Vertical Translations: f(x) = √x + k
Vertical translations of the square root function, written as f(x) = √x + k, shift the entire graph up or down without changing its shape or domain, a key skill in enVision Algebra 1 Chapter 10 for Grade 11. When k is positive, every output increases by k units — graphing f(x) = √x + 3 means the curve starts at (0, 3) instead of (0, 0). When k is negative, the graph shifts down by |k| units — f(x) = √x - 2 starts at (0, -2). Adding the constant directly changes every y-value while preserving the shape and left-to-right behavior of the original square root curve.
Key Concepts
The graph of $f(x) = \sqrt{x} + k$ shifts the graph of $f(x) = \sqrt{x}$ vertically $k$ units. If $k 0$, shift the square root function vertically up $k$ units. If $k < 0$, shift the square root function vertically down $|k|$ units.
Common Questions
What does adding k to √x do to the graph?
It shifts the entire graph of f(x) = √x vertically by k units. If k > 0 the graph moves up; if k < 0 it moves down. The shape and domain remain unchanged.
Where does the graph of f(x) = √x + 3 start?
It starts at the point (0, 3) because adding 3 raises the starting point from (0, 0) up by 3 units.
What is the starting point of f(x) = √x - 2?
The graph starts at (0, -2), shifted 2 units below the origin, because k = -2 lowers every output by 2.
Does a vertical translation change the domain of √x?
No. The domain stays x ≥ 0 for any vertical translation. Only the range changes: from [k, ∞) instead of [0, ∞).
How is f(x) = √x + 1 different from f(x) = √(x + 1)?
Adding 1 outside the square root (f(x) = √x + 1) shifts the graph up 1 unit. Adding 1 inside (f(x) = √(x+1)) shifts the graph left 1 unit — a horizontal translation.