State the domain and range. (Enter your answers using interval notation.)
\[
h(x)=\log _{2}(x-2)+2
\]
domain
range
State the $x$ - anday-intercepts, if they exist. (If an answer does not exist, enter DNE.)
$x$-intercept $(x, y)=$
$y$-intercept $(x, y)=$

Step 1 ：The domain of a logarithmic function is the set of all real numbers for which the argument of the logarithm is positive. In this case, the argument of the logarithm is \(x-2\), so the domain is the set of all real numbers \(x\) such that \(x-2>0\). This simplifies to \(x>2\). In interval notation, this is \((2, \infty)\).

Step 2 ：The range of a logarithmic function is the set of all real numbers. So, the range of this function is \((-\infty, \infty)\).

Step 3 ：The \(x\)-intercept of a function is the point where the function crosses the \(x\)-axis. This occurs when \(y=0\). So, to find the \(x\)-intercept, we need to solve the equation \(h(x)=0\) for \(x\).

Step 4 ：The \(y\)-intercept of a function is the point where the function crosses the \(y\)-axis. This occurs when \(x=0\). However, since the domain of this function is \((2, \infty)\), there is no \(y\)-intercept.

Step 5 ：\(x\) intercept is \(\left(\frac{9}{4}, 0\right)\)

Step 6 ：\(y\) intercept does not exist

Step 7 ：\(\boxed{\text{The domain of the function } h(x)=\log _{2}(x-2)+2 \text{ is } (2, \infty)}\)

Step 8 ：\(\boxed{\text{The range of the function } h(x)=\log _{2}(x-2)+2 \text{ is } (-\infty, \infty)}\)

Step 9 ：\(\boxed{\text{The } x \text{-intercept of the function } h(x)=\log _{2}(x-2)+2 \text{ is } \left(\frac{9}{4}, 0\right)}\)

Step 10 ：\(\boxed{\text{The function } h(x)=\log _{2}(x-2)+2 \text{ does not have a } y \text{-intercept}}\)