f(x)=-x^2+3mx-m-3.
The general equation of a parabola is f(x)=a(x-h)^2+k where (h,k) is the vertex.
a(x-h)^2+k=ax^2-2ahx+ah^2+k=-x^2+3mx-m-3.
a=-1; 2h=3m; -h^2+k=-m-3. h=3m/2 substituted in -h^2+k is -9m^2/4+k=-m-3.
Also the zeroes are separated by 4. f(x)=0=-(x-h)^2+k, so x=h+sqrt(k); so h+sqrt(k)-h+sqrt(k)=4, 2sqrt(k)=4, sqrt(k)=2, and k=4. -9m^2/4+4=-m-3. 9m^2-4m-28=0=(9m+14)(m-2), so m=-14/9 or 2.
f(x)=-x^2+6x-5=-(x-1)(x-5) seems to fit the picture, when m=2, where the zeroes are 1 and 5.
(h,k)=(3,4) and y intercept is -5. The other solution has (h,k)=(-7/3,4) which does not fit the picture.