Asked by Christian
The graph shows a square root function.
A curve starts on the y-axis in quadrant 4 and goes into quadrant 1. The curve opens down and to the right.
Use what you know about domain to select all of the following functions that could be the one graphed.
mr001-1.jpg
mr001-2.jpg
mr001-3.jpg
mr001-4.jpg
All Answers 1
Answered by
GPT-5 mini
AI
I don't have the text of the four answer choices (the mr001-*.jpg files), so I can't pick which of those four match. Could you either paste the four functions here or upload the images?
Meanwhile, here is how to decide which functions could match the given graph:
1. The curve starts on the y‑axis at x = 0, so the square‑root radicand must be zero at x = 0. That means the inside of the root should be something like x (or a constant times x with positive constant), not (x − h) with h ≠ 0 and not something like (−x + …) that would make the left endpoint somewhere else.
2. The starting point is in quadrant IV, so y(0) < 0. Thus the constant (vertical shift) k in y = a·√(x) + k must be negative.
3. The curve goes into quadrant I as x increases, so y increases with x (it must cross y = 0). That requires the coefficient a in front of the square root to be positive (a > 0), because √x by itself increases with x; a negative a would make the graph decrease to the right.
4. The graph is a right‑opening square‑root shape (concave down), which is the normal concavity for +√(x) (sign of a affects direction of vertical travel, not the horizontal opening).
So the matching functions are those of the form y = a·√(x) + k with a > 0, k < 0 (and no horizontal shift). Example matches: y = √x − 2, y = 3√x − 1, y = 0.5√x − 4. Non‑matches include any with a ≤ 0, any with a horizontal shift (√(x − 2) etc.), any with radicand making domain not start at x = 0 (like √(−x)), or any without a square root at all.
If you paste the four function options, I’ll mark which of those meet these criteria.
Meanwhile, here is how to decide which functions could match the given graph:
1. The curve starts on the y‑axis at x = 0, so the square‑root radicand must be zero at x = 0. That means the inside of the root should be something like x (or a constant times x with positive constant), not (x − h) with h ≠ 0 and not something like (−x + …) that would make the left endpoint somewhere else.
2. The starting point is in quadrant IV, so y(0) < 0. Thus the constant (vertical shift) k in y = a·√(x) + k must be negative.
3. The curve goes into quadrant I as x increases, so y increases with x (it must cross y = 0). That requires the coefficient a in front of the square root to be positive (a > 0), because √x by itself increases with x; a negative a would make the graph decrease to the right.
4. The graph is a right‑opening square‑root shape (concave down), which is the normal concavity for +√(x) (sign of a affects direction of vertical travel, not the horizontal opening).
So the matching functions are those of the form y = a·√(x) + k with a > 0, k < 0 (and no horizontal shift). Example matches: y = √x − 2, y = 3√x − 1, y = 0.5√x − 4. Non‑matches include any with a ≤ 0, any with a horizontal shift (√(x − 2) etc.), any with radicand making domain not start at x = 0 (like √(−x)), or any without a square root at all.
If you paste the four function options, I’ll mark which of those meet these criteria.
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