We know that this parabola intecepts the x-axis in 6 and -3. This means that is a vertical axis parabola. But also we know that our parabola passes through (1,10). Graphing these we conclude that is a concave down parabola
The general formula for this is:
[tex]y=ax^{2}+bx+c[/tex]
To solve this we need to form a 3 equations system with our three points
[tex]0=a6^{2}+b6+c\\0=a(-3)^{2}+b(-3)+c\\10=a1^{2}+b1+c[/tex]
We can solve for c in the first one
[tex]c=-a36-b6[/tex]
and replace in the second one and solve for b:
[tex]0=a9-b3-a36-b6\\0=-a27-b9\\b9=-a27\\b=-a(27/9)\\b=-3a[/tex]
And replace c and b in the third equation:
[tex]10=a-a3-a36+a18[/tex]
Solve for a:
[tex]10=-a20\\a=-0.5[/tex]
so b and c:
[tex]b=1.5[/tex]
[tex]c=9[/tex]
So:
[tex]y=-0.5x^{2}+1.5x+9[/tex]
