What is the starting point of the graph of the function \(y=5-\sqrt{x+2}\)?

Correct answer: \((-2,5)\)

For a square-root function, the radicand must satisfy \(x+2\geq 0\). The graph begins where the radicand is zero: \(x+2=0\), giving \(x=-2\). At this value, \(y=5-\sqrt{0}=5\), so the starting point is \((-2,5)\). Option B incorrectly uses \(x=2\), which is not the endpoint of the domain. Exam tip: a graph of \(\sqrt{x-a}\) starts at \((a,0)\), while an added constant shifts it vertically.