Compare the following functions.

How does [tex]\( f_2 \)[/tex] compare to [tex]\( f_1 \)[/tex]?

\begin{tabular}{|l|l|l|l|}
\hline
[tex]$f_1(x) = x$[/tex] & [tex]$f_2(x) = x + 4$[/tex] & [tex]$f_3(x) = x - 5$[/tex] & [tex]$f_4(x) = x - 7$[/tex] \\
\hline
\end{tabular}

A. Moved 4 units up

B. Moved 11 units down

C. Moved 7 units down

D. None of the other answers are correct

E. Moved 5 units up



Answer :

To determine how function [tex]\( f_2 \)[/tex] compares to [tex]\( f_1 \)[/tex], we examine their definitions:

- [tex]\( f_1(x) = x \)[/tex]
- [tex]\( f_2(x) = x + 4 \)[/tex]

Let's analyze these step by step:

1. Understand the base function [tex]\( f_1 \)[/tex]:
- [tex]\( f_1(x) = x \)[/tex] is a linear function with a slope of 1 and a y-intercept of 0.

2. Analyze the transformation applied to [tex]\( f_1 \)[/tex] to get [tex]\( f_2 \)[/tex]:
- [tex]\( f_2(x) = x + 4 \)[/tex] takes every output of [tex]\( f_1(x) = x \)[/tex] and adds 4 to it.

This transformation represents a vertical shift of [tex]\( f_1 \)[/tex]. Specifically:
- Adding 4 to the function [tex]\( f_1(x) \)[/tex] shifts it vertically upwards by 4 units.

Conclusion:
- [tex]\( f_2 \)[/tex] is the function [tex]\( f_1 \)[/tex] moved 4 units up.

Thus, the correct answer is:
- Moved 4 units up.