From the picture we can see that #f(x)# is a piecewise function.
I will assume that the domain of #f(x)# is entirely shown in the picture.
#:.# Domain of #f(x)# is #[-3,+5]# with subdomains #[-3,+3]# and #(+3,+5]#
By sight #f(x)# appears to be a parabola #in [-3,+3]#
and a straight line #in (+3,+5]#
I will assume to above to be true.
We can see that #f(x)# has zeros at #x=-3 and x=3#
Hence, we can say that #a(x-3)(x+3)=0# for some constant #a#
We can also observe that #f(0) = 9/2#
#:. a(+3)(-3) =9/2 -> a =-1/2#
Hence, #f(x) = -1/2(x-3)(x+3)# for #x in [-3,+3]#
Then, for #x in (3,5]# we have #f(x) = mx +c# (a straight line with slope m)
From the picture: #m = 1/(5-3) = 1/2#
Finally, we can see that #f(x) = 1 # at #x=+5#
#:. 1=1/2xx5 +c -> c =-3/2)#
#f(x) = 1/2x-3/2# for #x in (3,5]#
Hence, #f(x)=1/2(x-3)# for #x in (3,5]#
