How can I tell whether a graph transformation changes a function's domain or range?

Keep inputs and outputs in separate columns. The domain is the set of allowed x-values; the range is the set of resulting y-values. A transformation that changes which x-values work can change the domain. One that changes the y-values produced can change the range. Some transformations do both.

For a square-root function, first check the expression under the root: it must be nonnegative for the output to be real. For the parent function √x, both domain and range are [0, ∞). [S1] A horizontal reflection gives √(−x): its domain becomes (−∞, 0], while its range remains [0, ∞). A vertical reflection instead changes the range to (−∞, 0]. [S1] These examples show why you should check inputs and outputs separately, rather than assuming every transformation affects both.

Take f(x) = √(x − 2). For the root to be real, x − 2 must be at least zero, so x ≥ 2. Its domain is [2, ∞), and its range is [0, ∞). [S2]

Now shift the graph one unit right: h(x) = f(x − 1) = √(x − 3). The new radicand requires x − 3 ≥ 0, so the domain is [3, ∞). The smallest output is still zero, and outputs increase from there, so the range stays [0, ∞). This horizontal shift changes the allowed x-values but not the set of y-values.

For comparison, shift the original graph up three units: v(x) = f(x) + 3 = √(x − 2) + 3. The domain stays [2, ∞), because the expression under the root is unchanged. The outputs are now at least 3, so the range is [3, ∞). This vertical shift changes outputs but not allowed inputs. OpenStax gives this same domain-and-range pattern for a vertical shift of a restricted square-root function. [S2]

Track each part of the expression. Using the same starting function, shift it right one and up three: g(x) = f(x − 1) + 3 = √(x − 3) + 3.

First find the domain from the radicand: x − 3 ≥ 0, so x ≥ 3. The domain is [3, ∞). Then find the range: √(x − 3) starts at zero and increases, and adding 3 moves every output up three, so the range is [3, ∞).

A useful check is the boundary value. At x = 3, g(3) = 3, so the graph begins at (3, 3). That confirms the included endpoints. This combined transformation moves the domain boundary horizontally and the range boundary vertically.

Yes. Start with the original domain and range if they are known, but still check the transformed formula—especially when it contains a root or denominator. For a shift written as f(x − h) + k, the graph moves right by h and up by k (negative values mean left or down). For f(x) = √(x − 2), the original domain is [2, ∞) and range is [0, ∞). In f(x − 1) + 3, shifting right one changes the domain to [3, ∞), and shifting up three changes the range to [3, ∞).

Copyable working: Original: f(x) = √(x − 2) Original domain: x − 2 ≥ 0, so [2, ∞) Original range: [0, ∞) Transformed: g(x) = f(x − 1) + 3 = √(x − 3) + 3 New domain: x − 3 ≥ 0, so [3, ∞) New range: [3, ∞)

This procedure illustrates shifts. Other transformations, such as reflections or stretches, may affect endpoints or output values differently. Check the actual formula rather than relying on a shortcut.