Porisma

Worked solution

Take the modulus of a cubic output

Sketch y=∣(x+1)(x−2)(x−3)∣y=|(x+1)(x-2)(x-3)| from y=(x+1)(x−2)(x−3)y=(x+1)(x-2)(x-3).

CJC · Transformations · Example 13

Choose which arcs to reflect

Answer

Reflect x<−1x<-1 and 2<x<32<x<3; retain −1≤x≤2-1\le x\le2 and x≥3x\ge3.

Independently checking the signs of the three factors gives these intervals. Zeros remain −1,2,3-1,2,3. Negative outputs change sign, while positive outputs stay fixed.

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Retain the source graph labels

Answer

All new outputs are non-negative; the original negative trough becomes a positive arch.

The source labels the original extrema approximately (0.131,6.06)(0.131,6.06) and (2.53,−0.879)(2.53,-0.879). Retain these printed approximate readings, without treating rounded coordinates as exact. The first stays above the axis; the second reflects to approximately (2.53,0.879)(2.53,0.879). The original negative arcs remain dashed in the reconstruction.

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Cubic f=(x+1)(x−2)(x−3), dashed where negative, and blue |f|. Roots −1,2,3; source approximate extrema (0.131,6.06) and original (2.53,−0.879). The latter reflects to a positive arch of height about 0.879.
Cubic f=(x+1)(x−2)(x−3), dashed where negative, and blue |f|. Roots −1,2,3; source approximate extrema (0.131,6.06) and original (2.53,−0.879). The latter reflects to a positive arch of height about 0.879.
Cubic f=(x+1)(x−2)(x−3), dashed where negative, and blue |f|. Roots −1,2,3; source approximate extrema (0.131,6.06) and original (2.53,−0.879). The latter reflects to a positive arch of height about 0.879.
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