The form of quadratic is:
[tex]y=ax^2+bx+c[/tex]Since (0,6) is given, we know c = 6, thus we have:
[tex]\begin{gathered} y=ax^2+bx+c \\ ax^2+bx+6 \end{gathered}[/tex]Point 2 is (2,8), replace x and y and find equation:
[tex]\begin{gathered} y=ax^2+bx+6 \\ 8=a(2)^2+b(2)+6 \\ 8-6=4a+2b \\ 4a+2b=2 \end{gathered}[/tex]Putting point 2 (3,0), we have:
[tex]\begin{gathered} y=ax^2+bx+6 \\ 0=a(3)^2+b(3)+6 \\ 9a+3b=-6 \end{gathered}[/tex]Solving the 2 simulatenous equations for a and b, we get:
a = -3
b = 7
Now u have all the values, a, b, and c.
Just put it in the general form of parabola :
[tex]y=-3x^2+7x+6[/tex]y = -3x^2 + 7x + 6
