4.4 4 Practice Modeling Stretching And Compressing Functions Answers

Mastering Stretching and Compressing Functions: A Practice Guide with Answers

Function transformations are the artistic brushstrokes that allow mathematicians and scientists to model the real world. While shifting functions up, down, left, or right is intuitive, the concepts of stretching and compressing (or shrinking) functions add a dynamic layer of scale and rate change. These vertical and horizontal transformations are fundamental for accurately modeling phenomena from physics to economics. This guide provides a comprehensive breakdown of these concepts, followed by targeted practice problems with detailed, step-by-step answers to solidify your understanding.

The Core Concepts: Vertical vs. Horizontal Transformations

The first crucial distinction is the axis of transformation. A vertical transformation affects the output (the y-values) of the function, making the graph taller or shorter. A horizontal transformation affects the input (the x-values), making the graph wider or narrower. The notation for these operations is elegantly simple but requires careful interpretation.

Vertical Stretches and Compressions

A vertical transformation is applied outside the function's core rule. The general form is: g(x) = a * f(x) Here, the constant a is the vertical scaling factor.

• If |a| > 1, the graph of f(x) is vertically stretched. Every y-coordinate is multiplied by a number greater than 1, pulling the graph away from the x-axis. For example, 2*f(x) makes the graph twice as tall.
• If 0 < |a| < 1, the graph of f(x) is vertically compressed. Every y-coordinate is multiplied by a fraction, pushing the graph toward the x-axis. For example, (1/2)*f(x) makes the graph half as tall.
• If a is negative, the transformation includes a vertical reflection across the x-axis in addition to the stretch/compression. For instance, -3*f(x) reflects the graph and stretches it vertically by a factor of 3.

Key Insight: Vertical transformations are intuitive because they directly scale the output. You multiply the y-value you would normally get.

Horizontal Stretches and Compressions

A horizontal transformation is applied inside the function's argument (the x). The general form is: g(x) = f(b*x) Here, the constant b is the horizontal scaling factor. This is where intuition often fails.

• If |b| > 1, the graph of f(x) is horizontally compressed. The input x is multiplied by a number greater than 1 before being fed into f. To achieve the same output as f at a point c, we need a smaller input x = c/b. The graph appears to "speed up," becoming narrower. For example, f(2x) compresses the graph horizontally by a factor of 2.
• If 0 < |b| < 1, the graph of f(x) is horizontally stretched. The input x is multiplied by a fraction. To get the output f(c), we now need a larger input x = c/b (since dividing by a fraction increases the value). The graph appears to "slow down," becoming wider. For example, f((1/2)x) stretches the graph horizontally by a factor of 2.
• If b is negative, the transformation includes a horizontal reflection across the y-axis in addition to the stretch/compression.

The Golden Rule for Horizontal Transformations: The horizontal scaling factor is 1/|b|. The graph is stretched/compressed by the reciprocal of the number multiplying x. f(3x) compresses by a factor of 1/3. f(0.5x) stretches by a factor of 1/0.5 = 2.

Combined Transformations and Order of Operations

When multiple transformations are combined, such as g(x) = a * f(bx - h) + k, the order of application matters for understanding, though the algebraic order is fixed. The standard sequence to describe the transformation from the parent function f(x) is:

1. Horizontal: Start with f(bx). Apply the horizontal compression/stretch and reflection (if b is negative).
2. Horizontal Shift: Then apply f(b(x - h)) or f(bx - h). This shifts the graph horizontally by h units.
3. Vertical: Apply a * f(bx - h). This applies the vertical stretch/compression and reflection (if a is negative).
4. Vertical Shift: Finally, apply a * f(bx - h) + k, shifting the graph vertically by k units.

Critical Note: The horizontal shift h is affected by the horizontal scaling b. In the form f(bx - h), the actual shift is h/b units. To avoid this complication, it's often best to factor the b from the linear expression inside the function: f(b(x - h/b)). This makes the true horizontal shift explicit as h/b.

Practice

Practice

Problem 1

Given the parent function (f(x)=\sqrt{x}), describe the transformation and sketch the graph of
[ g(x)= -2,\sqrt{,3(x-4),}+5 . ]

Solution 1. Identify the constants: (a=-2), (b=3), (h=4), (k=5).
2. Horizontal work:

• Inside the radical we have (3(x-4)). - The factor (b=3) compresses the graph horizontally by (\frac{1}{|b|}=\frac{1}{3}).
• Because (b>0) there is no reflection across the (y)-axis.
• The term ((x-4)) shifts the graph right by (h=4) units (the shift is already isolated after factoring out (b)).

