Answer:
Therefore, domain of y = [tex]\sqrt{x-k}[/tex] ; Domain = [k, ∞)
Domain of y = [tex]\sqrt{x}+k[/tex] ; Domain = [0, ∞)
Step-by-step explanation:
Two functions are,
y = [tex]\sqrt{x-k}[/tex]
and y = [tex]\sqrt{x}-k[/tex]
Let the value of k = 1
Then we find the graph of the functions attached.
By analyzing these graphs,
y = [tex]\sqrt{x-1}[/tex] when shifted 1 unit right and 1 unit down we get a new function,
y = [tex]\sqrt{x}+1[/tex]
Therefore, domain of y = [tex]\sqrt{x-1}[/tex] ; Domain = [1, ∞)
Domain of y = [tex]\sqrt{x}+1[/tex] ; Domain = [0, ∞)
Similarly, domain of y = [tex]\sqrt{x-k}[/tex] ; Domain = [k, ∞)
And domain of y = √x + k ; Domain = [0, ∞)
