Question 1194535
A quadratic function: 

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


given:

{{{x}}}| {{{f(x)}}}
{{{0}}}| {{{3}}} 
{{{1}}}| {{{1.25}}}
{{{2}}} |{{{0}}}
{{{3}}}| {{{-0.75}}}
{{{4 }}}|{{{-1}}}
{{{5}}} |{{{-0.75}}}
{{{6}}} |{{{0}}}
{{{7 }}}|{{{1.25}}}
{{{8}}}| {{{3}}}

{{{f(x) = ax^2 + bx + c }}}......use given points to set up a system of three equations

{{{0}}}| {{{3}}}={{{x}}}| {{{f(x)}}}

{{{3 = a*0^2 + b*0 + c }}}

{{{c=3}}}...........eq.1

{{{2}}}| {{{0}}}={{{x}}}| {{{f(x)}}}

{{{0 = a*2^2 + b*2 + 3 }}}
{{{0 = 4a + 2b + 3 }}}
{{{-2b-3 = 4a  }}}
{{{a=-b/2-3/4 }}}........eq.2


{{{6}}}|{{{ 0}}}={{{x}}}| {{{f(x)}}}

{{{0 = a*6^2 + b*6 + 3 }}}
{{{0 = 36a + 6b + 3 }}}.......simplify
{{{0 = 12a + 2b + 1 }}}
{{{-2b-1 = 12a  }}}
{{{a=-b/6-1/12 }}}........eq.3

from eq.2 and eq.3 we have

{{{-b/2-3/4=-b/6-1/12  }}}......solve for {{{b}}}

{{{1/12-3/4=b/2-b/6  }}}
{{{-2/3=b/3 }}}

{{{b=-2 }}}

go to

{{{a=-b/2-3/4 }}}........eq.2, substitute {{{b}}}

{{{a=-(-2)/2-3/4 }}}

{{{a=1/4 }}}


your equation is:


{{{f(x) = (1/4)x^2 -2x + 3 }}}-> standard form

{{{f(x) =(1/4) (x - 2) (x - 6)}}}-> factored form

{{{f(x) =( 1/4) (x - 4)^2 - 1}}}-> vertex form