Porisma

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Recover a signed graph from its two modulus graphs

The supplied graphs are y=f(∣x∣)y=f(|x|) and y=∣f(x)∣y=|f(x)|. Both have maximum (4, a) and vertical tangents x=0, x=6. The first is even, with maxima (−4, a), (4, a), zeros (−6,0), (0,0), (6,0), and slant asymptotes y=x+2 on the left and y=−x+2 on the right. The second is nonnegative, with zeros (0,0), (6,0), and slant asymptotes y=−x+2 on the left and y=x−2 on the right.

Given f itself has only one oblique asymptote, sketch y=f(x)y=f(x), stating asymptotes, turning points and axial intercepts wherever possible. marks

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Two independent reconstructions of the supplied f(|x|) and |f(x)| graphs. Maximum height a remains symbolic and is not inferred numerically from the drawing. The first is even, with maxima (±4,a), zeros 0,±6, slant asymptotes y=x+2 left,y=-x+2 right. The second is nonnegative, with maximum (4,a), zeros 0,6, slant asymptotes y=-x+2 left,y=x-2 right. Vertical tangents are not asymptotes.
Two independent reconstructions of the supplied f(|x|) and |f(x)| graphs. Maximum height a remains symbolic and is not inferred numerically from the drawing. The first is even, with maxima (±4,a), zeros 0,±6, slant asymptotes y=x+2 left,y=-x+2 right. The second is nonnegative, with maximum (4,a), zeros 0,6, slant asymptotes y=-x+2 left,y=x-2 right. Vertical tangents are not asymptotes.
Two independent reconstructions of the supplied f(|x|) and |f(x)| graphs. Maximum height a remains symbolic and is not inferred numerically from the drawing. The first is even, with maxima (±4,a), zeros 0,±6, slant asymptotes y=x+2 left,y=-x+2 right. The second is nonnegative, with maximum (4,a), zeros 0,6, slant asymptotes y=-x+2 left,y=x-2 right. Vertical tangents are not asymptotes.

[2020/RI/Promo/8(ii)]

CJC · Review problems · Question 9

On x≥0, the graph of f(|x|) already equals the original f.

Read the worked solution