Question 1189357
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The axis of symmetry is parallel to the vertical y axis means we have a parabola in the form
{{{y = ax^2+bx+c}}}
This parabola either opens upward or opens downward (depending on if a > 0 or a < 0 respectively).


Plug in the coordinates of point A(1,1)
{{{y = ax^2+bx+c}}}
{{{1 = a(1)^2+b(1)+c}}}
{{{1 = a+b+c}}}
{{{a+b+c = 1}}}


Plug in the coordinates of point B(2,2)
{{{y = ax^2+bx+c}}}
{{{2 = a(2)^2+b(2)+c}}}
{{{2 = 4a+2b+c}}}
{{{4a+2b+c = 2}}}


Plug in the coordinates of point C(-1,5)
{{{y = ax^2+bx+c}}}
{{{5 = a(-1)^2+b(-1)+c}}}
{{{5 = a-b+c}}}
{{{a-b+c = 5}}}


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We have this system of equations
{{{system(a+b+c = 1,4a+2b+c = 2,a-b+c = 5)}}}
which I will refer to as equation (1), equation (2) and equation (3) in that order (from top to bottom).


Let's solve for 'a' in equation (3)
{{{a-b+c = 5}}}
{{{a = 5+b-c}}}
I'll refer to this as equation (4).


Then we'll do the same for equation (1) as well
{{{a+b+c = 1}}}
{{{a = 1-b-c}}}
Let's call this equation (5)


Equations (4) and (5) have 'a' on the left side and something with b's and c's on the right side. Let's equate the right hand sides of those two new equations.
{{{5+b-c = 1-b-c}}}


Next, we can add c to both sides
{{{5+b-c = 1-b-c}}}
{{{5+b-c+c = 1-b-c+c}}}
{{{5+b = 1-b}}}
and the c terms cancel.


We can solve for b
{{{5+b = 1-b}}}
{{{5+b+b = 1}}}
{{{5+2b = 1}}}
{{{2b = 1-5}}}
{{{2b = -4}}}
{{{b = -4/2}}}
{{{b = -2}}}


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With this b value in mind, let's go through equations (1), (2) and (3) and plug in b = -2.


Equation 1: 
{{{a+b+c = 1}}}
{{{a-2+c = 1}}}
{{{a+c = 1+2}}}
{{{a+c = 3}}}


Equation 2:
{{{4a+2b+c = 2}}}
{{{4a+2(-2)+c = 2}}}
{{{4a-4+c = 2}}}
{{{4a+c = 2+4}}}
{{{4a+c = 6}}}


Equation 3:
{{{a-b+c = 5}}}
{{{a-(-2)+c = 5}}}
{{{a+2+c = 5}}}
{{{a+c = 5-2}}}
{{{a+c = 3}}}
We get the <u>exact identical</u> result as before when we plugged b = -2 into equation (1). There's no need to repeat it again, so we ignore this.


The old system we formed earlier was
{{{system(a+b+c = 1,4a+2b+c = 2,a-b+c = 5)}}}
which is now equivalent to this somewhat reduced system
{{{system(a+c=3,4a+c=6)}}}
We've gone from 3 variables + 3 equations down to 2 variables + 2 equations.


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Solve {{{a+c = 3}}} for c
{{{a+c = 3}}}
{{{c = 3-a}}}


Plug this into the other equation and solve for 'a'
{{{4a+c = 6}}}
{{{4a+3-a = 6}}}
{{{3a+3 = 6}}}
{{{3a = 6-3}}}
{{{3a = 3}}}
{{{a = 3/3}}}
{{{a = 1}}}


We can now determine c
{{{c = 3-a}}}
{{{c = 3-1}}}
{{{c = 2}}}


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To recap everything, we have
a = 1
b = -2
c = 2


So the template of 
{{{y = ax^2+bx+c}}}
fully updates to
{{{y = 1x^2-2x+2}}}
{{{y = x^2-2x+2}}}


The last step is merely cosmetic in my opinion, but it's to subtract y from both sides to get 0 on its own side by itself.
{{{y = x^2-2x+2}}}
{{{0 = x^2-2x+2-y}}}
{{{0 = x^2-2x-y+2}}}
{{{x^2-2x-y+2 = 0}}}


Answer: Choice A
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