3.3 Function Models, Transformations, and Financial Applications

Quick Answer: Match a data set to a function family by examining differences and ratios. Then recognize transformations as shifts, reflections, and stretches of a parent graph, and apply the same modeling logic to financial situations such as interest and depreciation.

Choosing the Best Function Model

Choosing the best function model (MA.912.F.1.8) starts with the same diagnostics used for classification. Given a data set with evenly spaced x-values:

Real-world context also signals the model. A constant rate (cost per hour, distance per gallon) implies linear. A quantity that depends on area or a product of two changing quantities (area of a rectangle, projectile height under gravity) implies quadratic. A percent increase or decrease, population doubling, half-life, or repeated multiplication by the same factor implies exponential.

Worked example: A savings account starts with $200 and grows by 5% each year. The balance after t years is B(t) = 200(1.05)^t, an exponential model. After 3 years, B(3) = 200(1.157625) ≈ $231.53. The same account with a $10 flat deposit each year would be linear: B(t) = 200 + 10t.

Transformations of Parent Functions

Transformations (MA.912.F.2.1) modify a parent function f(x) to produce a new function. The EOC tests three types.

For f(x) transformed to f(x − h) + k, a positive h shifts the graph right by h units and a positive k shifts it up by k units. A negative h shifts left and a negative k shifts down. A common trap is reading f(x + 3) as a shift right by 3; the correct shift is left by 3 because the expression inside the function must equal zero at x = −3.

A reflection across the x-axis is −f(x), which changes the sign of every output: a point (x, y) on the parent graph moves to (x, −y). A reflection across the y-axis is f(−x), which changes the sign of every input: (x, y) moves to (−x, y). A vertical dilation by factor a multiplies every output by a, stretching the graph away from the x-axis when |a| > 1 and compressing it toward the x-axis when 0 < |a| < 1.

Worked example: The parent absolute value function y = |x| has its vertex at (0, 0). The transformed function y = |x − 4| + 2 shifts the vertex to (4, 2): the graph moves right 4 and up 2. The function y = −2|x| reflects the V across the x-axis (opening downward) and stretches it vertically by a factor of 2, so the slopes of the two rays are −2 and 2 instead of −1 and 1.

Identifying a transformation from a table requires comparing input-output pairs to the parent function. If the parent table has (0, 0), (1, 1), (2, 4) for y = x² and the new table has (0, 3), (1, 4), (2, 7), each output is 3 greater, indicating a vertical translation up by 3: y = x² + 3.

Financial Literacy Applications

Financial literacy applications (MA.912.FL.3.4) use function models to describe money situations. Simple interest is linear because the interest added each period is constant: I = Prt, where P is principal, r is the annual rate, and t is years. The total amount A = P + Prt = P(1 + rt) is a linear function of t with slope Pr.

Compound interest is exponential because interest is earned on prior interest: A = P(1 + r/n)^(nt), where n is the number of compounding periods per year. Compounded annually (n = 1), this becomes A = P(1 + r)^t, a pure exponential with base (1 + r). The EOC may ask which model fits a given scenario: a flat $50 fee per month is linear, while a 2% monthly growth in a fund balance is exponential.

Depreciation by a fixed percent is exponential decay: V(t) = P(1 − r)^t, where r is the depreciation rate. A car worth $20,000 losing 15% of its value each year has V(t) = 20000(0.85)^t; after 4 years V(4) = 20000(0.522) ≈ $10,440. Notice that the value never reaches exactly zero under this model, which is why exponential decay is sometimes described as approaching a horizontal asymptote at y = 0.

Common EOC Traps

Common EOC traps:

• Choosing a linear model for data with constant ratios. Always compute ratios when first differences are not constant.
• Reading f(x + 3) as a shift right by 3. The shift is left by 3.
• Forgetting that a vertical dilation by a negative factor (a < 0) includes a reflection across the x-axis in addition to a stretch.
• Treating compound interest as linear. The exponent nt makes the growth multiplicative, not additive.
• Confusing the depreciation rate r with the remaining-value factor (1 − r). A 15% depreciation leaves 85% of value, so the base is 0.85, not 0.15.