Graph and interpret key features of square-root, cube-root, absolute-value, and piecewise-defined functions, including domain restrictions and points of interest (MA.912.F.1.1, MA.912.AR.4.3).

What this topic is asking

Beyond lines, parabolas, and exponentials, MA.912.F.1 includes other nonlinear functions: square-root, cube-root, absolute-value, and piecewise-defined functions. The B.E.S.T. Algebra 1 EOC asks you to recognize their shapes, state their domains (square roots are restricted), and evaluate them, including choosing the right piece of a piecewise function.

Square-root functions

$f(x) = \sqrt{x}$ begins at $(0, 0)$ and rises slowly to the right, a half-curve. Its domain is restricted: the expression under the root must be nonnegative. For $\sqrt{x - 4}$, set $x - 4 \ge 0$, so the domain is $x \ge 4$, and the graph starts at $(4, 0)$. This domain restriction is the most-tested feature.

Cube-root functions

$f(x) = \sqrt[3]{x}$ is different in a key way: you can take the cube root of a negative number (since $(-2)^3 = -8$, $\sqrt[3]{-8} = -2$). So its domain is all real numbers, and the graph is a smooth S-shape passing through the origin, rising for all $x$.

Absolute-value functions

$f(x) = |x|$ is a V: it falls to a sharp corner (vertex) at the bottom, then rises symmetrically. The transformations from the functions topic apply: $|x - h| + k$ moves the vertex to $(h, k)$. The corner, not a smooth turn, is the absolute-value signature.

Piecewise functions

A piecewise function gives different rules on different input intervals:

To evaluate, find which interval the input falls in, then use that rule. To graph, draw each piece only over its interval, with open or closed endpoints matching the strict or inclusive inequality.

How the B.E.S.T. EOC examines this topic

• Multiple choice. State the domain of a square-root function, or identify a function's shape.
• Equation editor and number entry. Evaluate a piecewise function at given inputs.
• GRID and matching. Match a function to its graph, or plot a starting point.

A clarifying idea: each nonlinear family has a recognizable silhouette, a half-curve (square root), an S (cube root), a V (absolute value), assembled pieces (piecewise). Knowing the silhouette lets you match graphs to equations without plotting many points.

Why square roots are restricted but cube roots are not

The domain difference comes from what each root can undo. A square of a real number is always nonnegative ($3^2 = 9$ and $(-3)^2 = 9$), so no real number squares to a negative; therefore $\sqrt{\text{negative}}$ has no real value, and the square-root function only accepts inputs that keep the inside $\ge 0$. A cube, by contrast, preserves sign ($2^3 = 8$ but $(-2)^3 = -8$), so every real number, positive, negative, or zero, is the cube of some real number; therefore $\sqrt[3]{\text{anything}}$ is real, and the cube-root function accepts all inputs. This is the same even-versus-odd-root distinction behind rational exponents: even roots restrict the domain, odd roots do not. Recognizing it lets you state a radical function's domain without memorizing cases.

Try this

Q1. State the domain of $f(x) = \sqrt{x + 5}$. [1 point]

• Cue. $x + 5 \ge 0$, so $x \ge -5$.

Q2. For $f(x) = 2x$ if $x \le 0$ and $f(x) = x + 1$ if $x > 0$, find $f(-2)$ and $f(3)$. [2 points]

• Cue. $f(-2) = -4$ (first piece); $f(3) = 4$ (second piece).