Practice Maths

Square Root Functions

Key Terms

The basic square root function is y = √x. It is the top half of the horizontal parabola y² = x.
Domain: x ≥ 0   Range: y ≥ 0   Starts at the endpoint (0, 0).
The general form is y = a√(x − h) + k:
    — a > 0: opens right and up; a < 0: reflects in x-axis (opens right and down)
    — |a| > 1: vertical stretch; |a| < 1: vertical compression
    — h: horizontal shift (right if h > 0, left if h < 0)
    — k: vertical shift (up if k > 0, down if k < 0)
Endpoint of y = a√(x − h) + k is always (h, k).
Domain is always x ≥ h (for real a√, need x − h ≥ 0).
FunctionEndpointDomainRange
y = √x(0, 0)x ≥ 0y ≥ 0
y = √(x − 3)(3, 0)x ≥ 3y ≥ 0
y = √(x + 2) − 1(−2, −1)x ≥ −2y ≥ −1
y = 2√(x − 1) + 3(1, 3)x ≥ 1y ≥ 3
y = −√x + 4(0, 4)x ≥ 0y ≤ 4

Graph of y = √x (blue) and y = 2√(x − 1) + 1 (orange)

x y 1 4 1 2 y=√x (1,1) y=2√(x−1)+1
Hot Tip To find the equation of a square root function from its graph, first read the endpoint (h, k). Then substitute another point to find a. For example, if endpoint is (2, 3) and it passes through (6, 7): 7 = a√(6−2)+3 → 4 = 2a → a = 2. Equation: y = 2√(x−2)+3.

Worked Example 1 — Describing transformations

Question: Describe the transformations from y = √x to y = 3√(x + 2) − 4. State the domain and range.

• Horizontal shift left 2 (replace x with x + 2)

• Vertical dilation by factor 3 (multiply by 3)

• Vertical shift down 4

Endpoint: (−2, −4)   Domain: x ≥ −2   Range: y ≥ −4

Worked Example 2 — Finding the equation from a graph

Question: A square root function has endpoint (1, −2) and passes through (5, 2). Find the equation.

Start with: y = a√(x − 1) − 2

Substitute (5, 2): 2 = a√(5−1) − 2 → 4 = 2a → a = 2

Equation: y = 2√(x − 1) − 2

Why √x Only Works for x ≥ 0

The square root function √x asks: “what number, when multiplied by itself, gives x?” For x = 9, the answer is 3 (since 3² = 9). But there are actually two numbers whose square is 9: both 3 and −3. By mathematical convention, the symbol √x always refers to the non-negative (principal) square root. So √9 = 3, never −3.

For negative inputs, no real number squared gives a negative result. (Even (−3)² = 9, not −9.) So √x is undefined for x < 0 in the real number system. This is why the domain of y = √x is restricted to x ≥ 0 — it is not an arbitrary restriction but a consequence of what square roots mean.

Shape of y = √x and Its Connection to y = x²

The graph of y = √x starts at the origin and increases, but more and more slowly. This contrasts with y = x², which starts at the origin and increases more and more rapidly. In fact, these two functions are inverses of each other (when x ≥ 0): applying √x to x² gives x back, and vice versa.

Geometrically, the graph of y = √x is the reflection of y = x² (for x ≥ 0) in the line y = x. This symmetry between a function and its inverse is fundamental: every point (a, b) on y = x² corresponds to the point (b, a) on y = √x. The curve grows as x increases but with diminishing returns: from x = 0 to x = 1 the function increases by 1 unit, but from x = 1 to x = 4 it only increases by another 1 unit (from 1 to 2), even though x has increased by 3.

Transformations: Building Intuition for Each Parameter

In y = a√(x − h) + k, each parameter has a precise geometric effect:

The shift h moves the starting point (endpoint) horizontally. For y = √(x − 3), the expression under the root is non-negative when x − 3 ≥ 0, so x ≥ 3. The endpoint moves from (0, 0) to (3, 0) — a shift right by 3. If h is negative (i.e., y = √(x + 2)), the domain shifts LEFT and starts at x = −2.

The shift k moves the curve vertically: the entire graph, including the endpoint, shifts up or down. The endpoint for y = √(x − h) + k is always at (h, k).

The coefficient a stretches (|a| > 1) or compresses (0 < |a| < 1) the graph vertically. A negative a reflects the curve in the x-axis, making it decrease from its endpoint instead of increase. The domain remains x ≥ h, but the range becomes y ≤ k instead of y ≥ k.

Finding Domain and Range Precisely

The domain of y = a√(x − h) + k is found by solving x − h ≥ 0, giving x ≥ h. The range depends on the sign of a: if a > 0, then √(x−h) ≥ 0, so y = a√(x−h)+k ≥ 0+k = k. Range: y ≥ k. If a < 0, then a√(x−h) ≤ 0, so y ≤ k. Range: y ≤ k.

