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1. Rational functions: vertical and horizontal asymptotes

Warm-up

Three quick questions from earlier lessons.

Discover

Move the sliders and watch the graph of $y = \dfrac{a}{x - h}$ .

The graph of y = a over (x − h): two branches separated by a vertical asymptote at x = h, approaching the x-axis far left and far right.

Where is the vertical line that the graph never touches?

At $x = h$
At $x = a$

Key idea

A rational function is a quotient of polynomials.

Vertical asymptotes are where the denominator is 0 (and the numerator is not). Horizontal asymptote: if the top degree is smaller,

y = 0

; if the degrees are equal,

y = \dfrac{\text{leading coefficient on top}}{\text{leading coefficient below}}

; if the top degree is larger, there is none.

Worked example

Find the asymptotes of $f(x) = \dfrac{6x + 1}{2x - 8}$ .

Vertical asymptote:

x = 4

Denominator 0: 2x − 8 = 0.

Horizontal asymptote:

y = \dfrac{6}{2} = 3

Same degree: ratio of leading coefficients.

Your turn, guided

Your turn: $f(x) = \dfrac{5x - 2}{x + 3}$ .

Vertical asymptote:

x = -3

x + 3 = 0.

Horizontal asymptote:

y = 5

Same degree: 5 over 1.

Practice ★

$f(x) = \dfrac{ 4x + 6 }{x - 6}$ has a vertical asymptote $x = c$ . What is $c$ ?
$f(x) = \dfrac{ 6x + 4 }{x - 5}$ has a vertical asymptote $x = c$ . What is $c$ ?
$f(x) = \dfrac{ 2x + 5 }{x - 2}$ has a vertical asymptote $x = c$ . What is $c$ ?
$f(x) = \dfrac{ 5x + 7 }{x - 4}$ has a vertical asymptote $x = c$ . What is $c$ ?

Watch out!

Practice ★★

$f(x) = \dfrac{ 5x + 6 }{ 2x + 4 }$ has a horizontal asymptote $y = k$ . What is $k$ ?
$f(x) = \dfrac{ 2x + 2 }{ 4x + 7 }$ has a horizontal asymptote $y = k$ . What is $k$ ?
$f(x) = \dfrac{ 8x + 5 }{ 4x + 5 }$ has a horizontal asymptote $y = k$ . What is $k$ ?

What is the horizontal asymptote of $f(x) = \dfrac{ 4x + 1}{x^2 + 6}$ ?
1. $y = 0$
2. $y = 4$
3. There is none, because the degrees are different
What is the horizontal asymptote of $f(x) = \dfrac{ 8x + 1}{x^2 + 5}$ ?
1. $y = 0$
2. $y = 8$
3. There is none, because the degrees are different

Exit ticket

Three questions to check you have got it.

$f(x) = \dfrac{ 5x + 6 }{ 2x + 4 }$ has a horizontal asymptote $y = k$ . What is $k$ ?
$f(x) = \dfrac{ 2x + 2 }{ 4x + 7 }$ has a horizontal asymptote $y = k$ . What is $k$ ?
$f(x) = \dfrac{ 8x + 5 }{ 4x + 5 }$ has a horizontal asymptote $y = k$ . What is $k$ ?

Summary card

Asymptotes of rational functions

Vertical: zeros of the denominator.
Horizontal: compare degrees. Top smaller: $y = 0$ . Equal: ratio of leading coefficients. Top larger: none.

Watch out: use the leading coefficients, not the constants.

2. Rational functions: intercepts and holes

Warm-up

Three quick questions from earlier lessons.

Discover

$f(x) = \dfrac{(x - 2)(x + 1)}{x - 2}$ . Simplified, it looks like $x + 1$ . Is $f$ exactly the same function as $x + 1$ ?

