Grade by Grade
A. Characteristics of Functions · Functions and Equivalent Expressions

The Parent Functions: x, x², √x and 1/x

A1.3About 70 minutes
Learning goals
  • I can sketch the graphs of f(x) = x, f(x) = x², f(x) = √x and f(x) = 1/x using key points.
  • I can state the domain and range of each parent function and explain them.
  • I can find the domain of a function algebraically by avoiding division by zero and square roots of negatives.
  • I can identify the asymptotes of the reciprocal function.

Words to know

parent function
The simplest function of a family, such as f(x) = x²; other members are transformations of it.
principal square root
The non-negative square root, written √x. √25 = 5.
radicand
The expression under a square root sign.
reciprocal function
f(x) = 1/x.
asymptote
A line that a graph approaches more and more closely. For y = 1/x the asymptotes are x = 0 and y = 0.
hyperbola
The two-branch curve that is the graph of y = 1/x (and of xy = k).

Four functions to know by heart

Hockey players practise the same four or five moves until they are automatic, and then combine them in games. In this unit you will do the same with four basic graphs. Later in the unit you will slide, stretch and flip them to make whole families of new functions. Because every member of a family comes from one of these, they are called parent functions:

f(x) = x f(x) = x² f(x) = √x f(x) = 1/x

You already know the first two well. The square root function and the reciprocal function 1/x are new as functions, and they come with interesting domain and range restrictions.

Remember from Grade 9

In Grade 9 you graphed relations of the form xy = k, where the product of x and y is constant. With k = 1, xy = 1 is the same as y = 1/x: the reciprocal function in this lesson.

The square root function f(x) = √x

The symbol √x means the principal square root: the non-negative number whose square is x. So √9 = 3, not −3.

  • You cannot take the square root of a negative number and get a real number (no real number squared is negative). So the domain is {x ∈ ℝ | x ≥ 0}.
  • The outputs are never negative: range {y ∈ ℝ | y ≥ 0}.
  • Easy points come from perfect squares: (0, 0), (1, 1), (4, 2), (9, 3).

Notice that y = √x is the top half of the sideways parabola x = y² from Lesson 1. Taking only the top half is what makes it a function.

The reciprocal function f(x) = 1/x

  • Division by zero is undefined, so x = 0 is not allowed: domain {x ∈ ℝ | x ≠ 0}.
  • A fraction 1/x is never 0, because its numerator is 1: range {y ∈ ℝ | y ≠ 0}.
  • As x gets close to 0 from the right (0.1, 0.01, 0.001), 1/x gets huge (10, 100, 1000). As x gets very large (10, 100, 1000), 1/x gets close to 0.

The graph has two separate pieces called branches. It gets closer and closer to the y-axis (the line x = 0) and to the x-axis (the line y = 0) without ever touching them. A line like this is called an asymptote: x = 0 is the vertical asymptote and y = 0 is the horizontal asymptote.

x−2−1−0.500.51210
√xundefinedundefinedundefined0≈ 0.711≈ 1.41≈ 3.16
1/x−0.5−1−2undefined210.50.1
Tables of values for the square root and reciprocal functions. "Undefined" means the input is not in the domain.
xy−3−2−1123−3−2−1123f(x) = xD: {x ∈ ℝ}; R: {y ∈ ℝ}xy−3−2−112312345f(x) = x²D: {x ∈ ℝ}; R: {y ∈ ℝ | y ≥ 0}xy369−11234f(x) = √xD: {x ∈ ℝ | x ≥ 0}; R: {y ∈ ℝ | y ≥ 0}xy−424−424f(x) = 1/xD: {x ∈ ℝ | x ≠ 0}; R: {y ∈ ℝ | y ≠ 0}
The four parent functions with their key points. Notice the domain (how far left and right each graph goes) and the range (how far up and down).
Parent functionShapeKey pointsDomainRangeSpecial features
f(x) = xline(−1, −1), (0, 0), (1, 1){x ∈ ℝ}{y ∈ ℝ}slope 1; always increasing
f(x) = x²parabola(±1, 1), (±2, 4), (0, 0){x ∈ ℝ}{y ∈ ℝ | y ≥ 0}vertex (0, 0); symmetric in the y-axis
f(x) = √xhalf of a sideways parabola(0, 0), (1, 1), (4, 2), (9, 3){x ∈ ℝ | x ≥ 0}{y ∈ ℝ | y ≥ 0}starts at (0, 0); increasing
f(x) = 1/xhyperbola (two branches)(±1, ±1), (2, 0.5), (0.5, 2) and matching negatives{x ∈ ℝ | x ≠ 0}{y ∈ ℝ | y ≠ 0}asymptotes x = 0 and y = 0
Summary of the parent functions.

