Two different modulus moves: y = |f(x)| — reflect every part below the x-axis up (parts above stay put). The graph never dips below the axis.
y = f(|x|) — keep the graph for x ≥ 0 and mirror it across the y-axis (so the result is symmetric).
IB-style question — y = |f(x)|
The line y = 2x − 4 is given. Describe the graph of y = |2x − 4| and state its minimum value.
Step by step
- Where 2x − 4 ≥ 0 (x ≥ 2) the graph is unchanged; where it's negative (x < 2) reflect it up.
- It becomes a V with its corner where 2x − 4 = 0.
Final answer
A V-shape with vertex at (2, 0); minimum value 0.
IB-style question — y = f(|x|)
f(x) = x² − 2x. Describe how y = f(|x|) differs from y = f(x).
Step by step
- f(|x|) = |x|² − 2|x| = x² − 2|x|.
- For x ≥ 0 it matches f; for x < 0 it's the mirror image of the right side.
Final answer
Keep the x ≥ 0 part of f and reflect it across the y-axis (an even, symmetric curve).
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Flip every height: y = 1/f(x) takes the reciprocal of each y-value:
• where f = 0 → a vertical asymptote; • where f is large → 1/f is near 0; where f is near 0 → 1/f is large; • a maximum of f becomes a minimum of 1/f (and the sign is kept).
IB-style question — reciprocal graph
f(x) = x − 3. Describe the graph of y = 1/f(x).
Step by step
- f = 0 at x = 3 → vertical asymptote there.
- As x → ±∞, f is large so 1/f → 0 → horizontal asymptote.
Final answer
y = 1/(x − 3): vertical asymptote x = 3, horizontal asymptote y = 0.