Sketch $y = |f(x)|$ and $y = f(|x|)$ from the graph of $y = f(x)$. Two distinct transformations, two simple geometric rules, flip the negatives up for one, mirror the right side for the other. Get these clear and you'll handle any absolute value graph with confidence.
Describe in your own words what $y = |f(x)|$ does to the graph of $y = f(x)$. What about $y = f(|x|)$? Are they the same? Write your prediction before reading on.
These are two entirely different transformations. Do not confuse them. The absolute value either acts on the output (the $y$-value) or on the input (the $x$-value).
$y = |f(x)|$: the absolute value acts on the output. Any part of the graph below the x-axis is reflected up. Parts above stay the same.
$y = f(|x|)$: the absolute value acts on the input. The right half of the graph ($x \ge 0$) is kept; the left half is discarded and replaced by a mirror of the right half.
To sketch $y = |f(x)|$, apply this two-step process:
Result: The graph is always on or above the x-axis. All x-intercepts of the original curve are kept (they become points where $|f(x)| = 0$, i.e., cusps or x-intercepts). Any "valleys" below the x-axis become "hills" above it.
y = |f(x)|: keep all parts where f(x) 0; reflect all parts where f(x) < 0 upward; Result always 0, the graph never goes below the x-axis
Pause, copy the $y = |f(x)|$ rule into your book: keep all parts of the graph where $f(x) \ge 0$ unchanged; reflect any parts below the $x$-axis upward; the result is always $\ge 0$.
Quick check: If $f(x) < 0$ for $0 < x < 2$, what does $y = |f(x)|$ look like in this interval compared to $y = f(x)$?
We just saw that $y = |f(x)|$ reflects below-axis parts upward, keeping the same $x$-values but making all outputs non-negative. That raises a question: what if instead of taking the absolute value of the output, we feed in the absolute value of $x$, how does $y = f(|x|)$ transform the graph differently? This card answers it → keep the right half ($x ge 0$) of $f$ unchanged, discard the left half, then reflect the right half to create a $y$-axis symmetric graph.
To sketch $y = f(|x|)$, apply this three-step process:
Result: The graph is always symmetric about the y-axis. Since $|-x| = |x|$, we always have $f(|-x|) = f(|x|)$, which means the function value is the same for $x$ and $-x$. This is the definition of an even function.
Left: $y = |f(x)|$ reflects negative outputs upward. Right: $y = f(|x|)$ mirrors the right side across the y-axis.
y = f(|x|): keep x 0 unchanged; discard x < 0; reflect right half in y-axis; Result is always symmetric about the y-axis (even function)
Pause, copy the $y = f(|x|)$ rule into your book: keep $x \ge 0$ unchanged, discard $x < 0$, then reflect the right half in the $y$-axis; the result is always an even function symmetric about the $y$-axis.
Did you get this? True or false: the graph of $y = f(|x|)$ is always symmetric about the y-axis.
Worked examples · 3 in a row, reveal as you go
Given $f(x) = x - 2$, sketch $y = |f(x)| = |x - 2|$, showing all key features.
Given $f(x) = x - 2$, sketch $y = f(|x|) = |x| - 2$, showing all key features.
Given $f(x) = (x - 1)(x - 3)$, describe the key features of $y = |f(x)|$ and $y = f(|x|)$ without sketching.
Fill the gap: The graph of $y = |x^2 - 4|$ has cusps at $x = $ and $x = $ (enter the smaller value in the first box).
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the graph of $y = |f(x)|$ is always on or above the x-axis.
Activities · practice with the ideas
Sketch $y = |x^2 - 4|$, identifying where the original parabola is negative and what happens there.
Sketch $y = |x|^2 - 4 = f(|x|)$ where $f(x) = x^2 - 4$. Explain its symmetry.
For $f(x) = x - 1$, state the vertex of $y = |f(x)|$ and the vertex of $y = f(|x|)$.
Given $f(x) = (x-1)(x-3)$, state the x-intercepts of $y = f(|x|)$.
Explain why $y = f(|x|)$ always produces a graph symmetric about the y-axis, using the property $|-x| = |x|$.
Odd one out: Three of these are true about $y = |f(x)|$. Which is NOT true?
Earlier you predicted whether $y = |f(x)|$ and $y = f(|x|)$ are the same transformation. They are not the same:
Both involve absolute value, but they act on completely different parts of the function. One affects the output; the other affects the input.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Sketch $y = |x^2 - 4|$, showing all key features including x-intercepts, cusps and any turning points. (2 marks)
Q2. Sketch $y = |x|^2 - 4$ and explain its symmetry. (2 marks)
Q3. Given $f(x) = (x - 1)(x - 3)$, sketch $y = |f(x)|$ and $y = f(|x|)$ on separate axes, labelling all key features for each graph. (3 marks)
Activity answers: 1. $y = x^2 - 4$ is negative for $-2 < x < 2$; |·| reflects this part upward, creating a bump with max $(0, 4)$; cusps at $(\pm 2, 0)$ · 2. Since $|x|^2 = x^2$, this is $y = x^2 - 4$ which is already an even function, the graph is unchanged by the transformation · 3. Vertex of $|f(x)|$: $(1,0)$; vertex of $f(|x|)$: $(0, -1)$ · 4. Original roots $x = 1, 3$; for $f(|x|)$: $|x| = 1 \Rightarrow x = \pm 1$ and $|x| = 3 \Rightarrow x = \pm 3$ · 5. $f(|-x|) = f(|x|)$ means $y$-value at $-x$ equals $y$-value at $x$, which is the definition of y-axis symmetry (even function).
Q1 (2 marks): $y = x^2 - 4$: negative for $-2 < x < 2$ [0.5]. After $|·|$: cusps at $(\pm 2, 0)$, bump with maximum at $(0, 4)$ [1]. Arms for $|x| > 2$ unchanged: $y = x^2 - 4$ [0.5].
Q2 (2 marks): $|x|^2 = x^2$, so $y = x^2 - 4$ [1]. Symmetry: $f(|-x|) = (-x)^2 - 4 = x^2 - 4 = f(x)$, i.e., even function; graph symmetric about y-axis. Vertex $(0, -4)$, x-intercepts $x = \pm 2$ [1].
Q3 (3 marks): $y = |f(x)|$: Keep arms ($x < 1$ and $x > 3$) unchanged; reflect middle section upward; cusps at $(1,0)$ and $(3,0)$; maximum at $(2,1)$ [1.5]. $y = f(|x|)$: Right half of parabola through $(1,0)$, $(3,0)$, minimum $(3, 0)$ to $(1, 0)$; mirrored to left; x-intercepts at $\pm 1$, $\pm 3$; y-intercept $(0, 3)$; symmetric about y-axis [1.5].
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