Finding the domain algebraically

For the functions in this course, two rules do most of the work:

  1. No division by zero. Set the denominator equal to zero, solve, and exclude those x-values.
  2. No square root of a negative. Set the expression under the root (the radicand) ≥ 0 and solve.

If neither situation happens (as with linear and quadratic functions), the domain is {x ∈ ℝ}.

Worked example

Example 1: Domains from equations

State the domain of each function.

(a) f(x) = √(x − 3) (b) g(x) = 1/(x + 2) (c) h(x) = √(6 − 2x) (d) k(x) = 5/(x² − 9)

  1. (a) Radicand ≥ 0: x − 3 ≥ 0, so x ≥ 3. D = {x ∈ ℝ | x ≥ 3}. Check: f(3) = √0 = 0 works; f(2) = √(−1) does not.
  2. (b) Denominator ≠ 0: x + 2 ≠ 0, so x ≠ −2. D = {x ∈ ℝ | x ≠ −2}.
  3. (c) 6 − 2x ≥ 0, so 6 ≥ 2x, which means x ≤ 3. D = {x ∈ ℝ | x ≤ 3}. (Check x = 0: √6 works. x = 4: √(−2) does not.)
  4. (d) x² − 9 = 0 when x = 3 or x = −3. D = {x ∈ ℝ | x ≠ ±3}.

(a) x ≥ 3 (b) x ≠ −2 (c) x ≤ 3 (d) x ≠ 3 and x ≠ −3.

xy36912−11234startxy−62−4−224x = −2f(x) = √(x − 3): need x − 3 ≥ 0g(x) = 1/(x + 2): need x + 2 ≠ 0domain {x ∈ ℝ | x ≥ 3}domain {x ∈ ℝ | x ≠ −2}
Left: √(x − 3) starts at x = 3, where the radicand becomes 0. Right: 1/(x + 2) has a vertical asymptote at x = −2, where the denominator would be 0.
Worked example

Example 2: Ranges by reasoning

State the range of (a) y = √x + 2 (b) y = −√x (c) y = 1/x + 3.

  1. (a) √x ≥ 0 always, so √x + 2 ≥ 2. Every value 2 or more is reached (√x can be any non-negative number). R = {y ∈ ℝ | y ≥ 2}.
  2. (b) √x ≥ 0, so −√x ≤ 0. R = {y ∈ ℝ | y ≤ 0}.
  3. (c) 1/x can be any real number except 0. Adding 3 to it gives any real number except 3. R = {y ∈ ℝ | y ≠ 3}.

(a) y ≥ 2 (b) y ≤ 0 (c) y ≠ 3.

Worked example

Example 3: Both from one function

State the domain and range of f(x) = √(x + 4) and sketch it.

  1. Domain: x + 4 ≥ 0, so x ≥ −4.
  2. Range: a principal square root is never negative and can be as large as you like: y ≥ 0.
  3. Points: f(−4) = 0, f(−3) = 1, f(0) = 2, f(5) = 3. Plot them and join with a smooth curve that starts at (−4, 0) and rises more and more slowly to the right.
  4. This is the graph of √x moved 4 units left. You will study moves like this in the lesson on translations.

