Algebra 2
Radical functions
Graphing radical functions
20 practice questions
0 video lessons
Theory + worked examples
Graphing Radical Functions
California Algebra 2 • Standard F-IF.7b • Radical Functions
Graphing Radical Functions is a topic in Radical Functions in the California Common Core State Standards. It is aligned to Standard F-IF.7b, which requires students to graph square root and cube root functions.
The square root graph starts at a point with domain \(x\ge0\); the cube root graph is defined for all real numbers.
Theory
Radical parent functions come in two kinds:
- Square root \(\sqrt{x}\): domain \([0,\infty)\), starts at the origin.
- Cube root \(\sqrt[3]{x}\): domain all reals, passes through the origin.
They shift with \(\sqrt{x-h}+k\), moving the start point to \((h,k)\).
Even roots need a non-negative radicand; odd roots accept any real number.
\(\sqrt{x-2}+1\) starts at \((2,1)\).
The cube root is defined for all reals.
Radical parents:
\[f(x)=\sqrt{x-h}+k,\qquad f(x)=\sqrt[3]{x-h}+k\]
Set the radicand \(\ge 0\) to find a square-root domain.
How to graph a radical
- Find the domain (radicand \(\ge0\) for even roots).
- Locate the start point \((h,k)\).
- Plot a few points to the right.
- Draw the increasing curve.
Example 1 — Domain of a square root
Find the domain of \(f(x)=\sqrt{x-3}\).
Solution
The radicand must be \(\ge 0\).
| \(x-3\) | \(\ge\) | \(0\) |
| \(x\) | \(\ge\) | \(3\) |
Example 2 — Describe the transformation
Describe \(f(x)=\sqrt{x-2}+1\).
Solution
Shift the parent right \(2\), up \(1\).
| \(\text{start point}\) | \(=\) | \((2,1)\) |
Example 3 — Domain and range of √x
Give the domain and range of \(f(x)=\sqrt{x}\).
Solution
Both are non-negative.
| \(\text{domain}\) | \(=\) | \([0,\infty)\) |
| \(\text{range}\) | \(=\) | \([0,\infty)\) |
Example 4 — Cube root domain
What is the domain of \(f(x)=\sqrt[3]{x}\)?
Solution
An odd root accepts negatives too.
| \(\text{domain}\) | \(=\) | \((-\infty,\infty)\) |
Common pitfalls
Even-root domains are restricted; odd roots are not.
The start point is \((h,k)\) for \(\sqrt{x-h}+k\).
\((x-h)\) shifts right, as always.
Frequently asked questions
What is the domain of \(\sqrt{x}\)?
\([0,\infty)\) — the radicand must be non-negative.
What is the domain of \(\sqrt[3]{x}\)?
All real numbers.
Where does \(\sqrt{x-2}+1\) start?
At the point \((2,1)\).
Why do square roots have restricted domains?
You can't take an even root of a negative number in the reals.
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