The table below contains four statements. for which function below are all four statements true? the domain of the function is x ≤ 3. the range of the function is y ≥ 0. the x-intercept of the function is 3. when x decreases, the function increases.

A. If the domain is x<=3, a typical case is the square-root function, such as
f(x)=sqrt(3-x).  When x=3, f(x)=0, when x=0, x=sqrt(3)...
B. testing f(x) for range of y.  Since sqrt(x) function has a range of y>=0 for x real, so f(x)>=0.
C. at x=3, f(x)=f(3)=0, so satisfied.
D. f'(x)=-1/sqrt(3-x)<0 for all x, therefore when x decreases, function increases.

Therefore f(x)=sqrt(3-x) satisfies all four given conditions.
Note that this is not the only possible choice.
All functions of the form f(x)=k*sqrt(3-x) for all k>0 will satisfy the given conditions as well.

Ithink it's b sorry if i'm wrong