Answer: Option B
[tex]k> 0[/tex]
Step-by-step explanation:
The graph shows a radical function of the form [tex]f(x)=a(x+k)^{\frac{1}{n}}+c[/tex]
Where n is a even number.
This type of function has its vertex at the origin when [tex]k = 0[/tex] and [tex]c = 0[/tex]
If [tex]k> 0[/tex] the graph moves horizontally k units to the left
If [tex]k <0[/tex] the graph moves horizontally k units to the right.
Note that in this case the vertex of the function is horizontally shifted 5 units to the left. Therefore we know that [tex]k = 5> 0[/tex]
The correct answer is option B
