Answer
(a)
The graph of function $f(x)=|x-2|$ is symmetric about the line $x=2$. See graph below.
(b)
Let $f(x)$ be symmetric about line $x=a$, which means $f(a-x)=f(a+x)$.
Now, for function $g(x)=f(x+a)$, we have $g(-x)=f(-x+a)=f(a-x)=f(a+x)=f(x+a)=g(x)$ using the symmetry of $f(x)$ about line $x=a$. Since, $g(-x)=g(x)$, so, $g(x)$ is even function. Hence, proved.
Work Step by Step
(a)
Note that for the function $f(x)=|x-2|$, we have $f(2-x)=|(2-x)-2|=|x|=|(x+2)-2|=f(2+x)$, so, the function $f(x)$ is symmetric function about line $x=2$.
(b)
This part can be done as follows:
Step 1: Write the property we get from symmetry of function $f(x)$ about line $x=a$.
Step 2: Evaluate $g(-x)$ and if $g(-x)=g(x)$, then, the proof is done and function $g(x)$ is even function.
