Let
[tex]y=a(x-h)^2[/tex]be the quadratic function that passes through the points (6,-3) and (3,0), then we get that:
[tex]\begin{gathered} -3=a(6-h)^2, \\ 0=a(3-h)\text{.} \end{gathered}[/tex]Now, notice that a≠0 because y=a(x-h)² is a quadratic equation, therefore, from the last equation we get:
[tex]3-h=0.[/tex]Adding h to the above equation we get:
[tex]\begin{gathered} 3-h+h=0+h, \\ 3=h\text{.} \end{gathered}[/tex]Substituting h=3 in -3=a(6-h)² we get:
[tex]-3=a(6-3)^2\text{.}[/tex]Simplifying the above equation we get:
[tex]\begin{gathered} -3=a(3)^2, \\ -3=9a\text{.} \end{gathered}[/tex]Dividing the above equation by 9 we get:
[tex]\begin{gathered} -\frac{3}{9}=\frac{9a}{9}, \\ -\frac{1}{3}=a\text{.} \end{gathered}[/tex]Answer:
[tex]y=-\frac{1}{3}(x-3)^2\text{.}[/tex]