Please help thx

What is the equation of the parabola passing through (1.5), (0.6), and (2,3)?

Is it a parable of 2 or 3 potency?
Also a circle cuts the three points.

(1,5), (0,6), and (2,3)

We have this form

y = a(x - h)^2  + k       where "a" determines the width (and direction - "up" or "down" ) of the parabola, and (h,k) is the vertex

So  we know that

5  =  a ( 1 - h)^2 + k     →  5 = a(1 -2h + h^2) + k  →  5 = a -2ah + ah^2 + k   (1)

6  = a(0 - h)^2 + k  →  6 = ah^2 + k   (2)

3 = a(2 - h)^2 + k   →  3 = a(4 - 4h + h^2) + k → 3 =  4a -4ah +ah^2 + k     (3) 

Sub ( 2) into  (1)  and (3)

5 =  a - 2ah + 6     →  -1  =  a - 2ah   ( 4)

3  = 4a - 4ah + 6  →  -3 = 4a - 4ah     (5)

Multiply  (4) by -2 and add it to (5)

-1  =  2a    →  a  = -1/2

Using (4)  to find h, we have

-1 = (-1/2) - 2 (-1/2)h

-1/2  =  h

Using (2)  to find k, we have

 6 = (-1/2)(1/4) + k

k = 6 + 1/8  =  49/8

So..........our equation is

y = (-1/2)(x + 1/2)^2 + 49/8

Here's the graph with the points of interest  : https://www.desmos.com/calculator/vnyrl52lp8

What is the equation of the parabola passing through (1,5), (0,6), and (2,3)?

Formula parabola:

\(\begin{array}{|rcll|} \hline y = ax^2+bx+c \\ \hline \end{array} \)

a, b, c = ?

\(\begin{array}{|lrcll|} \hline P(0,6): & 6 &=& 0^2\cdot a+0\cdot b+c \\ & 6 &=& c \\\\ P(1,5): & 5 &=& 1^2\cdot a+1\cdot b+c \\ & 5 &=& a+b+c \quad & | \quad c=6 \\ (1) & 5 &=& a+b+6 \\\\ P(2,3): & 3 &=& 2^2\cdot a+2\cdot b+c \\ & 3 &=& 4a+2b+c \quad & | \quad c=6 \\ (2) & 3 &=& 4a+2b+6 \\ \hline \end{array} \)

a, b = ?

\(\begin{array}{|rcll|} \hline (2) & 3 &=& 4a+2b+6 \quad & | \quad : 2\\ & 1.5 &=& 2a+b+3 \\\\ (1) & 5 &=& a+b+6 \\ \hline (2)-(1): & 1.5-5 &=& 2a+b+3- (a+b+6) \\ & -3.5 &=& 2a+b+3- a-b-6 \\ & -3.5 &=& a-3 \\ & -0.5 &=& a \\\\ & 5 &=& a+b+6 \quad & | \quad a=-0.5 \\ & 5 &=& -0.5+b+6 \\ & 5 &=& 5.5+b \\ & 5-5.5 &=& b \\ & -0.5 &=& b \\ \hline \end{array} \)

Formula parabola:

\(\begin{array}{|rcll|} \hline y &=& ax^2+bx+c \quad & | \quad a=-0.5 \quad b=-0.5 \quad c=6 \\ \mathbf{y} & \mathbf{=} & \mathbf{-0.5x^2-0.5x+6} \\ \hline \end{array}\)