Question 1060914
The equation would be of the form {{{y=ax^2+bx+c}}} .
Substituting the coordinates of {{{A(1,1)}}} , we get
{{{1=a*1^2+b*1+c}}} --> {{{a+b+c=1}}} .
Substituting the coordinates of {{{B(2,2)}}} , we get
{{{2=a*2^2+b*2+c}}} --> {{{4a+2b+c=2}}} .
Substituting the coordinates of {{{C(-1,5)}}} , we get
{{{5=a*(-1)^2+b*(-1)+c}}} --> {{{a-b+c=5}}} .
The 3 equations above give us the system
{{{system(a+b+c=1,4a+2b+c=2,a-b+c=5)}}}
Subtracting the first equation from each ot the other two equations, we get the equivalent system
{{{system(a+b+c=1,4a+2b+c-a-b-c=2-1,a-b+c-a-b-c=5-1)}}}--->{{{system(a+b+c=1,3a+b=1,-2b=4)}}}-->{{{system(a+b+c=1,3a+b=1,b=4/(-2))}}}-->{{{system(a+b+c=1,3a+b=1,highlight(b=-2))}}} .
Substituting {{{-2}}} for {{{b}}} in {{{3a+b=1}}} we get
{{{3a-2=1}}} --> {{{3a=1+2}}} --> {{{3a=3}}} --> {{{a=3/3}}} --> {{{highlight(a=1)}}} .
Then substituting {{{-2}}} for {{{b}}} and {{{1}}} for {{{a}}} in {{{a+b+c=1}}} , we get
{{{1-2+c=1}}} --> {{{-1+c=1}}} --> {{{c=1+1}}} --> {{{highlight(c=2)}}} .
So, the equation of the parabola is
{{{highlight(y=x^2-2x+2)}}} .
 
We could also write it as
{{{y=x^2-2x+1+1}}} --> {{{y=(x^2-2x+1)+1}}} --> {{{highlight(y=(x-1)^2+1)}}} .