Porisma

Worked solution

Start from the actual given graph f(x)+3

The supplied graph is y=f(x)+3y=f(x)+3, with maximum (−1,5), y-intercept 3 and a right-hand horizontal asymptote y=1.

Sketch on separate diagrams:

(i) y=f(x)y=f(x);

(ii) y=−f(x)+3y=-f(x)+3;

(iii) y=f(−x+3)y=f(-x+3);

(iv) y=f(2x+1)y=f(2x+1).

Clearly indicate asymptotes, turning points and intercepts where possible.

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Independent qualitative reconstruction of the supplied graph y=f(x)+3: maximum (-1,5),y intercept (0,3), right-hand horizontal asymptote y=1. Unlabelled crossings and interpolation are schematic; no exact f formula or additional intercept coordinate is inferred.
Independent qualitative reconstruction of the supplied graph y=f(x)+3: maximum (-1,5),y intercept (0,3), right-hand horizontal asymptote y=1. Unlabelled crossings and interpolation are schematic; no exact f formula or additional intercept coordinate is inferred.
Independent qualitative reconstruction of the supplied graph y=f(x)+3: maximum (-1,5),y intercept (0,3), right-hand horizontal asymptote y=1. Unlabelled crossings and interpolation are schematic; no exact f formula or additional intercept coordinate is inferred.

CJC · Discussion · Question 2

Answer

(i) Maximum (−1,2), O, y=−2.

(ii) Minimum (−1,1), y-intercept 3, y=5, no x-intercept.

(iii) Maximum (4,2), known zero 3, y=−2.

(iv) Maximum (−1,2), known zero −1/2, y=−2.

(i) Remove the given vertical shift first

Translate the supplied graph 3 units down to obtain f. Its maximum is (−1,2), known intercept is O=(0,0), and right horizontal asymptote is y=−2. The other x-intercept is shown but unlabelled; call it α<−1 rather than assigning a value.

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(ii) Retain both source stages, including the coloured final annotation

Reflect f in the x-axis to get −f: minimum (−1,−2), origin and horizontal asymptote y=2. Then translate 3 units up to get −f+3: minimum (−1,1), y-intercept (0,3), horizontal asymptote y=5. Since the supplied f has maximum 2, this final curve has no x-intercept.

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(iii) Translate, then reflect, as printed

f(x)→f(x+3)→f(−x+3)f(x)\to f(x+3)\to f(-x+3). The intermediate has maximum (−4,2), known zero −3, and asymptote −2. Reflection in the y-axis gives final maximum (4,2), known zero (3,0), other zero (3−α,0), and left horizontal asymptote y=−2. Its y-intercept f(3) is not numerically determined by the source labels.

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(iv) Translate, then compress, including the coloured final step

f(x)→f(x+1)→f(2x+1)f(x)\to f(x+1)\to f(2x+1), with a horizontal scale 1/2 in the last step. The intermediate maximum (−2,2) and known zero (−1,0) become maximum (−1,2) and known zero (−1/2,0). The other zero is ((α−1)/2,0); the right horizontal asymptote stays y=−2. The y-intercept f(1) is not numerically specified.

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Compare your sketch

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Seven source-derived panels retain the printed intermediate graphs and coloured final annotations. Final f has maximum (-1,2), origin and right asymptote y=-2. Final -f+3 has minimum (-1,1),y intercept 3, asymptote 5. Final f(3-x) has maximum (4,2), known x intercept 3, asymptote -2 on the left. Final f(2x+1) has maximum (-1,2), known x intercept -1/2, right asymptote -2. Other unlabelled intercepts are not assigned numbers.
Seven source-derived panels retain the printed intermediate graphs and coloured final annotations. Final f has maximum (-1,2), origin and right asymptote y=-2. Final -f+3 has minimum (-1,1),y intercept 3, asymptote 5. Final f(3-x) has maximum (4,2), known x intercept 3, asymptote -2 on the left. Final f(2x+1) has maximum (-1,2), known x intercept -1/2, right asymptote -2. Other unlabelled intercepts are not assigned numbers.
Seven source-derived panels retain the printed intermediate graphs and coloured final annotations. Final f has maximum (-1,2), origin and right asymptote y=-2. Final -f+3 has minimum (-1,1),y intercept 3, asymptote 5. Final f(3-x) has maximum (4,2), known x intercept 3, asymptote -2 on the left. Final f(2x+1) has maximum (-1,2), known x intercept -1/2, right asymptote -2. Other unlabelled intercepts are not assigned numbers.

Common mistakes

Do not stop at the black printed intermediate sketches: the coloured annotations supply the final +3, reflection and horizontal-scale steps. Unlabelled intercepts cannot be assigned numeric values.

Solution notes

All four asks, all 7 intermediate/final source sketches and the given graph retained. Coloured annotations absent from native text extraction were visually compared and integrated into the relevant parts. Curves are qualitative reconstructions of supplied landmarks, not exact recovered formulas.

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