Find the quadratic function $f(x)=a x^{2}+b x+c$ that goes through $(3,0)$ and has a local maximum at $(0,1)$

Step 1 ：We are given that the quadratic function goes through the point (3,0) and has a local maximum at (0,1). This suggests that the function opens downwards, meaning the coefficient 'a' is negative.

Step 2 ：The vertex form of a quadratic function is \(f(x) = a(x-h)^2 + k\), where (h,k) is the vertex of the parabola. In this case, the vertex is also the maximum point, which is (0,1). So, h=0 and k=1.

Step 3 ：We can use the point (3,0) to find the value of 'a'. Substituting h=0, k=1 and x=3, y=0 into the vertex form and solving for 'a', we get \(a = -\frac{1}{9}\).

Step 4 ：Converting the vertex form to the standard form \(f(x) = ax^2 + bx + c\), we find that 'b' is 0 and 'c' is 1.

Step 5 ：\(\boxed{f(x)=-\frac{1}{9}x^{2}+1}\) is the quadratic function that goes through (3,0) and has a local maximum at (0,1).