Porisma

Worked solution

Reflect a two-branch graph and handle a reciprocal hole

The given graph y=f(x)y=f(x) intersects the axes at A=(0,−2), B=(−1,0), has minimum C=(4,0.5), and asymptotes x=1, y=3.

Sketch separately:

(i) y=f(−1−x)y=f(-1-x); marks

(ii) y=1/f(x)y=1/f(x); marks

Clearly indicate asymptotes, axial intercepts and corresponding A,B,C where appropriate.

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Source schematic with A(0,-2),B(-1,0), minimum C(4,1/2), asymptotes x=1,y=3. Branch shapes are qualitative and reproduce the stated landmarks and limiting behavior.
Source schematic with A(0,-2),B(-1,0), minimum C(4,1/2), asymptotes x=1,y=3. Branch shapes are qualitative and reproduce the stated landmarks and limiting behavior.
Source schematic with A(0,-2),B(-1,0), minimum C(4,1/2), asymptotes x=1,y=3. Branch shapes are qualitative and reproduce the stated landmarks and limiting behavior.

[2010/HCI/Promo/5]

CJC · Review problems · Question 3

Answer

(i) A(−1,−2), B(0,0), C(−5,1/2), asymptotes x=−2, y=3.

(ii) A(0,−1/2), C(4,2), asymptotes x=−1, y=1/3; open point (1,0), no x-intercept.

(i) Apply the horizontal point map

The input replacement −1−x maps each source point (x,y)(x,y) to (−1−x,y)(−1−x,y). Thus A=(−1,−2), B=(0,0), C=(−5,1/2). The vertical asymptote x=1 maps to x=−2; horizontal y=3 remains. B is both the x- and y-intercept of the transformed curve, as shown in the source.

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(ii) Reciprocate finite nonzero ordinates

A becomes (0,−1/2), the y-intercept. C becomes (4,2), a maximum on its positive branch. The zero at B becomes a vertical asymptote x=−1. The horizontal asymptote y=3 becomes y=1/3.

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(ii) Preserve the excluded original asymptote input

The original is undefined at x=1. Its unbounded one-sided values yield reciprocal limits of 0, but (1,0) remains excluded. Draw an open point there; no actual x-intercept occurs. The source sketch runs through that limiting location without marking the domain exclusion, so the authored sketch and qualification make it explicit.

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Compare your sketch

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First panel f(-1-x): A(-1,-2),B(0,0),C(-5,1/2), asymptotes x=-2,y=3. Second reciprocal: A(0,-1/2),C(4,2), pole x=-1, horizontal asymptote y=1/3, and open point (1,0), since the original f is undefined at x=1. No actual reciprocal x intercept exists.
First panel f(-1-x): A(-1,-2),B(0,0),C(-5,1/2), asymptotes x=-2,y=3. Second reciprocal: A(0,-1/2),C(4,2), pole x=-1, horizontal asymptote y=1/3, and open point (1,0), since the original f is undefined at x=1. No actual reciprocal x intercept exists.
First panel f(-1-x): A(-1,-2),B(0,0),C(-5,1/2), asymptotes x=-2,y=3. Second reciprocal: A(0,-1/2),C(4,2), pole x=-1, horizontal asymptote y=1/3, and open point (1,0), since the original f is undefined at x=1. No actual reciprocal x intercept exists.

Common mistakes

A finite zero limit of 1/f at an original vertical asymptote does not make that input part of the reciprocal domain.

Solution notes

All source landmarks, branch directions and asymptotes retained in independent qualitative sketches. Domain exclusion at x=1 is explicit rather than silently adopting the source’s unmarked zero-limit crossing.

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