How do you find the domain and range of # sqrt(25-(x-2)^2) +3#?

1 Answer
Jan 30, 2017

Answer:

The domain is #x in [-3,7]#
The range is #y in [3,8]#

Explanation:

Let #g(x)=sqrt(25-(x-2)^2)+3#

What is under the #sqrt# sign is #>=0#. this is the domain

So,

#25-(x-2)^2>=0#

#25-(x^2-4x+4)>=0#

#x^2-4x+4-25<=0#

#x^2-4x-21<=0#

Let's factorise

#(x-7)(x+3)<=0#

Let #f(x)=(x-7)(x+3)#

Let 's do a sign chart to solve this inequality

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##7##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##x-7##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<=0# when #x in [-3,7]#, this is the domain

To calculate the range,

When #x=-3#, #=>#, #g(-3)=3#

When #x=7#, #=>#, #g(7)=3#

When #x=2#, #=>#, #g(2)=8#

Let #y=sqrt(25-(x-2)^2)+3#

The range is #y in [3,8]#

graph{(sqrt(25-(x-2)^2)+3) [-9.74, 12.76, -2.055, 9.195]}