}}=> it is a point ({{{0}}},{{{-1}}})=({{{x}}},{{{f(x)}}})
{{{f(3)=-16}}}=> it is a point ({{{3}}},{{{-16}}})
{{{f(-1)=-8}}}=> it is a point ({{{-1}}},{{{-8}}})

form the system using given points to find the coefficients {{{a}}},{{{b}}} and {{{c}}}

{{{-1=a*0^2+b*0+c}}}

{{{highlight(c= -1)}}}.......eq.1

{{{-16=a*3^2+b*3-1}}}

{{{-16=9a+3b-1}}}

{{{-16+1=9a+3b}}}

{{{-15=9a+3b}}}......simplify, both sides divide by {{{3}}}

{{{-5=3a+b}}}.......solve for {{{b}}}

{{{b= -5-3a}}}............eq.2


{{{-8=a(-1)^2+b(-1)-1}}}

{{{-8=a-b-1}}}

{{{b=a+8-1}}}

{{{b=a+7}}}...........eq.3

since left sides in eq.2 and eq.3 are same, equal right sides

{{{-5-3a=a+7}}}...........solve for {{{a}}}

{{{-5-7=a+3a}}}

{{{-12=4a}}}

{{{-12/4=a}}}

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

go to {{{b=a+7}}}...........eq.3, plug in {{{a}}} and find {{{b}}}

{{{b= -3+7}}}

{{{highlight(b=4)}}}


so, you have {{{highlight(a=-3)}}},{{{highlight(b=4)}}} and {{{highlight(c=-1)}}} 

a quadratic model is:

{{{f(x)=-3x^2+4x-1}}}


{{{drawing( 600, 600, -10, 10, -20, 10,
circle(0,-1,.12),locate(0,-1,p(0,-1)),
circle(3,-16,.12),locate(3,-16,p(3,-16)),
circle(-1,-8,.12),locate(-1,-8,p(-1,-8)),

 graph( 600, 600, -10, 10, -20, 10, -3x^2+4x-1)) }}}