AP Precalc 1.12B-1.13 Dilations & ModelsAP Precalc 1.12B-1.13 Dilations & Models

A P P R E C A L C U L U S · UNIT1

Dilations and Function Model Selection

Topics 1.12B and 1.13 Study Guide

In one sentence: multiplying the output stretches the graph vertically, multiplying the input compresses it horizontally, and finite differences tell you whether data is linear, quadratic, or cubic. Why it works, in plain language. Multiplying the output by a stretches the picture up and down, so a tall point gets taller. Multiplying the input by b changes how fast you move along the x-axis. With f(2x), you reach every f-value in half the distance, so the graph squishes toward the y-axis by ½. The factor is the reciprocal, which is why f(2x) compresses and f(½x) stretches. For data, a linear function adds the same amount each step, a quadratic adds an amount that itself grows steadily (so the second difference is constant), and a cubic needs three layers of differences before it levels out. Step by step: finite differences. (1) Check that the x-values go up by the same amount each time. (2) Subtract each y from the next. (3) If all those differences match: linear. (4) If not, subtract the differences again. If those match: quadratic. (5) If not, once more: cubic. (6) The constant you land on tells you the model's leading behavior (positive or negative, and how large).

Step by step: domain and range of a transformed function. (1) Domain: undo the inside. For f(bx − ...): divide x-values by b. For f(x − h): add h. (2) Range: apply outside operations in order, multiply by a and then add k. (3) If a is negative, swap the endpoints (and their open or closed ends). 1.12B Dilations (multiplicative transformations)

General form: g(x) = a·f(b(x − h)) + k.

Change Name Effect

a·f(x) vertical dilation (multiplies the

output)

stretch by |a| if |a| > 1, shrink if |a| < 1. If a < 0, also reflect over

the x-axis.

f(bx) horizontal dilation (multiplies the

input)

compress by 1/|b| if |b| > 1, stretch if |b| < 1. If b < 0, also reflect

over the y-axis.

Vertical dilation: (x, y) → (x, a·y). Horizontal dilation: (x, y) → (x/b, y).

f(2x): every x-value is halved (compressed by ½). f(½x): every x-value is doubled (stretched by 2). The horizontal factor is the reciprocal.

Zeros and x-intercepts change with horizontal dilations but not with vertical ones. The y-intercept changes with vertical dilations but not with horizontal ones.

AP Precalculus 1 of 5

Algebraic transformations

f(x) = 4x + 3, g(x) = 3f(x) + 5: 3(4x + 3) + 5 = 12x + 14.

f(x) = 2x + 6, g(x) = f(½(x + 2)) − 1: f(½(x + 2)) = 2·½(x + 2) + 6 = x + 8, so g(x) = x + 7. f(x) = 4x − 5, g(x) = −2f(x + 1) + 5: −2(4x − 1) + 5 = −8x + 7.

f(x) = 2x² − 3x + 1, g(x) = f(2x) + 3: 2(2x)² − 3(2x) + 1 + 3 = 8x² − 6x + 4.

Numeric transformations

g(x) = f(3x) + 1, g(2) = f(6) + 1. With f(6) = −2: g(2) = −1.

g(x) = 2f(x + 2) − 3, g(1) = 2f(3) − 3. With f(3) = 8: g(1) = 13.

g(x) = 4f(−x), g(2) = 4f(−2). With f(−2) = 6: g(2) = 24.

Method: plug the x-value into g, rewrite in terms of f at a new input, then look that input up in the table. Domain and range

f: domain [−6, 8], range [−4, 8]. g(x) = −3f(2x): domain [−6, 8] ÷ 2 = [−3, 4]. Range: times −3 flips it, [−24, 12].

f: domain (0, 5), range [−10, 4]. g(x) = 2f(x − 3) + 4: domain (3, 8). Range: [−20, 8] + 4 = [−16, 12]. f: domain [−2, 4], range (−6, 8). g(x) = f(½x) + 5: domain doubles, [−4, 8]. Range (−1, 13). Graph question: g(x) = 2f(x − 3) − 1 where the graph of f has domain [−5, 3] and range [1, 4]. Then g has domain [−2, 6] and range [1, 7]. The key also gives g(−2) = 3, no zeros, and y-intercept 7. Test-prep. The graph of y = f(2x) is f compressed horizontally by ½.

Three transformations in order: horizontal dilation ½, vertical dilation 4, vertical translation 6. Then g(x) = 4f(2x) + 6, and g(4) = 4f(8) + 6. If f(8) = −12, g(4) = −42.

Find a constant: g(x) = 3f(2x) − b, f(2) = a, f(4) = 8, g(1) = 7, g(2) = 19. From g(2) = 3f(4) − b = 24 − b = 19: b = 5. From g(1) = 3f(2) − 5 = 3a − 5 = 7: a = 4.

Traps. (1) Horizontal factors are reciprocals. (2) Apply dilation to domain/range before shifting. (3) A negative factor outside means flip over x-axis (it also flips the order of the range endpoints). (4) Keep open and closed endpoints through every transformation.

1.13 Function model selection

Finite differences

For data with equally spaced x-values, subtract consecutive y-values repeatedly.

Constant after… Model

1st differences Linear

2nd differences Quadratic

3rd differences Cubic

AP Precalculus 2 of 5

Examples.

y = 4, 15, 50, 121, 240, 419 gives differences 11, 35, 71, 119, 179, then 24, 36, 48, 60, then 12, 12, 12. Cubic, third difference 12.

y = 49, 46, 43, 40, 37, 34: first difference −3 every time. Linear.

y = 119, 110, 95, 74, 47, 14: differences −9, −15, −21, −27, −33, then −6 every time. Quadratic, second difference −6.

Selected values x = 2, 4, …, 12: f = 97, 405, 977, 1861, 3105, 4757. Differences 308, 572, 884, 1244, 1652; then 264, 312, 360, 408; then 48, 48, 48. Cubic because the third differences are a nonzero constant.

The x-values must be equally spaced. If they are not, compute the average rates of change (Δy/Δx) first.

If the rate of change in a graph is roughly linear, the function is quadratic. (The second difference is roughly constant.)

Geometry: perimeter is linear in the side length, area is quadratic, volume is cubic. Restricted domain and range in context

A model can be a perfectly good polynomial, but the context restricts the inputs and outputs (no negative time, no negative length).

Domain = inputs that make sense. Range = outputs that come out of that domain.

Always explain answers in context with units.