3. Vertical work: - The coefficient (a=-2) stretches the graph vertically by (|a|=2) and reflects it across the (x)-axis (negative sign).
4. Vertical shift: (k=5) moves the entire graph up 5 units.

Resulting description: Start with (y=\sqrt{x}); compress horizontally by a factor of (\frac13); shift right 4 units; stretch vertically by a factor of 2 and flip over the (x)-axis; finally shift up 5 units.

Key points for the sketch:

• The original domain ([0,\infty)) becomes, after the horizontal compression and shift, ([4,\infty)).
• The point ((0,0)) on (f) maps to (\bigl(4,; -2\cdot\sqrt{3\cdot0}+5\bigr) = (4,5)) after all steps.
• The point ((1,1)) on (f) maps to (\bigl(4+\tfrac13,; -2\cdot\sqrt{3\cdot1}+5\bigr) = \bigl(\tfrac{13}{3},; -2\sqrt{3}+5\bigr)).
  Plot a few transformed points and connect them with the characteristic half‑parabola shape of the square‑root function.

Problem 2 For (f(x)=|x|), write the equation of the function that results from: - a horizontal stretch by a factor of 4,

• a reflection across the (y)-axis,
• a shift left 3 units,
• a vertical compression by a factor of (\frac12),
• and a shift down 2 units.

Solution
Apply the transformations in the order prescribed (horizontal first, then vertical).

1. Horizontal stretch by 4: replace (x) with (\frac{x}{4}) → (f\bigl(\frac{x}{4}\bigr)=\bigl|\frac{x}{4}\bigr|).
2. Reflection across the (y)-axis: multiply the input by (-1) → (f\bigl(-\frac{x}{4}\bigr)=\bigl|-\frac{x}{4}\bigr|=\bigl|\frac{x}{4}\bigr|) (the absolute value removes the sign, so the reflection does not change the expression; we keep the negative sign to show the step).
3. Shift left 3: replace (x) with (x+3) inside the argument → (f\bigl(-\frac{x+3}{4}\bigr)=\bigl|-\frac{x+3}{4}\bigr|=\frac{|x+3|}{4}).
4. Vertical compression by (\frac12): multiply the whole function by (\frac12) → (\frac12\cdot\frac{|x+3|}{4}= \frac{|x+3|}{8}). 5. Shift down 2: subtract 2 → (\displaystyle g(x)=\frac{|x+3|}{8}-2).

Thus the final equation is
[ \boxed{g(x)=\frac{|x+3|}{8}-2}. ]

Problem 3 Given (f(x)=\sin x), determine the transformation represented by

[ g(x)=3,\sin!\bigl(2x-\pi\bigr)+1 . ]
State the amplitude, period, phase shift, and vertical shift.

Solution
Rewrite the inner term to expose the shift:
[ 2x-\pi = 2\Bigl(x-\frac{\pi}{2}\Bigr). ]
Hence (g(x)=3\sin\bigl[2(x-\frac{\pi}{2})\bigr]+1).

• Amplitude: (|a| = 3).
• Period: For (\sin(bx)), period (=\frac{2\pi}{|b|}). Here (|b|=2) → period (=\frac{2\pi}{2

Continuing from the sinefunction transformation discussion:

Solution
Rewrite the argument to match the standard form (a \sin(b(x - c)) + d):
[ g(x) = 3 \sin(2x - \pi) + 1 = 3 \sin\left(2\left(x - \frac{\pi}{2}\right)\right) + 1. ]

• Amplitude: The coefficient (a = 3), so the amplitude is (|3| = 3).
• Period: For (a \sin(b(x - c)) + d), the period is (\frac{2\pi}{|b|}). Here (|b| = 2), so the period is (\frac{2\pi}{2} = \pi).
• Phase Shift: The term ((x - \frac{\pi}{2})) indicates a phase shift of (\frac{\pi}{2}) units to the right.
• Vertical Shift: The constant (+1) indicates a vertical shift of 1 unit upward.

Conclusion
The transformation (g(x) = 3 \sin(2x - \pi) + 1) represents a sine wave with an amplitude of 3, a period of (\pi) (half the standard period), a phase shift of (\frac{\pi}{2}) to the right, and a vertical shift of 1 unit upward. This results in a wave that oscillates between (-2) and (4), completes one full cycle every (\pi) units, starts its cycle (\frac{\pi}{2}) units to the right of the origin, and is uniformly elevated by 1 unit.