Porisma

Worked solution

Compare modulus, negative scaling and reciprocal graphs

The given graph y=f(x)y=f(x) has asymptotes x=−1x=-1, y=−2y=-2, turning point A=(1,3)A=(1,3), and zeros O=(0,0)O=(0,0), B=(−2,0)B=(-2,0), C=(4,0)C=(4,0).

(a) State the turning point of y=1+13f(x)y=1+\frac13f(x). mark

(b) Sketch y=∣f(x)∣y=|f(x)|, labelling asymptotes and the points corresponding to A,B,C. marks

(c) Sketch y=−f(2x)y=-f(2x) with those labels. marks

(d) Sketch y=1/f(x)y=1/f(x), labelling asymptotes, turning points and axis crossings. marks

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Qualitative reconstruction checked against the actual given graph in lecture Student PDF 38 / printed 110 and Teacher PDF 39 / printed 111: A(1,3), B(−2,0), O(0,0), C(4,0), asymptotes x=−1,y=−2, with the source branch directions. The standalone solution PDF omits that original figure; no unique formula is inferred.
Qualitative reconstruction checked against the actual given graph in lecture Student PDF 38 / printed 110 and Teacher PDF 39 / printed 111: A(1,3), B(−2,0), O(0,0), C(4,0), asymptotes x=−1,y=−2, with the source branch directions. The standalone solution PDF omits that original figure; no unique formula is inferred.
Qualitative reconstruction checked against the actual given graph in lecture Student PDF 38 / printed 110 and Teacher PDF 39 / printed 111: A(1,3), B(−2,0), O(0,0), C(4,0), asymptotes x=−1,y=−2, with the source branch directions. The standalone solution PDF omits that original figure; no unique formula is inferred.

[2022/YIJC/Promo/6]

CJC · Review problems · Question 5

Answer

(a)(1,2). (b) A(1,3), B(−2,0), C(4,0), asymptotes x=−1, y=2. (c) A(1/2,−3), B(−1,0), C(2,0), x=−1/2, y=2. (d) Minimum (1,1/3), poles x=−2,0,4, horizontal y=−1/2; open point (−1,0), no axis intercept.

(a) Map the turning-point ordinate

1+13(3)=21+\frac13(3)=2, so A′=(1,2), as printed.

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(b) Reflect only negative ordinates upwards

∣f(x)∣|f(x)| preserves A=(1,3), B=(−2,0), C=(4,0) and O. All values become nonnegative; the zero crossings become sharp local minima where the branches change sign. Vertical asymptote x=−1 remains, with both sides upward. The horizontal asymptote is y=2.

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(c) Compress horizontally and reflect vertically

The point map is (x,y)↦(x/2,−y)(x,y)\mapsto(x/2,-y). Thus A′=(1/2,−3), B′=(−1,0), C′=(2,0), and O remains. The asymptotes become x=−1/2 and y=2.

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(d) Convert zeros to poles and the maximum to a minimum

The zeros at x=−2,0,4 give reciprocal vertical asymptotes there. A becomes the minimum (1,1/3) on its positive branch. The horizontal asymptote becomes y=−1/2. The original vertical-asymptote input x=−1 stays excluded: its reciprocal zero limit must be drawn as an open point (−1,0). There is no true x-intercept and no y-intercept, since f(0)=0.

How did this part go?

Compare your sketch

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Three source-solution panels. Absolute value keeps A(1,3),B(-2,0),C(4,0),O and has asymptotes x=-1,y=2. Negative f(2x) has A(1/2,-3),B(-1,0),C(2,0), asymptotes x=-1/2,y=2. Reciprocal has poles x=-2,0,4, minimum (1,1/3), horizontal asymptote y=-1/2, and open point (-1,0), not a true axis crossing.
Three source-solution panels. Absolute value keeps A(1,3),B(-2,0),C(4,0),O and has asymptotes x=-1,y=2. Negative f(2x) has A(1/2,-3),B(-1,0),C(2,0), asymptotes x=-1/2,y=2. Reciprocal has poles x=-2,0,4, minimum (1,1/3), horizontal asymptote y=-1/2, and open point (-1,0), not a true axis crossing.
Three source-solution panels. Absolute value keeps A(1,3),B(-2,0),C(4,0),O and has asymptotes x=-1,y=2. Negative f(2x) has A(1/2,-3),B(-1,0),C(2,0), asymptotes x=-1/2,y=2. Reciprocal has poles x=-2,0,4, minimum (1,1/3), horizontal asymptote y=-1/2, and open point (-1,0), not a true axis crossing.

Common mistakes

The source reciprocal sketch labels (−1,0) as a crossing, but the original f is undefined at x=−1. Retain the domain exclusion.

Solution notes

The standalone solution PDF omits the original graph referred to in its prompt. That original is present in the lecture Student PDF 38 / printed 110 and Teacher PDF 39 / printed 111; direct comparison confirms the reconstruction’s labelled features and branch directions. The original absence is retained as scoped historical evidence, not claimed for the lecture copies. The reciprocal axis-crossing ambiguity remains explicitly corrected with an open point.

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