In Algebra 1 you moved and stretched graphs (Shifting and Stretching Graphs) and found domains and ranges (Domain and Range). Most graphs in Algebra 2 are built from a short list of basic shapes called parent functions. Knowing them by heart works like knowing your times tables: once you see the parent inside a formula, the graph is half drawn.
Key idea
Seven parents to know: $y = x$, $y = x^2$, $y = x^3$, $y = |x|$, $y = \sqrt{x}$, $y = \sqrt[3]{x}$ and $y = \frac{1}{x}$. For each one, learn its shape, three or four anchor points, its domain and its range.
Why it works
Each parent does one thing to $x$, and its shape follows from that one thing. Squaring and taking absolute value erase the sign, so $x^2$ and $|x|$ are mirror-symmetric and never negative. Cubing keeps the sign, so $x^3$ runs from the lower left to the upper right. No real number squares to a negative, so $\sqrt{x}$ starts at the origin and only goes right. And $\frac{1}{x}$ can't use $x = 0$; as $x$ grows, $\frac{1}{x}$ shrinks toward 0, so the graph hugs both axes without touching them.
| Parent | Anchor points | Domain | Range |
|---|---|---|---|
| $y = x$ | $(-1, -1)$, $(0, 0)$, $(1, 1)$ | $(-\infty, \infty)$ | $(-\infty, \infty)$ |
| $y = x^2$ | $(-2, 4)$, $(-1, 1)$, $(0, 0)$, $(1, 1)$, $(2, 4)$ | $(-\infty, \infty)$ | $[0, \infty)$ |
| $y = x^3$ | $(-2, -8)$, $(-1, -1)$, $(0, 0)$, $(1, 1)$, $(2, 8)$ | $(-\infty, \infty)$ | $(-\infty, \infty)$ |
| $y = \lvert x\rvert$ | $(-2, 2)$, $(0, 0)$, $(2, 2)$ | $(-\infty, \infty)$ | $[0, \infty)$ |
| $y = \sqrt{x}$ | $(0, 0)$, $(1, 1)$, $(4, 2)$, $(9, 3)$ | $[0, \infty)$ | $[0, \infty)$ |
| $y = \sqrt[3]{x}$ | $(-8, -2)$, $(-1, -1)$, $(0, 0)$, $(1, 1)$, $(8, 2)$ | $(-\infty, \infty)$ | $(-\infty, \infty)$ |
| $y = \frac{1}{x}$ | $(-1, -1)$, $(\frac{1}{2}, 2)$, $(1, 1)$, $(2, \frac{1}{2})$ | $x \ne 0$ | $y \ne 0$ |
Worked examples
Worked example: Naming a parent from its points
A parent function passes through $(-8, -2)$, $(0, 0)$ and $(27, 3)$. Which one is it?
- The outputs are much smaller than the inputs, and a negative input gives a negative output. That rules out $x^2$, $|x|$ and $\sqrt{x}$.
- Try the cube root: $\sqrt[3]{-8} = -2$ because $(-2)^3 = -8$, and $\sqrt[3]{27} = 3$ because $3^3 = 27$.
- Every point matches, and $y = x$, $y = x^3$ and $y = \frac{1}{x}$ each miss at least one.
Answer: $y = \sqrt[3]{x}$
Worked example: Domain and range in interval notation
Give the domain and range of $y = \sqrt{x}$ and of $y = \frac{1}{x}$.
- $\sqrt{x}$ needs $x \ge 0$, so the domain is $[0, \infty)$. Its outputs are never negative and reach every $y \ge 0$, so the range is $[0, \infty)$.
- $\frac{1}{x}$ refuses only $x = 0$: domain $(-\infty, 0) \cup (0, \infty)$.
- It can never output 0 (a fraction with numerator 1 is never 0), and any other $y$ comes from $x = \frac{1}{y}$: range $(-\infty, 0) \cup (0, \infty)$.
Answer: $\sqrt{x}$: domain and range $[0, \infty)$. $\frac{1}{x}$: domain and range $(-\infty, 0) \cup (0, \infty)$.
Common mistake
Treating the cube root like the square root and saying $\sqrt[3]{x}$ is undefined for negatives. Only even roots need a nonnegative input. Check: $(-2)^3 = -8$, so $\sqrt[3]{-8} = -2$ is a perfectly good real number.