D = {x ∈ ℝ | x ≥ −4}, R = {y ∈ ℝ | y ≥ 0}.

Just for fun

The graph of y = 1/x spends its whole life getting closer and closer to the axes without ever touching them. Mathematicians call that an asymptote. Everyone else calls it "playing it cool".

Hands-on activity

Calculator chase: sneaking up on an asymptote

You need: a scientific calculator (or a phone calculator), paper and a pencil.

  1. Make a table for y = 1/x with x = 1, 0.1, 0.01, 0.001 and 0.0001. What happens to y?
  2. Now try x = −1, −0.1, −0.01, −0.001. What happens?
  3. Try x = 10, 100, 1000, 10 000. What happens?
  4. Type 1 ÷ 0 into the calculator. What does it say?
  5. Type √(−4). What does it say?
  6. Write one sentence for each parent function explaining why its domain is what it is.

What to look for: y = 1, 10, 100, 1000, 10 000 (growing without limit); then −1, −10, −100, −1000 (growing without limit in the negative direction); then 0.1, 0.01, 0.001, 0.0001 (approaching 0 but never reaching it). 1 ÷ 0 and √(−4) give an error message, because those inputs are not in the domain.

Your turn

1

Evaluate f(x) = √x at x = 25 and at x = 0.16.

2

Evaluate g(x) = 1/x at x = −4 and at x = 0.2.

3

State the domain of f(x) = √(x + 7).

4

State the domain of g(x) = 3/(x − 5).

5

State the domain of h(x) = √(10 − x).

6

Name the asymptotes of y = 1/x.

7

Which parent functions have domain {x ∈ ℝ}?

More practice: warm-up

1

What is √49?

2

Why is x = 0 not in the domain of y = 1/x?

3

What is the range of y = √x?

4

Name the parent function whose graph is a parabola.

5

Give two points on y = √x other than (0, 0).

More practice: on your own

1

State the domain and range of f(x) = √x − 5.

2

State the domain of p(x) = 1/(2x − 7).

3

State the domain of q(x) = √(3x + 12).

4

State the range of y = 1/x − 2 and explain.

5

State the domain of r(x) = 4/(x² − 16) and explain.

Challenge

1

State the domain of f(x) = √(x − 2) / (x − 5).

2

State the domain of g(x) = 1/√(x + 1).

3

A square has area A cm². Its side is s(A) = √A. Its perimeter is P(A) = 4√A. State the domain of P in context and find the area of a square with perimeter 36 cm.

Check your understanding

Quiz

1.

What is the domain of f(x) = √x?

2.

Which line is a vertical asymptote of g(x) = 1/(x − 4)?

3.

What is the range of y = 1/x?

4.

What is the domain of h(x) = √(8 − 2x)?

5.

Which point is not on the graph of y = √x?

6.

Which parent function has range {y ∈ ℝ | y ≥ 0} and domain {x ∈ ℝ}?

For parents and teachers

Big idea: Expectation A1.3 names four functions, f(x) = x, f(x) = x², f(x) = √x and f(x) = 1/x, and asks students to explain their domains and ranges numerically, graphically and algebraically. These parent functions are the starting point for every transformation later in the unit.

How to help: Ask your teen to sketch all four parent functions from memory with three or more exact points each, and to say why √x and 1/x have restricted domains ("no square roots of negatives", "no division by zero").

Common difficulties:

  • Thinking √9 = ±3. The symbol √ means the non-negative root only.
  • Including the asymptote value in the domain or range (x = 0 and y = 0 for 1/x).
  • Solving inequalities like 6 − 2x ≥ 0 incorrectly. Moving terms carefully (6 ≥ 2x, so x ≤ 3) avoids the sign-flip trap.
  • Drawing 1/x so that it touches or crosses the axes.