This is the original function; y=f(x) with a domain of -3<_ x <_ 3 and a range of 0 <_ y <_ 3.
Now I was asked to find the range and domain of y=f(x-2).
So I said -3<_ x <_ 3 and 1<_ y <_ 5.
But the answer was -1<_ x <_ 5 and 0 <_ y <_ 3.
I don't get it. Am I supposed to keep the same range, not the domain? Is that what it is? Why is my answer wrong?
On
Key Point: For domain we need condition on $x$ and for range we need condition on $y$.
Given that domain of $f(x)$ is $-3 \leq x \leq 3$, it means whatever is written in place of '$?$' in $f(?)$ should lie between $-3$ and and $3$
Now if I write $x-2$ in place of '$?$', then I get
$$-3 \leq x-2 \leq3$$
But for domain we need condition of on x, hence add $2$ in inequality, which gives
$$-3+2 \leq x-2+2 \leq3+2$$
$$-1 \leq x\leq5$$ which is the required domain.
Is it clear now?
The domain is the set of numbers you plug into $f$. Here, the number you plug into $f$ needs to be between $-3$ and $3$. Since the number you're plugging into $f$ is $x-2$, this means you need $-3 \le x-2 \le 3$, which is equivalent to $-1 \le x \le 5$.
The range is the set of values that the function $f$ takes; since $f(x-2)$ is simply a value of $f$ for each input $x$, the range of the new function is the same as the range of the old function.