Porisma

Worked solution

Scale input distances about the y-axis

Use a graphing calculator to sketch:

  1. y=(x+1)(x−3)y=(x+1)(x-3)
  2. y=(12x+1)(12x−3)y=(\frac12x+1)(\frac12x-3)
  3. y=(2x+1)(2x−3)y=(2x+1)(2x-3)

Copy the curves. What do you observe about the curves?

CJC · Transformations · Horizontal scaling activity

Sketch and retain the coordinate table

Answer

Vertices (1,−4)(1,-4), (2,−4)(2,-4), (1/2,−4)(1/2,-4); common y-intercept (0,−3)(0,-3).

Let f(x)=(x+1)(x−3)f(x)=(x+1)(x-3). Draw the original, f(x/2)f(x/2) and f(2x)f(2x) on the same axes. The Teacher completes the Student table:

CurveMinimumy-interceptx-intercepts
f(x)f(x)(1,−4)(1,-4)(0,−3)(0,-3)(−1,0),(3,0)(-1,0),(3,0)
f(x/2)f(x/2)(2,−4)(2,-4)(0,−3)(0,-3)(−2,0),(6,0)(-2,0),(6,0)
f(2x)f(2x)(1/2,−4)(1/2,-4)(0,−3)(0,-3)(−1/2,0),(3/2,0)(-1/2,0),(3/2,0)
Expand diagram

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Overlaid f=(x+1)(x−3), f(x/2), f(2x). Minimum coordinates (1,−4),(2,−4),(1/2,−4); roots −1 and 3,−2 and 6,−1/2 and 3/2. Shared y-intercept −3. Horizontal distances double or halve.
Overlaid f=(x+1)(x−3), f(x/2), f(2x). Minimum coordinates (1,−4),(2,−4),(1/2,−4); roots −1 and 3,−2 and 6,−1/2 and 3/2. Shared y-intercept −3. Horizontal distances double or halve.
Overlaid f=(x+1)(x−3), f(x/2), f(2x). Minimum coordinates (1,−4),(2,−4),(1/2,−4); roots −1 and 3,−2 and 6,−1/2 and 3/2. Shared y-intercept −3. Horizontal distances double or halve.
How did this part go?

Explain the scale factors

Answer

f(x/2)f(x/2) doubles horizontal distances; f(2x)f(2x) halves them.

Replace x with x/2 to scale parallel to the x-axis by factor 2. Replace x with 2x to scale parallel to that axis by factor 1/2. The source says the shape is preserved: all three remain parabolas. This is not rigid congruence; their widths change. The y-coordinates stay fixed, while each point’s x-coordinate is doubled or halved.

How did this part go?