Porisma

Worked solution

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

Answer

Asymptote y=−x+2; minimum (0,0), maximum (4, a); axial intercepts (0,0), (6,0). Positive for x<0, negative for x>6; a remains symbolic.

Recover the nonnegative-x portion first

For x≥0,f(∣x∣)=f(x)f(|x|)=f(x), so use that supplied right-hand branch unchanged. It starts at (0,0), rises to maximum (4, a), crosses at (6,0), then becomes negative and approaches y=−x+2. The vertical tangents at x=0,6 are retained; they are not vertical asymptotes.

How did this part go?

Choose the left-hand sign using the single oblique asymptote

The original has only one oblique asymptote, already identified as y=−x+2 from its right tail. Hence its left tail also approaches y=−x+2, which is positive as x→−∞. The supplied |f(x)| left branch has no further zeros, so the continuous branch stays positive for x<0. Use that branch without reflecting it downward.

How did this part go?

State the labelled features without inventing a

The recovered graph has minimum (0,0), maximum (4, a), x-intercepts (0,0), (6,0), and y-intercept (0,0). Its only oblique asymptote is y=−x+2. The height a is not numerically determined; the schematic leaves it symbolic. The reflected zero at (−6,0) belongs to f(|x|), not to the recovered original f.

How did this part go?

Compare your sketch

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Original f has only one slant asymptote y=-x+2 at both ends. It is positive for x<0, has minimum (0,0), maximum (4,a), crosses at (6,0) with a vertical tangent, and is negative after 6. Maximum height a is symbolic; no numerical value is assigned by the schematic. The only axial intercepts are (0,0),(6,0).
Original f has only one slant asymptote y=-x+2 at both ends. It is positive for x<0, has minimum (0,0), maximum (4,a), crosses at (6,0) with a vertical tangent, and is negative after 6. Maximum height a is symbolic; no numerical value is assigned by the schematic. The only axial intercepts are (0,0),(6,0).
Original f has only one slant asymptote y=-x+2 at both ends. It is positive for x<0, has minimum (0,0), maximum (4,a), crosses at (6,0) with a vertical tangent, and is negative after 6. Maximum height a is symbolic; no numerical value is assigned by the schematic. The only axial intercepts are (0,0),(6,0).

Common mistakes

A vertical tangent is not an asymptote. The maximum height a cannot be read as a numerical value from a schematic.

Solution notes

Both original modulus figures and the printed recovered curve reconstructed. All source labels, tangent conditions, one-asymptote reasoning and positivity conclusion retained. The rendered vertical scale is schematic with a symbolic, not a claimed fixed number.

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