A critical subtlety: students often confuse the sign in the domain. For y = √(x − 5), the domain is x ≥ 5 (solve x − 5 ≥ 0). For y = √(x + 3), the domain is x ≥ −3 (solve x + 3 ≥ 0). The domain threshold is where the expression under the root equals zero, not where it is positive.

Connection to Inverse Functions

Square root functions are the inverses of quadratic functions (restricted to non-negative values). The graph of y = √(x − h) + k is the reflection of the upper half of y = (x − k)² + h in the line y = x. Understanding this connection gives deeper insight: finding the inverse of a quadratic function always produces a square root function, and vice versa. In Year 12, this relationship becomes especially important when studying inverse functions.

Exam Tip: The domain of y = √(x − h) is x ≥ h, NOT x ≥ −h. Solve the inequality x − h ≥ 0 algebraically: add h to both sides to get x ≥ h. For y = √(x + 4), the domain is x ≥ −4 (not x ≥ 4). Many students incorrectly negate the sign when reading the domain from the equation.
Exam Tip: If you are working through an algebra problem and obtain a negative number under a square root in your final answer, this signals an error. Check whether you have restricted the domain correctly, or whether you have made an algebra mistake. A negative radicand means the expression is undefined for real numbers, which should cause you to review your working.

Mastery Practice

See Answers ➔
  1. Fluency

    State the domain, range and endpoint of each square root function.

    1. (a) y = √(x − 4)
    2. (b) y = √(x + 3) + 2
    3. (c) y = −√x + 5
    4. (d) y = 4√(x − 1) − 3
  2. Fluency

    Evaluate each function at the given x-value.

    1. (a) f(x) = √(2x + 1) at x = 4
    2. (b) g(x) = 3√(x − 2) + 1 at x = 11
    3. (c) h(x) = −2√x + 6 at x = 9
  3. Fluency

    Describe the transformations from y = √x to each of the following. State domain and range.

    1. (a) y = √(x − 5) + 3
    2. (b) y = 2√(x + 1)
    3. (c) y = −√(x − 2) + 4
  4. Fluency

    Find the x-intercept (if it exists) of each square root function.

    1. (a) y = √x − 3
    2. (b) y = 2√(x + 4) − 6
    3. (c) y = −√(x − 1) + 2
    4. (d) y = √(x − 9) + 3 (no x-intercept — explain why)
  5. Understanding

    Finding the equation of a square root function.

    Method: Identify the endpoint (h, k), write y = a√(x−h)+k, then substitute another known point to find a.
    1. (a) Endpoint (0, 0), passes through (4, 6).
    2. (b) Endpoint (2, 1), passes through (6, 5).
    3. (c) Endpoint (−1, 3), passes through (3, −1). Note: a will be negative.
  6. Understanding

    Solving equations involving square root functions.

    Note: To solve √(x−h) = c, square both sides: x−h = c². Always check the solution is in the domain.
    1. (a) √(2x + 3) = 5
    2. (b) 3√(x − 1) + 2 = 8
    3. (c) √(x + 5) = x − 1 (square both sides and check solutions)
  7. Understanding

    Connecting square root functions and parabolas.

    1. (a) The function y = √(x − 2) + 1 is the top half of a horizontal parabola. Write the equation of the full horizontal parabola (as a relation in the form (y − k)² = x − h).
    2. (b) Write the equation of the bottom half of the same parabola as a function.
    3. (c) A square root function y = a√(x) has the property that f(9) = 6. Find a and determine f(25).
  8. Understanding

    Range with restricted domain.

    1. (a) For f(x) = √(x + 4) defined on −4 ≤ x ≤ 12, state the range.
    2. (b) For g(x) = −2√x + 8 defined on 0 ≤ x ≤ 16, state the range.
    3. (c) Find the value of x where f(x) from part (a) equals g(x) from part (b). Show full working.
  9. Problem Solving

    Speed of a pendulum.

    Challenge. The period T (in seconds) of a pendulum is given by T = 2π√(L/9.8), where L is the length in metres.
    1. (a) Find the period of a pendulum of length 2.45 m. Give your answer to 2 decimal places.
    2. (b) What length is needed for a period of exactly 2 seconds? Give an exact answer in terms of π.
    3. (c) Describe how the period changes as L increases. Is this relationship linear? Explain using the shape of the square root function.
  10. Problem Solving

    Modelling with square root functions.

    Challenge. The speed v (m/s) of a skateboarder rolling down a ramp from a height of h metres is modelled by v = √(20h).
    1. (a) Find the speed when h = 5 m. Give your answer in simplified exact form.
    2. (b) Sketch a graph of v against h for 0 ≤ h ≤ 5. Label the endpoint and one other point.
    3. (c) What height would be needed to reach a speed of 15 m/s? Show working and interpret your answer in context.