Question

14 of 46

E Sa
Function Transformations: Tutorial
Part B
Open the graphing tool, and compare the functions using the sliders. What can be said about the domain of
y= VX - kand the domain of y = VX + k?

Answer 1

Therefore, domain of y =
`√(x-k)` ; Domain = [k, ∞)

Domain of y =
`√(x)+k` ; Domain = [0, ∞)

Explanation:

Two functions are,

y =
`√(x-k)`

and y =
`√(x)-k`

Let the value of k = 1

Then we find the graph of the functions attached.

By analyzing these graphs,

y =
`√(x-1)` when shifted 1 unit right and 1 unit down we get a new function,

y =
`√(x)+1`

Therefore, domain of y =
`√(x-1)` ; Domain = [1, ∞)

Domain of y =
`√(x)+1` ; Domain = [0, ∞)

Similarly, domain of y =
`√(x-k)` ; Domain = [k, ∞)

And domain of y = √x + k ; Domain = [0, ∞)