Almost: $f$ is not defined at $x = 2$
Yes, exactly the same

Key idea

The numerator and a cancelled factor tell you where the graph meets the axis and where it has a gap.

x-intercepts: zeros of the numerator that are not zeros of the denominator. Hole: a factor that cancels from top and bottom; the hole is at that

x

, and its height is the value of the simplified rule there. y-intercept:

f(0)

Worked example

Analyze $f(x) = \dfrac{(x - 3)(x + 4)}{(x - 3)(x - 1)}$ .

f(x) = \dfrac{x + 4}{x - 1}

for

x \ne 3

x − 3 cancels.

Hole at

\left(3, \dfrac{7}{2}\right)

The simplified rule at x = 3: 7/2.

x

-intercept:

x = -4

Numerator zero that remains.

Vertical asymptote:

x = 1

Denominator zero that remains.

Your turn, guided

Your turn: $f(x) = \dfrac{(x + 2)(x - 5)}{x + 2}$ .

Hole at

x = -2

x + 2 cancels: f(x) = x − 5 for x ≠ −2.

The hole is at

(-2, -7)

The simplified rule at x = −2: −2 − 5.

Practice ★

Where does the graph of $f(x) = \dfrac{x - 8}{x + 4}$ cross the $x$ -axis? Give $x$ .
Where does the graph of $f(x) = \dfrac{x - 5}{x + 1}$ cross the $x$ -axis? Give $x$ .
Where does the graph of $f(x) = \dfrac{x - 8}{x + 8}$ cross the $x$ -axis? Give $x$ .
Where does the graph of $f(x) = \dfrac{x - 3}{x + 7}$ cross the $x$ -axis? Give $x$ .

Watch out!

Practice ★★

$f(x) = \dfrac{(x - 5)(x + 6)}{x - 5}$ . At which $x$ does the graph have a hole?
$f(x) = \dfrac{(x - 6)(x + 4)}{x - 6}$ . At which $x$ does the graph have a hole?
$f(x) = \dfrac{(x - 2)(x + 4)}{x - 2}$ . At which $x$ does the graph have a hole?

$f(x) = \dfrac{(x - 1)(x + 3)}{x - 1}$ has a hole at $x = 1$ . What is the $y$ -coordinate of the hole?
$f(x) = \dfrac{(x - 7)(x + 7)}{x - 7}$ has a hole at $x = 7$ . What is the $y$ -coordinate of the hole?

Exit ticket

Three questions to check you have got it.

$f(x) = \dfrac{(x - 1)(x + 3)}{x - 1}$ has a hole at $x = 1$ . What is the $y$ -coordinate of the hole?
$f(x) = \dfrac{(x - 7)(x + 7)}{x - 7}$ has a hole at $x = 7$ . What is the $y$ -coordinate of the hole?
$f(x) = \dfrac{(x - 9)(x + 1)}{x - 9}$ has a hole at $x = 9$ . What is the $y$ -coordinate of the hole?

Summary card

Intercepts and holes

Cancel common factors: each one leaves a hole.
Hole height: the simplified rule at that $x$ .
$x$ -intercepts: remaining zeros of the numerator; $y$ -intercept: $f(0)$ .

Watch out: zeros of the denominator give asymptotes, not intercepts.

3. Simplifying, multiplying and dividing rational expressions

Warm-up

Three quick questions from earlier lessons.

Discover

Simplify 6/9.

2/3
6/9 cannot be simplified

Key idea

Rational expressions work like fractions.

Factor numerators and denominators, then cancel common factors (never terms). Multiply numerators and denominators; to divide, multiply by the reciprocal. Keep the values that make an original denominator zero excluded.

Worked example

Simplify (x² − 9)/(x² + 3x).

(x − 3)(x + 3) / (x(x + 3))

Factor.

(x − 3)/x, for x ≠ 0 and x ≠ −3

Cancel x + 3.

Your turn, guided

Your turn: the value of (x² − 4)/(x − 2), simplified, at x = 3.

5

x + 2 at x = 3.

