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Math · Algebra 2

Chapter 5: Radicals and Rational Exponents

Graphing Radical Functions

Half a parabola, lying on its side.

Lesson
3
Time
About 21 minutes
0 of 12 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

The graph of y = √x starts at the origin and rises to the right, forever slower. It is the top half of a sideways parabola.

Domain

A square root refuses negative inputs, so the domain of y = √x is x ≥ 0 and the range is y ≥ 0.

Shifted domains

For y = √(x − 3) the inside must be at least zero, so the domain is x ≥ 3 and the graph starts there.

Same rules as everything else

The transformations are the ones from Chapter 1. Outside is vertical, inside is horizontal and backwards.

Cube roots differ

A cube root accepts negatives, since (−2)³ = −8. Its domain and range are all real numbers, and its graph passes through the origin in an S shape.

Why the difference

Squaring destroys the sign, so a square root cannot recover one. Cubing keeps it, so a cube root can.

Half a parabola on its side

y = √x is the upper half of the sideways parabola x = y². The square root symbol means the principal root, so only the non-negative half appears — otherwise it would fail the vertical line test.

The domain is restricted

y = √(x − 3) requires x ≥ 3, since the inside cannot be negative. Solving the inside-is-non-negative inequality gives the domain, and it determines where the graph starts.

Transformations work as usual

y = −2√(x − 3) + 1 shifts right 3, stretches by 2, reflects and raises 1. The general transformation rules apply to radical graphs exactly as to any other, which is the point of learning them generally.

Cube roots behave differently

y = ∛x accepts negative inputs, so its domain is all reals and its graph runs through all four relevant quadrants. Odd roots have no domain restriction, which is the same asymmetry seen with cube roots in Grade 8.

Step 2: Try It Yourself

Tap and try it out.

Change the coefficient and follow the starting point, then read where the curve is allowed to exist.
-8-8-6-6-4-4-2-222446688
y = 1√x + 0
  • Point(4, 2)

Step 3: Watch an Example

One step at a time.

Watch Yusuf Describe y = √(x + 4) − 2

Yusuf needs the starting point, the domain and the range.

  1. Step 1

    The inside must be at least zero, so x + 4 ≥ 0 and the domain is x ≥ −4.

Step 4: Your Turn

Practice makes it stick.

The Domain

Problem 1 of 2

For y = √(x − 5), what is the smallest allowed value of x?

The Pendulum

Problem 2 of 2

A pendulum period is T = 2√L seconds, with L in metres. What is T when L = 9?

seconds

Where the Curve Lives

1 of 8

y = √(x − 2). Smallest allowed x?

2 of 8

y = √(x + 7). Smallest allowed x?

3 of 8

y = √x + 3. What is y when x = 16?

4 of 8

y = √(x − 1) + 5. What is the smallest value of y?

5 of 8

What is the cube root of −27?

6 of 8

y = 2√x. What is y when x = 25?

7 of 8

Sort each function by whether negatives are allowed as inputs.

Tap something to move it.

  • Empty
  • Empty

8 of 8

y = √(2x − 6). Smallest allowed x?

Step 5: Quick Check

Show what you know.

Question 1 of 2

y = √(x − 8). What is the smallest allowed value of x?

Question 2 of 2

Why does a cube root accept negative inputs?

What You Learned

  • y = √x starts at the origin and rises ever more slowly.
  • Set the inside to zero to find where a radical graph begins.
  • Cube roots accept every real number; square roots do not.