Answer:
Left 2 units
Step-by-step explanation:
Without knowing what f(x), it is impossible to answer this question. I'll answer assuming a parent function of [tex]f(x)=x^2[/tex].
In the function [tex]y=(x-c)^2[/tex], [tex]c[/tex] represents the phase shift from parent function [tex]f(x)=x^2[/tex]. If [tex]c[/tex] is positive (e.g. [tex](x-3)^2[/tex]), then the function shifts to the right however many units [tex]c[/tex] is. If [tex]c[/tex] is negative (e.g. [tex](x+5)^2[/tex]), the function shifts to the left however many units the absolute value of [tex]c[/tex] is.
In the function [tex]g(x)=(x+2)^2[/tex], let's find out the value of [tex]c[/tex]:
Format: [tex]y=(x-c)^2[/tex]
[tex]g(x)=(x+2)^2=(x-(-2))^2[/tex]
Therefore, [tex]c=-2[/tex].
Since [tex]c[/tex] is negative, the function must shift to the left. To find out how many units it shifts to the left, take the absolute value of [tex]c[/tex]:
[tex]|-2|=\boxed{2\text{ units to the left}}[/tex].
Thus, the graph of [tex]g(x)=(x+2)^2[/tex] is a translation of the graph [tex]f(x)=x^2[/tex] by 2 units to the left.
*Note: Once again, this answer is assuming [tex]f(x)=x^2[/tex] as it is not clarified in the question. If [tex]f(x)\neq x^2[/tex] and is already shifted, you will need to account for shift. If you believe this is the case, feel free to let me know in the comments.
