Radical Functions

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Radical functions have the variable inside a root, like f(x) = \sqrt{x} or g(x) = \sqrt[3]{x}. A square root graph starts at a point and curves to one side. A cube root graph is an S shape through a center point. To graph y = a\sqrt{x - h} + k, start at (h, k) and use a few easy points.

Before this lesson: You should know Parent Functions, Transformations of Functions, Domain and Range and nth Roots.

The Two Parent Graphs

The square root function y = \sqrt{x} starts at (0, 0) and rises slowly to the right. It has no graph for negative x, because the square root of a negative number is not real. Easy points to plot are (0, 0), (1, 1), (4, 2) and (9, 3).

The cube root function y = \sqrt[3]{x} works for every x, since negative numbers have cube roots. Its graph passes through (-8, -2), (-1, -1), (0, 0), (1, 1) and (8, 2).

FunctionDomainRangeKey point
y = \sqrt{x}x \ge 0y \ge 0Starts at (0, 0)
y = \sqrt[3]{x}All real numbersAll real numbersCenter at (0, 0)

Transformations of Radical Functions

y = a\sqrt{x - h} + k \qquad y = a\sqrt[3]{x - h} + k

These follow the same rules as other functions (see transformations of functions). The graph moves right h and up k. If |a| \gt 1 it stretches, and if a \lt 0 it flips over the x-axis. For a square root, (h, k) is the starting point.

Coordinate grid with the parent graph y = √x as a dashed gray curve starting at (0, 0), and the graph of g(x) = √(x − 2) + 1 as a solid blue curve starting at (2, 1) and passing through (3, 2) and (6, 3). An arrow shows the shift: right 2 and up 1.
Shifting y = √x right 2 and up 1 moves the starting point from (0, 0) to (2, 1).

Worked Examples

Example 1: Domain and range

Find the domain and range of f(x) = \sqrt{x + 4}.

  1. The radicand must not be negative: x + 4 \ge 0
  2. Subtract 4 from both sides (see solving inequalities): x \ge -4
  3. A principal square root is never negative, so f(x) \ge 0.

Answer: Domain x \ge -4, or [-4, \infty). Range y \ge 0, or [0, \infty) (see domain and range).

Example 2: A shifted square root

Graph g(x) = \sqrt{x - 2} + 1 and give its domain and range.

  1. Compare with y = a\sqrt{x - h} + k: h = 2 and k = 1, so the graph starts at (2, 1).
  2. Pick x-values that make x - 2 a perfect square: 2, 3, 6 and 11.
  3. g(3) = \sqrt{3 - 2} + 1 = \sqrt{1} + 1 = 2
  4. g(6) = \sqrt{6 - 2} + 1 = \sqrt{4} + 1 = 3
  5. g(11) = \sqrt{11 - 2} + 1 = \sqrt{9} + 1 = 4
  6. Plot (2, 1), (3, 2), (6, 3) and (11, 4), then draw a smooth curve to the right.

Answer: The graph above. Domain x \ge 2; range y \ge 1.

Example 3: A reflection and a stretch

Graph h(x) = -2\sqrt{x} + 3.

  1. The start point is (0, 3), since h = 0 and k = 3.
  2. The −2 stretches by 2 and flips the curve downward.
  3. h(1) = -2\sqrt{1} + 3 = -2 + 3 = 1
  4. h(4) = -2\sqrt{4} + 3 = -4 + 3 = -1
  5. h(9) = -2\sqrt{9} + 3 = -6 + 3 = -3
  6. Plot (0, 3), (1, 1), (4, -1) and (9, -3), and draw a curve falling to the right.

Answer: Domain x \ge 0; range y \le 3, because the curve starts at height 3 and goes down.

Example 4: A cube root function

Graph f(x) = \sqrt[3]{x + 1} - 2.

  1. The center point moves from (0, 0) to (-1, -2): left 1, down 2.
  2. Pick x-values that make x + 1 a perfect cube: −9, −2, −1, 0 and 7.
  3. f(-9) = \sqrt[3]{-8} - 2 = -2 - 2 = -4
  4. f(-2) = \sqrt[3]{-1} - 2 = -1 - 2 = -3
  5. f(-1) = \sqrt[3]{0} - 2 = -2
  6. f(0) = \sqrt[3]{1} - 2 = 1 - 2 = -1
  7. f(7) = \sqrt[3]{8} - 2 = 2 - 2 = 0

Answer: An S-shaped curve through (-9, -4), (-2, -3), (-1, -2), (0, -1) and (7, 0). Domain and range are all real numbers.

Example 5: Word problem with falling time

If an object is dropped from a height of h feet, the time it takes to hit the ground is about t = \frac{\sqrt{h}}{4} seconds. How long does it take to fall 100 feet? Why is the domain h \ge 0?

  1. Substitute h = 100: t = \frac{\sqrt{100}}{4}
  2. Take the square root (see square root): t = \frac{10}{4}
  3. Divide: t = 2.5
  4. A height can’t be negative, and \sqrt{h} is not real for negative h, so the domain is h \ge 0.

