Finding the general and standard form equations for the circle and parabola passing through thee points (4,6), (-6,2), and (1,-3).

I used the y=Ax^2+Bx+C to find a,b,c.

(4,6)=> 6=A(4)^2+B(4)+C

(-6,2)=> 2=A(-6)^2+B(-6)+C

(1,-3)=> -3A(1)^2+B(1)+C

then,

a=13/15, b=8/7, and c=-158/35

so the equation for parabola:

\(\displaystyle y=(13/35)x^2+(8/7)x-(158/35)\)

how do I make this equation into this equation f(x)a(x-h)^2+k without the vertex?

I used the y=Ax^2+Bx+C to find a,b,c.

(4,6)=> 6=A(4)^2+B(4)+C

(-6,2)=> 2=A(-6)^2+B(-6)+C

(1,-3)=> -3A(1)^2+B(1)+C

then,

a=13/15, b=8/7, and c=-158/35

so the equation for parabola:

\(\displaystyle y=(13/35)x^2+(8/7)x-(158/35)\)

how do I make this equation into this equation f(x)a(x-h)^2+k without the vertex?

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