Precalculus-I · Unit 1 · Video 5 · Interactive Practice

Vertical and Horizontal Line Tests and the Toolkit Functions

IKey Formulas

Idea	Statement	Video example
Vertical line test	If some vertical line $x = a$ meets the graph more than once, the graph is not a function.	$x^2 + y^2 = 9$ at $x = 2$ : $y = \pm\sqrt{5}$
Horizontal line test	If some horizontal line $y = k$ meets the graph of a function more than once, the function is not one-to-one.	$y = x^2 - 1$ at $y = 3$ : $x = \pm 2$
One-to-one by algebra	Solve for the input: each output must give back exactly one input.	$A = \pi r^2,\ r > 0 \;\Rightarrow\; r = \sqrt{A/\pi}$

Key Insight: Being a function is a condition on the inputs (each input has one output). Being one-to-one adds the same condition on the outputs (each output has one input), so the horizontal line test only makes sense after the vertical line test has passed.

IIFunction or One-to-One? Arrow Diagrams

A function limits the arrows leaving each input; a one-to-one function also limits the arrows arriving at each output.

p

↦

q

↦

r

↦

IIIThe Two Line Tests

A vertical line marks one input, a horizontal line one output; meeting the graph twice breaks the rule.

Graph Test Line at x = 2

IVThe Nine Toolkit Functions

All nine toolkit graphs pass the vertical line test; one horizontal line at height $k$ splits them.

Height k 1

Constant $f(x) = c$ , here $c = 2$

Identity $f(x) = x$

Absolute value $f(x) = |x|$

Quadratic $f(x) = x^2$

Cubic $f(x) = x^3$

Reciprocal $f(x) = \frac{1}{x}$

Reciprocal squared $f(x) = \frac{1}{x^2}$

Square root $f(x) = \sqrt{x}$

Cube root $f(x) = \sqrt[3]{x}$

💡 Each thumbnail shows only $-3 \le x \le 3$ , but the crossings are counted over the whole domain: the cube root meets $y = 2$ at $x = 8$ , off the picture.

VQuiz Questions

Problem 1 · A Vertical Line Through a Circle

Given: the circle $x^2 + y^2 = 25$ and the vertical line $x = 3$ . Find every point where the line meets the circle.

✅ Correct! One input,

x = 3

, has two outputs,

4

and

-4

, so the circle is not a function.

❌ Not quite.

y^2 = 16

has two solutions,

y = 4

and

y = -4

; the line also meets the lower half of the circle.

❌ Not quite.

y^2 = 16

means

y

is a square root of

16

, not

16

itself;

(3, 16)

is not even on the circle, since

9 + 256 \ne 25

❌ Not quite. On the line

x = 3

every point has first coordinate

3

; these points have

y = 3

instead.

Show solution

Substitute $x = 3$ into the equation of the circle:

3^2 + y^2 = 25 \;\Rightarrow\; 9 + y^2 = 25 \;\Rightarrow\; y^2 = 16 \;\Rightarrow\; y = \pm 4

The line meets the circle at $(3, 4)$ and $(3, -4)$ . One input gives two outputs, so by the vertical line test the circle is not a function.

Problem 2 · Which Test Comes First?

Given: the sideways parabola $x = y^2 - 2$ . Every horizontal line $y = k$ meets it exactly once, at $(k^2 - 2,\ k)$ . Is this graph a one-to-one function?

✅ Correct! One-to-one is a property of functions, and this graph fails the vertical line test first.

❌ Not quite. The horizontal line test is used only after the vertical line test has passed; check the vertical lines first.

❌ Not quite. The stem says each horizontal line meets it once, at

(k^2 - 2,\ k)

; the failure is somewhere else.

❌ Not quite. A function of

x

needs each input

x

to give one output

y

; here

x = 2

gives

y = 2

and

y = -2

Show solution

Step 1: vertical line test. At $x = 2$ :

2 = y^2 - 2 \;\Rightarrow\; y^2 = 4 \;\Rightarrow\; y = \pm 2

The vertical line $x = 2$ meets the graph at $(2, 2)$ and $(2, -2)$ , so the graph is not a function of $x$ .

Step 2. Because it is not a function, the question "is it one-to-one?" does not arise; passing the horizontal line test does not rescue it. The answer is no.

Problem 3 · Area Back to Radius

Given: the area of a circle, $A = \pi r^2$ with $r > 0$ . A circle has area $A = 9\pi$ . Find its radius, then the inputs that would share this output if $r$ could be any real number.

What is the radius?

