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Worked solution

Two reflections of the same straight line

Given f(x)=x+1f(x)=x+1, sketch on separate diagrams (a) y=−f(x)y=-f(x) and (b) y=f(−x)y=f(-x).

CJC · Transformations · Self-Practice 3

(a) Sketch −f(x)

Answer

y=−x−1y=-x-1, through (−1,0)(-1,0) and (0,−1)(0,-1).

Independently derived: −f(x)=−(x+1)=−x−1-f(x)=-(x+1)=-x-1. Reflect in the x-axis, changing every output sign. Neither copy prints a self-practice solution on this page.

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Separate independently derived straight lines: −x−1 through (−1,0),(0,−1), and −x+1 through (1,0),(0,1). Both have gradient −1 and distinct intercepts.
Separate independently derived straight lines: −x−1 through (−1,0),(0,−1), and −x+1 through (1,0),(0,1). Both have gradient −1 and distinct intercepts.
Separate independently derived straight lines: −x−1 through (−1,0),(0,−1), and −x+1 through (1,0),(0,1). Both have gradient −1 and distinct intercepts.
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(b) Sketch f(−x)

Answer

y=−x+1y=-x+1, through (1,0)(1,0) and (0,1)(0,1).

Independently derived: f(−x)=−x+1f(-x)=-x+1. Reflect in the y-axis, changing every input coordinate sign. These two reflections have equal slopes for this line, but different y-intercepts; draw separate diagrams as requested.

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