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

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Answer 1 · 2018-07-08T06:17:32

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Given: #y = 1/2 sqrt(x + 2)#

parent function #y = sqrt(x)#

horizontal shift 2 left, horizontal stretch by 1/2

Find the domain:

The value under the radical must be #>= 0#:

#x + 2 >= 0; " "x >= -2#

domain: #x >= -2 " or " [-2, oo)#

Find the range:

The range depends on the domain values. Since #x >= -2#,

#y = 1/2 sqrt(-2 + 2) = 0#

range: #y >= 0 " or " [0, oo)#

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