The square root and cube root are two of the Parent Functions, and you transform them with the same moves as any other parent. What makes them special is where they start (square root) and where they are centered (cube root). You will use nth Roots to read their domains.
Key idea
$f(x) = a\sqrt{x - h} + k$ starts at the point $(h, k)$; its domain is $[h, \infty)$ and its range is $[k, \infty)$ if $a > 0$ or $(-\infty, k]$ if $a < 0$. $g(x) = a\sqrt[3]{x - h} + k$ has domain and range all real numbers and is symmetric about its center $(h, k)$.
Why it works
A square root needs a nonnegative input, so the graph starts exactly where the inside is 0 and goes one way from there. Its outputs start at 0 and grow, so after stretching by $a$ and shifting by $k$ they start at $k$ and go up (or down, if $a < 0$). A cube root accepts every input and gives every output, so its graph runs across the whole plane.
Worked examples
Worked example: Domain, range and starting point
Describe $f(x) = -2\sqrt{x + 3} + 4$.
- Inside $x + 3 \ge 0$, so $x \ge -3$: domain $[-3, \infty)$. The graph starts at $(-3, 4)$.
- $\sqrt{x + 3} \ge 0$, and multiplying by $-2$ makes it $\le 0$; adding 4 gives $f(x) \le 4$.
Answer: Starts at $(-3, 4)$, goes down and to the right; domain $[-3, \infty)$, range $(-\infty, 4]$.
Worked example: A reversed square root
Find the domain of $f(x) = \sqrt{6 - 2x}$.
- Require $6 - 2x \ge 0$, so $-2x \ge -6$.
- Dividing by $-2$ flips the inequality: $x \le 3$.
Answer: $(-\infty, 3]$
Worked example: A cube root
Describe $g(x) = \sqrt[3]{x - 1} + 2$.
- Shift $\sqrt[3]{x}$ right 1 and up 2. Its center moves from $(0, 0)$ to $(1, 2)$.
- Cube roots accept all inputs, so the domain and range are all real numbers.
Answer: Center $(1, 2)$; domain and range $(-\infty, \infty)$.
Common mistake
Giving a cube root function a restricted domain, as if it were a square root. $\sqrt[3]{x - 1}$ is defined for every $x$, including $x = -7$, where it equals $-2$. Only even roots require a nonnegative inside.