Modulus and transformations of graphs flashcards

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The modulus |x|
x if x ≥ 0 and −x if x < 0: the size of x without its sign, its distance from 0. Never negative.
The graph y = |f(x)|
The graph of y = f(x) with every part below the x-axis reflected in the x-axis. Never below the x-axis.
The graph y = f(|x|)
The part of y = f(x) for x ≥ 0, together with its reflection in the y-axis. Always symmetrical about the y-axis.
False root (of a modulus equation)
A value from one of the two cases that does not satisfy the original equation, e.g. it makes the side equal to a modulus negative.
Critical values of a modulus inequality
The solutions of the corresponding equation; they split the number line into intervals to test or read from a sketch.
Stretch parallel to the x-axis, scale factor k
Every x-coordinate is multiplied by k; y-coordinates unchanged. y = f(x/k).
|A| = B (B an expression in x)
A = B or A = −B; keep only roots with B ≥ 0. (No — learn it)
|A| = |B|
A² = B², or A = ±B. No false roots, since both sides are non-negative. (No — derive it)
|A| < c and |A| > c (c > 0)
|A| < c ⟺ −c < A < c. |A| > c ⟺ A < −c or A > c. (No — learn it)
Vertex of y = |ax + b|
(−b/a, 0), on the x-axis; y-intercept (0, |b|). (No — derive it)
Image of (p, q) under y = af(bx) + c
(p/b, aq + c): divide x by b; multiply y by a, then add c. (No — derive it)
Asymptotes x = h and y = v under y = af(bx) + c
x = h/b and y = av + c. (No — derive it)
Sketch y = |2x − 3|
V shape, vertex (3/2, 0), y-intercept (0, 3), arms of gradient ±2.
Sketch y = |x² − 2x − 3|
Roots −1 and 3 kept; the part between them reflected, so the minimum (1, −4) becomes a local maximum (1, 4); cusps at the roots.
Sketch y = f(|x|) for f(x) = (x + 1)(x − 2)²
W shape symmetrical in the y-axis: maximum (0, 4), touching the x-axis at (±2, 0); the root at −1 disappears.
How many solutions has |f(x)| = k? (graph method)
Draw the horizontal line y = k on the sketch of y = |f(x)| and count the intersections.
y = f(x) has asymptote y = −3: what is the asymptote of y = |f(x)|?
y = 3: the part of the curve near the asymptote is reflected in the x-axis.
Sketch y = 3 − f(x) from y = f(x)
Reflect in the x-axis, then translate 3 up: (p, q) → (p, 3 − q). Maxima become minima.
Why can |x − 4| = 2x + 1 not have the root x = −5?
At x = −5, 2x + 1 = −9, but a modulus is never negative. The root comes from the wrong case and is rejected.
Why is squaring safe for |A| = |B| but not for |A| = B?
Both moduli are non-negative, so A² = B² loses nothing. If B can be negative, squaring adds the roots of |A| = −B.
Why does y = f(3x) divide x-coordinates by 3?
The input is tripled, so each output value is reached at one third of the original x.
In y = 2f(x) − 1, which comes first: the stretch or the translation?
The stretch: multiply y by 2, then subtract 1. (2(f(x) − 1) would be the other order.)
Why can y = f(|x|) lose a vertical asymptote?
Only the part with x ≥ 0 is kept, so an asymptote x = h with h < 0 is discarded with that part.
Which transformations move a vertical asymptote x = h?
Only horizontal ones (translations, stretches and reflections acting on x). Vertical changes leave it fixed.
Solve |x − 2| < 5.
−5 < x − 2 < 5, so −3 < x < 7.
Solve |x − 4| = 2x + 1.
x − 4 = 2x + 1 gives x = −5 (RHS −9: reject). 4 − x = 2x + 1 gives x = 1. Answer x = 1.
Solve |3x − 1| = |x + 5|.
Square: 8x² − 16x − 24 = 0, so x² − 2x − 3 = 0, x = 3 or x = −1.
Solve |2x − 3| > x + 1.
Critical values x = 2/3 and x = 4; the V is above the line outside them: x < 2/3 or x > 4.
Image of the maximum (1, 4) on y = 2f(3x)?
(1/3, 8): x ÷ 3, y × 2.
Image of (1, 4) on y = f(−x) + 1?
(−1, 5): change the sign of x, add 1 to y.
f(x) = 2 + 3/(x − 1). Asymptotes of y = 2f(½x) − 1?
x = 1 → x = 2; y = 2 → y = 2 × 2 − 1 = 3.
For which k does |x² − 2x − 3| = k have exactly four solutions?
The reflected minimum gives a maximum at (1, 4), so 0 < k < 4.
Solving |A| = B without checking
Substitute every root back and write “reject” for any that makes B negative; the check is part of the answer.
Writing a two-interval answer as 2/3 > x > 4
Impossible statement. Write x < 2/3 or x > 4 (two separate inequalities).
Confusing |f(x)| and f(|x|)
Ask where the modulus is: round the whole function (reflect in the x-axis) or round x (reflect in the y-axis).
Describing a transformation as just “a stretch”
Give the direction and scale factor: “stretch parallel to the x-axis, scale factor 1/3”; give vectors for translations.
Solving f(|x|) = k
Solve f(x) = k for x ≥ 0 only, discard negative roots, then include ± each positive root.

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