Function Notation, Composition, and Transformations — SAT Math

Introduction

Function notation, composition, and transformations fall under Advanced Math, which makes up roughly 35% of the digital SAT Math section. You'll be asked to evaluate a function at a given input, work backward from an output to find the input, build a composed function like $f(g(x))$ , and connect an algebraic transformation to a graph. These questions appear across multiple-choice and student-produced response (SPR) formats, and the built-in Desmos calculator can save significant time on several of them.

Core Concept

Evaluating and Inverting Functions

Evaluating means substituting a known $x$ into $f(x)$ to find the output. Inverting means you're given the output and must find the input — it's solving an equation.

Task	Given	Find	Method
Evaluate	$x = a$	$f(a)$	Substitute
Invert	$f(x) = b$	$x$	Solve algebraically

For a quadratic $f(x) = x^2 - 3x$ , if you need to find $x$ when $f(x) = 28$ :

x^2 - 3x = 28 \implies x^2 - 3x - 28 = 0 \implies (x+4)(x-7) = 0

So $x = -4$ or $x = 7$ . If the problem restricts $x > 0$ , the answer is $x = 7$ . Always check whether the domain restricts which root is valid.

For an exponential $f(x) = 3^x - 5$ , setting $f(x) = 22$ gives:

3^x = 27 \implies x = 3

For a radical $f(x) = \sqrt{x+4} - 1$ , setting $f(x) = 3$ :

\sqrt{x+4} = 4 \implies x + 4 = 16 \implies x = 12

Composition of Functions

$f(g(x))$ means: first apply $g$ , then feed the result into $f$ .

f(g(a)) \neq g(f(a)) \quad \text{in general}

Order matters. Always evaluate the inner function first.

Transformations of Graphs

Starting from a base graph $y = f(x)$ , three core shifts are tested:

Transformation	Equation	Effect on graph
Vertical shift up $k$	$y = f(x) + k$	Every point moves up $k$ units
Horizontal shift right $h$	$y = f(x - h)$	Every point moves right $h$ units
Reflection over $x$ -axis	$y = -f(x)$	Every point's $y$ -value is negated

Critical trap: $f(x - h)$ shifts right (not left) when $h > 0$ . The minus sign inside the function is counterintuitive.

Combined example: if $f(x) = x^2$ has vertex $(0, 0)$ , then $y = -(x-3)^2 + 4$ :

Reflects over $x$ -axis
Shifts right 3 (vertex $x$ -coordinate: $0 + 3 = 3$ )
Shifts up 4 (vertex $y$ -coordinate: $0 + 4 = 4$ )
New vertex: $(3, 4)$

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