A horizontal dilation is a multiplicative transformation g(x) = f(bx), where b ≠ 0, that stretches or compresses the graph of f horizontally by a factor of 1/b; if b < 0 it also reflects the graph over the y-axis (AP Precalc EK 1.12.A.5).
A horizontal dilation happens when you multiply the input of a function by a constant. If g(x) = f(bx) with b ≠ 0, the graph of f gets stretched or compressed horizontally by a factor of 1/b. That 1/b is the part everyone trips on. If b = 3, the graph does NOT get three times wider. It gets squeezed to one-third the width, because every input now reaches f three times faster. A point at x = 6 on f lands at x = 2 on g.
Think of it as a speed change on the x-axis. When |b| > 1, the function 'plays faster,' so the graph compresses toward the y-axis. When 0 < |b| < 1, it 'plays slower,' so the graph stretches out. And if b is negative, you get a reflection over the y-axis on top of the dilation. Under the hood, this is just function composition. The CED frames g(x) = f(bx) as composing f with the inner function kx (EK 2.7.C.3), which is why the dilation acts on x-values, not y-values.
Horizontal dilation lives in Topic 1.12 (Transformations of Functions) under learning objective 1.12.A, but it refuses to stay there. In Unit 3 it becomes the period machine. The b in f(θ) = a sin(b(θ + c)) + d sets the period to 2π/b, and identifying that period is exactly what LO 3.6.A asks you to do. In Unit 2 it pulls a disappearing act. For exponentials, b^(cx) = (b^c)^x, so a horizontal dilation is secretly just a change of base (LO 2.4.A). For logarithms, log_b(kx) = log_b k + log_b x, so a horizontal dilation is secretly a vertical translation (LO 2.12.A.1). If you only memorize 'multiply inside means squish,' you miss half of what the exam tests, which is recognizing the same transformation wearing different costumes across function families.
Keep studying AP® Precalculus Unit 1
Multiplicative transformation and vertical dilation (Unit 1)
Horizontal dilation is one of the two multiplicative transformations in EK 1.12.A. Its sibling, vertical dilation a·f(x), multiplies the output instead of the input. Vertical dilation scales by exactly a; horizontal dilation scales by 1/b. Inside the parentheses, transformations always act 'backwards.'
Period of sinusoidal and tangent functions (Unit 3)
The b in sin(bθ) is a horizontal dilation, and that is precisely why the period becomes 2π/b for sine and cosine (Topic 3.6) and π/b for tangent (Topic 3.8). When an FRQ asks for the period of a sinusoidal model, you are really being asked to read off a horizontal dilation factor.
Exponent rules and exponential manipulation (Unit 2)
The power property (b^m)^n = b^(mn) means b^(cx) = (b^c)^x. So horizontally dilating an exponential function is the same graph as changing its base. One transformation, two algebraic disguises, and Topic 2.4 expects you to move between them.
Function composition (Unit 2)
Per EK 2.7.C.3, f(bx) is the composition of f with the inner function g(x) = bx. Seeing dilations as compositions explains the 1/b factor. The inner function rescales the x-axis before f ever sees the input.
Horizontal dilation shows up most often in multiple-choice questions that give you a feature of f (a zero, an x-intercept, a maximum) and ask where that feature lands on g(x) = f(bx). The move is always the same. Divide the original x-value by b. For example, if f has an x-intercept at x = 6, then g(x) = f(3x) has one at x = 2. If f has a maximum at x = 4, then g(x) = f(-2x) has one at x = -2, since the negative b also flips the graph over the y-axis. If f has zeros at x = -4, 0, and 3, then h(x) = f(x/2) has zeros at -8, 0, and 6, because b = 1/2 doubles every x-value. In Unit 2, expect equivalence questions instead, such as recognizing that 4^x equals 2^(2x), a horizontal dilation rewritten as a base change. In Unit 3, the skill becomes pulling the period 2π/b out of a sinusoidal model.
Vertical dilation a·f(x) multiplies the output, so the graph stretches vertically by exactly a. Horizontal dilation f(bx) multiplies the input, so the graph stretches horizontally by 1/b, the reciprocal. So a·f(x) with a = 2 makes the graph twice as tall, but f(2x) makes it half as wide. Outputs behave intuitively; inputs behave backwards. Quick test: vertical dilations move y-coordinates and leave x-intercepts alone, while horizontal dilations move x-coordinates and leave y-intercepts alone.
A horizontal dilation g(x) = f(bx) stretches or compresses the graph of f horizontally by a factor of 1/b, not b.
If |b| > 1 the graph compresses toward the y-axis, and if 0 < |b| < 1 the graph stretches away from it.
If b is negative, the transformation also includes a reflection over the y-axis.
To find where a feature of f lands on f(bx), divide the original x-value by b, so an intercept at x = 6 on f moves to x = 2 on f(3x).
In sinusoidal functions, the horizontal dilation factor b determines the period, which is 2π/b for sine and cosine and π/b for tangent.
Horizontal dilations can be rewritten using algebra rules. For exponentials, b^(cx) = (b^c)^x is a change of base, and for logs, log_b(kx) = log_b k + log_b x is a vertical translation.
It's the multiplicative transformation g(x) = f(bx), where b ≠ 0, that stretches or compresses the graph of f horizontally by a factor of 1/b. It's part of EK 1.12.A.5 in Topic 1.12, Transformations of Functions.
No, it compresses the graph to half its width. Multiplying the input by 2 makes the function reach every output twice as fast, so all x-coordinates get divided by 2. The dilation factor is 1/b, not b.
Vertical dilation a·f(x) multiplies outputs and scales the graph vertically by exactly a. Horizontal dilation f(bx) multiplies inputs and scales horizontally by the reciprocal, 1/b. That reciprocal flip only happens with input transformations.
In f(θ) = a sin(b(θ + c)) + d, the b is a horizontal dilation, and it makes the period 2π/b. So sin(2θ) completes a full cycle in just π instead of 2π. For tangent, the period becomes π/b.
You get a horizontal dilation by 1/|b| plus a reflection over the y-axis. For example, if f has a maximum at x = 4, then g(x) = f(-2x) has its maximum at x = -2, since 4 divided by -2 is -2.
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