The vertex form of quadratic equation is #y=a(x-h)^2+k# and its vertex is #(h,k)# and axis of symmetry is #x-h=0#.

Hence we should first convert #y=(x^2+21x+22)/16# into vertex form.

#y=(x^2+21x+22)/16#

= #1/16(x^2+2xx21/2xx x+(21/2)^2-(21/2)^2+22)#

= #1/16(x+21/2)^2-441/64+22/16#

= #1/16(x+21/2)^2-(441-88)/64#

= #1/16(x-(-21/2))^2-353/64#

Therefore, its vertex is #(-21/2,-353/64)# and axis of symmetry is #x-(-353/64)=0# or #x+353/64=0#.

#y#-intercept can be found by putting #x=0# and #x#-intercept can be found by putting #y=0#.

Hence #y#-intercept is #22/16=11/8#

and #x#-intercept are given by using quadratic formula on #x^2+21x+22=0# i.e. #x=(-21+-sqrt(21^2-4xx22))/2=(-21+-sqrt353)/2# i.e. #-1.1# and #-19.9#.

While domain is #(-oo,oo)#, range is #[-353,64,oo)#

graph{(y-(x^2+21x+22)/16)=0 [-27.33, 12.67, -10.16, 9.84]}