Question 1153438

{{{x}}}| {{{-2}}},{{{ -1}}}, {{{0}}}, {{{1}}},{{{ 2}}}
{{{y}}}|{{{ 9}}}, {{{4}}}, {{{1}}},{{{ 0}}}, {{{1}}}

{{{y=ax^2+bx+c}}}..........use given {{{x}}} and {{{y}}} values to calculate coefficients {{{a}}}, {{{b}}}, and constant {{{c}}}


{{{9=a(-2)^2+b(-2)+c}}}
{{{9=4a-2b+c}}}............eq.1


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

{{{1=a(0)^2+b(0)+c}}}
{{{1=c}}}............eq.3


go to

{{{9=4a-2b+c}}}............eq.1, plug in {{{c}}}
{{{9=4a-2b+1}}}............solve for {{{b}}}
{{{2b=4a-9+1}}}
{{{2b=4a-8}}}
{{{b=2a-4}}}..........1a

go to

{{{4=a-b+c}}}............eq.2, plug in {{{c}}}
{{{4=a-b+1}}}...........solve for {{{b}}}
{{{b=a-4+1}}}
{{{b=a-3}}}..........2a


from 1a and 2a we have

{{{2a-4=a-3}}}.......solve for{{{ a}}}
{{{2a-a=4-3}}}
{{{a=1}}}

go to

{{{b=a-3}}}..........2a,plug in {{{a}}}
{{{b=1-3}}}
{{{b=-2}}}


your equation is:

{{{y=x^2-2x+1}}}


determining the vertex: write it in vertex form

{{{y=(x-1)^2+0}}}=> the vertex is at ({{{1}}},{{{0}}})

and axis of symmetry is {{{x=1}}}

the general form of the equation of the quadratic function:

{{{ax^2 + bx + c = 0}}}=> in your case {{{x^2 -2x + 1= 0}}}



{{{drawing( 600,600, -10, 10, -10, 10,
circle(-2,9,.12),locate(-2,9,p(-2,9)),
circle(-1,4,.12),locate(-1,4,p(-1,4)),
circle(0,1,.12),locate(0,1,p(0,1)),
circle(1,0,.12),locate(1,0,p(1,0)),
circle(2,1,.12),locate(2,1,p(2,1)),
 graph( 600,600, -10, 10, -10, 10, x^2-2x+1)) }}}