Question 929710
to find the quadratic function {{{y = ax ^ 2 + bx + c}}} whose graph passes through the given points ({{{1}}},{{{-1}}})({{{-1}}},{{{-9}}})({{{-2}}},{{{-10}}}), we need to find {{{a}}},{{{b}}},and {{{a}}} and we can do it using given points

{{{y = ax ^ 2 + bx + c}}} ...plug in {{{x=1}}} and {{{y=-1}}} from  ({{{1}}},{{{-1}}})

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

{{{-1 = a + b + c}}}.........eq.1


{{{y = ax ^ 2 + bx + c}}} ...plug in {{{x=-1}}} and {{{y=-9}}} from ({{{-1}}},{{{-9}}})

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

{{{-9= a -b + c}}}.........eq.2


{{{y = ax ^ 2 + bx + c}}} ...plug in {{{x=-2}}} and {{{y=-10}}} from ({{{-2}}},{{{-10}}})

{{{-10= a(-2) ^ 2 + b(-2) + c}}}

{{{-10= 4a -2b + c}}}..........eq.3


you have a system:

{{{-1 = a + b + c}}}.........eq.1
{{{-9= a -b + c}}}.........eq.2
{{{-10= 4a -2b + c}}}..........eq.3
_______________________________

start with

{{{-1 = a + b + c}}}.........eq.1
{{{-9= a -b + c}}}.........eq.2
___________________________subtract 2 from 1

{{{-1-(-9) = a-a + b-(-b) + c-c}}}

{{{-1+9 = cross(a)-cross(a) + b+b + cross(c)-cross(c)}}}

{{{8 = 2b}}}

{{{8/2 = b}}}

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

go to {{{-10= 4a -2b + c}}}..........eq.3, plug in {{{b}}}

{{{-10= 4a -2*4 + c}}}

{{{-10= 4a -8 + c}}}

{{{-10+8= 4a+ c}}}

{{{-2= 4a+ c}}}

{{{-2-4a=c}}}

{{{c=-2-4a}}}

go to {{{-1 = a + b + c}}}.........eq.1 substitute {{{b}}} and {{{c}}}

{{{-1 = a + 4 -2-4a}}}

{{{-1 = 2 -3a}}}

{{{3a =2+1}}}

{{{3a = 3}}}

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

go to {{{c=-2-4a}}} plug in {{{a}}} and find {{{c}}}

{{{c=-2-4(1)}}}

{{{c=-2-4}}}

{{{highlight(c=-6)}}}

so, your equation is: {{{y = x ^ 2 + 4x -6}}}



(1,-1)(-1,-9)(-2,-10)
{{{drawing( 600, 600, -10, 10, -15, 10,locate(1,-1,p(1,-1)),locate(-1,-9,p(-1,-9)),locate(-2,-10,p(-2,-10)),circle(-2,-10,.12),circle(-1,-9,.12),circle(1,-1,.12), graph( 600, 600, -10, 10, -15, 10, x ^ 2 + 4x -6)) }}}