Answer: About 2.5 seconds. Notice that 4 times the height (400 feet) only doubles the time, to 5 seconds. That is the slow growth of a square root.

Common Mistakes

MistakeExample of the errorFix
Shifting the wrong way\sqrt{x - 2} starts at x = -2x - 2 = 0 at x = 2, so it starts at x = 2.
Giving a square root graph a left sideDrawing \sqrt{x} for negative x\sqrt{x} is not real for x \lt 0. The graph starts at the origin.
Limiting the cube root’s domainDomain of \sqrt[3]{x} is x \ge 0Cube roots of negatives are real, so the domain is all real numbers.
Wrong range after a flipRange of -2\sqrt{x} + 3 is y \ge 3The flip makes the curve go down from 3: y \le 3.
Picking hard x-valuesUsing x = 5 for \sqrt{x - 2}Pick values that make the radicand a perfect square, like 3, 6 and 11.

Practice Problems

  1. Find the domain of f(x) = \sqrt{x - 5}.
  2. Find the domain and range of f(x) = \sqrt{3 - x}.
  3. Find the domain of f(x) = \sqrt[3]{x - 7}.
  4. Describe how g(x) = \sqrt{x + 3} - 4 is shifted from y = \sqrt{x}, and give its starting point.
  5. Make a table for f(x) = 3\sqrt{x} at x = 0, 1, 4, 9.
  6. Graph f(x) = -\sqrt{x - 1} using three points, and give its range.
  7. Find f(-8), f(0) and f(8) for f(x) = \sqrt[3]{x} + 2.
  8. Write the equation of y = \sqrt{x} shifted right 4 and up 2.
  9. Find the range of f(x) = 2\sqrt{x + 1} - 5.
  10. Use t = \frac{\sqrt{h}}{4} to find the falling time from 256 feet and from 36 feet.
Show answers
  1. Radicand not negative: x - 5 \ge 0
    Add 5: x \ge 5
    Answer: [5, \infty)
  2. Radicand not negative: 3 - x \ge 0
    Add x: 3 \ge x, so x \le 3
    A square root is never negative: y \ge 0
    Answer: domain (-\infty, 3], range [0, \infty)
  3. A cube root works for any real number
    Answer: all real numbers
  4. x + 3 means left 3; -4 means down 4
    Answer: left 3, down 4, starting at (-3, -4)
  5. f(0) = 3\sqrt{0} = 0
    f(1) = 3\sqrt{1} = 3
    f(4) = 3\sqrt{4} = 3 \cdot 2 = 6
    f(9) = 3\sqrt{9} = 3 \cdot 3 = 9
    Answer: (0, 0), (1, 3), (4, 6), (9, 9)
  6. Start: (1, 0)
    f(2) = -\sqrt{1} = -1
    f(5) = -\sqrt{4} = -2
    Points (1, 0), (2, -1), (5, -2); the curve falls to the right
    Answer: range y \le 0
  7. f(-8) = \sqrt[3]{-8} + 2 = -2 + 2 = 0
    f(0) = \sqrt[3]{0} + 2 = 2
    f(8) = \sqrt[3]{8} + 2 = 2 + 2 = 4
  8. Right 4 replaces x with x - 4; up 2 adds 2
    Answer: f(x) = \sqrt{x - 4} + 2
  9. 2\sqrt{x + 1} is never negative, so it is at least 0
    Subtract 5: 2\sqrt{x + 1} - 5 \ge -5
    Answer: y \ge -5
  10. t = \frac{\sqrt{256}}{4} = \frac{16}{4}
    Divide: 4 seconds
    t = \frac{\sqrt{36}}{4} = \frac{6}{4}
    Divide: 1.5 seconds

FAQ

What is a radical function?

It is a function with the variable inside a root, such as f(x) = \sqrt{x - 3} or g(x) = \sqrt[3]{2x}.

How do you find the domain of a square root function?

Set the radicand greater than or equal to 0 and solve. For \sqrt{x + 4}, x + 4 \ge 0, so x \ge -4.

What is the domain of a cube root function?

All real numbers. You can take the cube root of any number, positive, negative or zero.

How do you graph y = √(x − h) + k?

Start at the point (h, k). Then add 1, 4 and 9 to h to get easy x-values, find their y-values, and connect the points with a smooth curve.

Summary

  • y = \sqrt{x} starts at (0, 0), has domain x \ge 0 and range y \ge 0.
  • y = \sqrt[3]{x} is an S shape with domain and range all real numbers.
  • In y = a\sqrt{x - h} + k, (h, k) is the start point; a stretches and, if negative, flips the graph.
  • For a square root, set the radicand \ge 0 to find the domain.
  • Choose x-values that make the radicand a perfect square or cube.

Previous lesson: Rational Exponents

Next lesson: Solving Radical Equations

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