If $r$ could be any real number, which two inputs would $\pi r^2$ send to $9\pi$ ?

✅ Correct! With

r > 0

each area gives back one radius, so

A = \pi r^2

is one-to-one; on all real numbers

3

and

-3

share the output

9\pi

❌ Check the radius. Divide by

\pi

, then take the square root:

r^2 = 9

❌ Close.

r^2 = 9

has the solutions

\pm 3

, but a radius is a length, so

r > 0

keeps only one of them.

❌ Not quite. Dividing by

\pi

gives

r^2 = 9

; one more step, the square root, gives

r

❌ Not quite. That is

\sqrt{9\pi}

: divide both sides by

\pi

before taking the square root.

❌ Check the pair. Substitute each number into

\pi r^2

; both must give exactly

9\pi

❌ Not quite.

9

is the value of

r^2

, not a second radius:

\pi (9)^2 = 81\pi

❌ Not quite.

\pi \left(\tfrac{1}{3}\right)^2 = \tfrac{\pi}{9}

; the second input is the opposite of

3

, not its reciprocal.

❌ Not quite.

\pi (9)^2 = 81\pi

, not

9\pi

; the inputs are the square roots of

9

Show solution

Step 1: solve for the input.

9\pi = \pi r^2 \;\Rightarrow\; r^2 = \frac{9\pi}{\pi} = 9 \;\Rightarrow\; r = \pm 3

A radius is positive, so $r = 3$ . Each area gives back exactly one radius, $r = \sqrt{A/\pi}$ , so the area is a one-to-one function of the radius.

Step 2: drop the condition $r > 0$ .

\pi (3)^2 = 9\pi \qquad \pi (-3)^2 = 9\pi

The two inputs $3$ and $-3$ share the output $9\pi$ , so on all real numbers the formula $\pi r^2$ is not one-to-one. The restriction $r > 0$ is what makes it one-to-one.

Problem 4 · Sorting Toolkit Functions

Given: $g(x) = |x|$ , $h(x) = \sqrt[3]{x}$ , $k(x) = \frac{1}{x^2}$ and $m(x) = \frac{1}{x}$ . Decide which are one-to-one, then find two inputs that show $k$ is not.

Which of the four are one-to-one?

Which pair of inputs does $k(x) = \frac{1}{x^2}$ send to the same output $\tfrac{1}{4}$ ?

✅ Correct!

h

and

m

pass the horizontal line test;

k(2) = k(-2) = \tfrac{1}{4}

shows

k

fails it, as does

g(2) = g(-2) = 2

for

g

❌ Check the set. Look for opposite inputs that give the same output.

❌ Not quite.

g(-2) = g(2) = 2

: the horizontal line

y = 2

meets the V twice.

❌ Not quite. The reciprocal sends each nonzero input to its own output: if

\frac{1}{a} = \frac{1}{b}

then

a = b

, so

m

is one-to-one too.

❌ Not quite. Squaring erases the sign:

k(-1) = k(1) = 1

, so both branches of

k

sit above the axis and a horizontal line meets both.

❌ Check the pair. Substitute each input into

\frac{1}{x^2}

; both must give

\tfrac{1}{4}

❌ Not quite.

k(4) = \frac{1}{16}

, not

\frac{1}{4}

: the square of the input must be

4

❌ Not quite.

k\!\left(\tfrac{1}{2}\right) = \frac{1}{1/4} = 4

, the reciprocal of the target.

❌ Not quite.

k(2) = \tfrac{1}{4}

, but

k\!\left(\tfrac{1}{2}\right) = 4

; the two inputs must have the same square.

Show solution

Step 1: test each function.

$g(x) = |x|$ : $g(-2) = g(2) = 2$ , so not one-to-one.
$h(x) = \sqrt[3]{x}$ : each output $y$ comes only from $x = y^3$ , so one-to-one.
$k(x) = \frac{1}{x^2}$ : $k(-1) = k(1) = 1$ , so not one-to-one.
$m(x) = \frac{1}{x}$ : each nonzero output $y$ comes only from $x = \frac{1}{y}$ , so one-to-one.

Step 2: solve $k(x) = \tfrac{1}{4}$ .

\frac{1}{x^2} = \frac{1}{4} \;\Rightarrow\; x^2 = 4 \;\Rightarrow\; x = 2 \text{ or } x = -2

Check: $k(2) = \frac{1}{4}$ and $k(-2) = \frac{1}{(-2)^2} = \frac{1}{4}$ . The horizontal line $y = \tfrac{1}{4}$ meets the graph of $k$ twice.

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