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The graph of a quadratic function is a parabola.
The standard (or vertex) form is `f(x) = a(x-h)^2+k`
where h and k are the co-ordinates of the vertex (3;-6).
The general form is `f(x)=ax^2+bx+c`
As we have the vertex (3;-6) substitute into that equation:
`f(x) = a(x-3)^2 +(-6)` . Note that as x=3, the factor is x-3.
We also have the point (-1;10) where x=-1 and y or f(x)=10
So to find a, we can say `f(-1) =10`
Substituting `f(x) = a(x-3)^2-6` becomes
`therefore 10= a(-4)^2-6`
So in standard (vertex) form : `f(x)=1(x-3)^2-6`
To convert to general form, expand the brackets:
`f(x)=x^2 -6x +9 -6`
`therefore f(x) = x^2-6x+3`
Ans: f(x) = x^2-6x+3
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