How do you graph #y=1/4sqrt(x-1)+2#, compare to the parent graph, and state the domain and range?

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Answer 1 · 2018-01-04T16:57:47

#D_f=[1,+oo)# , #R_f=[2,+oo)#

graph{sqrt(x-1)/4+2 [-10, 10, -5, 5]}

The graph is #sqrtx# shifted #1# to the right and what comes out of that shifted #2# times upwards.

#D_f##=##{AAx##in##RR#: #x-1>=0##}# #=# #[1,+oo)#

I will find the range using Monotony and continuity.

#f(x)=1/4sqrt(x-1)+2# ,
#x>=1#

#f'(x)=1/4*((x-1)')/(2sqrt(x-1))# #=#

#1/(8sqrt(x-1))# #>0# , #x##in##(1,+oo)#

Therefore #f# is strictly increasing #uarr# in #[1,+oo)#

#=# #[2,+oo)#

because #lim_(xrarr+oo)f(x)=lim_(xrarr+oo)(sqrt(x-1)/4+2)# #=# #+oo#

NOTE: #lim_(xrarr+oo)(x-1)=+oo# therefore, #lim_(xrarr+oo)sqrt(x-1)=+oo#