Functions: Notation, Transformations, and Rate of Change
What this domain tests
Functions is the language the rest of the exam is written in. Expect function notation and evaluation, domain and range, transformations of parent graphs, average rate of change, comparison of linear, quadratic, absolute value, piecewise, and exponential models, inverse and composite functions, and sequences treated as functions on the integers. Georgia items frequently ask you to compare two representations - a table against an equation, a graph against a verbal description - and justify which grows faster.
Core rules
- Function notation is substitution. f(3) means replace every x in the rule with 3; f(x + h) means replace every x with the entire expression x + h.
- Transformations of y = f(x): a·f(x) stretches vertically by |a| and reflects across the x-axis when a < 0; f(x − h) shifts right by h (the sign is counterintuitive); f(x) + k shifts up by k; f(bx) compresses horizontally by 1/b.
- Average rate of change on [a, b] is (f(b) − f(a))/(b − a), the slope of the secant line between two points on the graph. It is an average, not the slope at a single instant.
- Domain excludes values that make a denominator zero or a radicand negative. Range is the set of outputs actually produced.
- Inverses undo each other: swap x and y, then solve. The graph of f⁻¹ is the reflection of f across y = x, and the domain of f becomes the range of f⁻¹. Write f⁻¹(x), never confuse it with 1/f(x).
- Exponential versus linear growth: linear models add a constant amount per unit; exponential models multiply by a constant factor per unit, so an exponential eventually overtakes any linear model.
Worked example
Find the average rate of change of f(x) = x² − 4x + 1 on the interval [1, 5].
Evaluate the endpoints: f(1) = 1 − 4 + 1 = −2 and f(5) = 25 − 20 + 1 = 6. Then (6 − (−2))/(5 − 1) = 8/4 = 2. Interpretation: across the whole interval, the function rises an average of 2 output units per input unit - even though the graph falls, then rises. A common error is evaluating a single point, or dividing by f(b) − f(a) instead of b − a.
Traps that cost marks
- Computing f(x + h) as f(x) + h.
- Reversing the direction of a horizontal shift.
- Treating f⁻¹(x) as the reciprocal of f(x).
- Reporting the open interval or missing endpoints in domain and range.
- Comparing growth rates using a single data point instead of the rate over an interval.