You can sketch $y = f(x)$. But what happens when you apply a square root or squaring operation to the whole function? The graph changes in non-obvious ways, some parts are cut off, others are flipped above zero, and key points stay fixed. In this lesson you'll master exactly how $y = \sqrt{f(x)}$ and $y = [f(x)]^2$ transform any graph.
Where is $y = \sqrt{f(x)}$ defined? Without looking anything up what happens to the graph when $f(x) = 0$? When $f(x) = 1$? Write your prediction before reading on.
There are two transformations in this lesson, and they behave in opposite ways. Lock these rules into memory.
$y = \sqrt{f(x)}$ is defined only where $f(x) \ge 0$. It flattens the curve, the square root grows more slowly than $f(x)$ itself, so values above 1 are pulled down and values between 0 and 1 are pushed up.
$y = [f(x)]^2$ is defined everywhere $f(x)$ is defined. It reflects negative parts above zero and stretches values above 1 while shrinking values between 0 and 1.
Domain: $\sqrt{f(x)}$ is only defined where $f(x) \ge 0$. You must solve the inequality $f(x) \ge 0$ to find the domain.
Key graphing rules:
Domain of f(x): solve f(x) 0 first; Where f(x) = 0: the graph touches the x-axis
Pause, copy the $y = \sqrt{f(x)}$ domain rule into your book: solve $f(x) \ge 0$ first to find the domain; wherever $f(x) = 0$ the graph touches the $x$-axis; wherever $f(x) > 0$ the graph is above it.
Quick check: For which values of $x$ is $y = \sqrt{x - 3}$ defined?
We just saw that $y = \sqrt{f(x)}$ requires $f(x) ge 0$ as its domain and squashes tall values down toward the $x$-axis. That raises a question: squaring is the reverse of the square root, does $y = [f(x)]^2$ have the same domain restriction, and how does it treat negative parts of $f$? This card answers it → $[f(x)]^2$ has no domain restriction; squaring raises the graph above the axis everywhere, including where $f$ was negative.
Squaring $f(x)$ has no domain restriction it is defined wherever $f(x)$ is defined.
Key graphing rules:
![For f(x) = x², √f = |x| lies below and [f]² = x⁴ rises above f on the working interval.](/assets/diagrams-pilot-v3/function-sqrt-square-compare.mobile.svg)
Both $\sqrt{f}$ and $[f]^2$ pass through the same point where $f(x)=1$ (invariant point).
[f(x)]^2 0 always, no values below the x-axis; Negative parts of f get reflected up (like modulus but scaled differently)
Pause, copy the $y = [f(x)]^2$ rule into your book: no domain restriction; always $\ge 0$; where $f(x)$ crosses zero, $[f(x)]^2 = 0$; where $f(x) < 0$, squaring brings it above the axis, similar to $|f(x)|$ but quadratic in scale.
Did you get this? True or false: $y = [f(x)]^2$ can produce negative $y$-values.
Worked examples · 3 in a row, reveal as you go
Given $f(x) = x + 2$, sketch $y = \sqrt{f(x)}$ and state its domain.
Given $f(x) = x + 2$, sketch $y = [f(x)]^2$ and mark any invariant points.
State the domain of $y = \sqrt{x^2 - 1}$ and identify key points for a sketch.
Fill the gap: For $f(x) = x - 5$, the domain of $y = \sqrt{f(x)}$ is $x \ge$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: where $f(x) = 1$, the graphs of $y = f(x)$ and $y = \sqrt{f(x)}$ intersect.
Activities · practice with the ideas
State the domain of $y = \sqrt{2x - 6}$.
For $f(x) = x - 1$, find all invariant points of $y = \sqrt{f(x)}$ and $y = [f(x)]^2$.
For $f(x) = 4 - x^2$, find the domain of $y = \sqrt{f(x)}$ and name the key points on the graph.
If $f(x) = x^2 - 4$, at which $x$-values does $[f(x)]^2 = f(x)$? Explain why.
Compare and contrast $y = |f(x)|$ and $y = [f(x)]^2$ for a function that takes both positive and negative values. In what ways are they similar? In what ways do they differ?
Odd one out: Which of the following is the odd one out? Select the one that does NOT share a property with the others.
Earlier you predicted where $\sqrt{f(x)}$ is defined and what happens at $f(x) = 0$ and $f(x) = 1$.
The domain is $\{x : f(x) \ge 0\}$. At $f(x) = 0$: the graph touches the $x$-axis. At $f(x) = 1$: both $f$ and $\sqrt{f}$ pass through $y = 1$, an invariant point. Squaring also shrinks values between 0 and 1 (e.g. $0.5^2 = 0.25 < 0.5$) and stretches values above 1 (e.g. $2^2 = 4 > 2$).
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 $y = \sqrt{x^2 - 1}$, showing the domain and key points. (2 marks)
Q2. Given $f(x) = x - 2$, sketch $y = [f(x)]^2 = (x-2)^2$ and mark any invariant points. (2 marks)
Q3. Compare and contrast $y = |f(x)|$ and $y = [f(x)]^2$ for a function that takes both positive and negative values. (3 marks)
Activity 1: 1. Domain: $x \ge 3$ · 2. Invariant at $x=1$ (f=0) and $x=2$ (f=1) · 3. Domain: $-2 \le x \le 2$; key points $(\pm 2, 0)$, invariant at $f=1$ · 4. At $f=0$ (i.e. $x=\pm 2$) and $f=1$; these are the points where $t^2 = t$ which is $t(t-1)=0$, so $t=0$ or $t=1$ · 5. Both always non-negative; $|f|$ simply reflects negatives, same magnitude; $[f]^2$ scales by the square, much larger for $|f|>1$, much smaller for $0<|f|<1$.
Q1 (2 marks): Domain: $x \le -1$ or $x \ge 1$ [1]. Key points: $(\pm 1, 0)$, touches $x$-axis; $(\pm\sqrt{2}, 1)$, invariant. Two branches rising from $x=\pm 1$ [1].
Q2 (2 marks): Parabola $(x-2)^2$ with vertex at $(2,0)$ [1]. Invariant at $x=2$ (f=0) and $x=3$ (f=1) [1].
Q3 (3 marks): Both always $\ge 0$ [1]; $|f|$ reflects negative parts up with same magnitude; $[f]^2$ reflects AND scales, values $>1$ become larger, $0<f<1$ become smaller [1]; at $f=0$ and $f=1$ (also $f=-1$ for squaring), the graphs meet [1].
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
Work through this topic 1-on-1 with an experienced HSC tutor.
Book a free session →