What is the starting point of (f(x)=\sqrt{x+1}-3)?

Correct answer: ((-1,-3))

The direct answer is A: (-1,-3). Start with the basic graph y=√x, whose starting point is (0,0). In y=√(x+1)-3, replacing x by x+1 shifts the graph one unit left, so the starting x-coordinate becomes -1. Subtracting 3 outside the root shifts every point three units down, so the y-coordinate becomes -3. Therefore the starting point is (-1,-3). Direct verification is also simple: the square root begins when its radicand is zero, so x+1=0 gives x=-1; then y=√0-3=-3. Option A is correct. Option B, (1,-3), would correspond to √(x-1)-3, a shift right. Option C, (-1,3), has the correct horizontal location but the wrong vertical shift; the graph moves down, not up. Option D, (0,-3), ignores the left shift caused by x+1. The domain begins at x=-1 and the range begins at y=-3. Memory cue: inside plus 1 means left 1, outside minus 3 means down 3.