Horizontal Translations: f(x) = √(x - h)
Horizontal translations of square root functions with the form f(x) = sqrt(x - h) are covered in Grade 11 Algebra 1 enVision Chapter 10. Subtracting h inside the square root shifts the graph h units to the right if h > 0, or |h| units to the left if h < 0. The starting point moves from (0,0) to (h, 0), and the domain shifts to x >= h. For example, f(x) = sqrt(x - 3) shifts right 3 units with starting point (3, 0), while f(x) = sqrt(x + 4) rewrites as sqrt(x - (-4)), shifting left 4 units. The shift is counterintuitive: subtracting moves right.
Key Concepts
The graph of $f(x) = \sqrt{x h}$ shifts the graph of $f(x) = \sqrt{x}$ horizontally $h$ units. If $h 0$, shift the graph horizontally right $h$ units. If $h < 0$, shift the graph horizontally left $|h|$ units.
Common Questions
How does f(x) = sqrt(x - 3) differ from f(x) = sqrt(x)?
It shifts the graph 3 units to the right. The starting point moves from (0,0) to (3,0), and the domain becomes x >= 3.
Why does subtracting h shift the graph to the right, not left?
To get the same output as sqrt(x), you need the input to be h units larger. This means the graph starts h units further right than the original.
How do you graph f(x) = sqrt(x + 4)?
Rewrite as sqrt(x - (-4)). This shifts the graph left 4 units. Starting point moves to (-4, 0), and domain is x >= -4.
What happens to the domain when you apply a horizontal shift?
The domain shifts by h units. For f(x) = sqrt(x - h), the domain is x >= h instead of x >= 0.
Does a horizontal translation change the shape of the square root curve?
No. The shape remains identical. Only the starting position changes. All points on the original graph shift horizontally by h units.
What is the starting point of g(x) = sqrt(x - 1)?
(1, 0). The graph of sqrt(x) shifted right 1 unit starts at (1, 0) instead of the origin.