Practice ★

Simplify (x² − 4) ÷ (x − 2), that is, the fraction with numerator x² − 4 and denominator x − 2. Which answer is right?
1. x + 2, for x ≠ 2
2. x − 2
3. x + 2, for every x
Simplify (x² − 25) ÷ (x − 5), that is, the fraction with numerator x² − 25 and denominator x − 5. Which answer is right?
1. x + 5, for x ≠ 5
2. x − 5
3. x + 5, for every x

Multiply and simplify: (x + 8)/(x − 6) · (x − 6)/(x + 9).

Watch out!

Practice ★★

Divide and simplify: x/(x + 3) ÷ x²/(x + 3).
Divide and simplify: x/(x + 6) ÷ x²/(x + 6).

Multiply and simplify: (x + 8)/(x − 6) · (x − 6)/(x + 9).

Exit ticket

Three questions to check you have got it.

Simplify (x² − 4) ÷ (x − 2), that is, the fraction with numerator x² − 4 and denominator x − 2. Which answer is right?
1. x + 2, for x ≠ 2
2. x − 2
3. x + 2, for every x
Simplify (x² − 25) ÷ (x − 5), that is, the fraction with numerator x² − 25 and denominator x − 5. Which answer is right?
1. x + 5, for x ≠ 5
2. x − 5
3. x + 5, for every x
Simplify (x² − 9) ÷ (x − 3), that is, the fraction with numerator x² − 9 and denominator x − 3. Which answer is right?
1. x + 3, for x ≠ 3
2. x − 3
3. x + 3, for every x

Summary card

Multiplying and dividing rational expressions

Factor, cancel common factors, multiply; divide by multiplying by the reciprocal.

Watch out: cancel factors, never terms.

4. Adding and subtracting rational expressions

Warm-up

Three quick questions from earlier lessons.

Discover

What is 1/2 + 1/3?

Key idea

Add rational expressions over a common denominator.

Factor the denominators, find the least common denominator, rewrite each fraction over it, then add or subtract the numerators only. Simplify at the end.

Worked example

Write 2/x + 3/(x + 1) as one fraction.

x(x + 1)

LCD.

2(x + 1)/(x(x + 1)) + 3x/(x(x + 1))

Rewrite.

(5x + 2)/(x(x + 1))

Add.

Your turn, guided

Your turn: the value of 1/x + 1/x at x = 4.

\tfrac12

2/x at x = 4.

Practice ★

Simplify 6/(x − 1) − 9/(x − 1).
Simplify 9/(x − 1) − 2/(x − 1).

What is the least common denominator of 1/(x² − 16) and 1/(x + 4)?

Watch out!

Practice ★★

Write 1/x + 1/(x + 8) as a single fraction.
Write 1/x + 1/(x + 1) as a single fraction.

Simplify 6/(x − 1) − 9/(x − 1).

Exit ticket

Three questions to check you have got it.

Write 1/x + 1/(x + 8) as a single fraction.
Write 1/x + 1/(x + 1) as a single fraction.
Write 1/x + 1/(x + 4) as a single fraction.

Summary card

Adding rational expressions

Common denominator, then add the numerators.

Watch out: never add the denominators.

5. Solving rational equations

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $\dfrac{6}{x} = 2$ . Which value works?

$x = 3$
$x = 12$

Key idea

Clear the denominators, solve, then check.

Multiply both sides by the common denominator, solve the resulting polynomial equation, then check every solution in the original equation: a value that makes a denominator 0 is extraneous and must be rejected.

Worked example

$\dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 1$ .

x = 2 + (x - 2)

, so

x = x

Multiply by x − 2.

…except where the denominator is 0

True for every x.

Solutions: every

x \ne 2

x = 2 is excluded.

Your turn, guided

Your turn: $\dfrac{x}{x + 1} = \dfrac{2}{3}$ .

x = 2

3x = 2x + 2.

Practice ★

Solve $\dfrac{x}{x + 2} = \dfrac{ 2 }{ 7 }$ .
Solve $\dfrac{x}{x + 2} = \dfrac{ 4 }{ 6 }$ .
Solve $\dfrac{x}{x + 6} = \dfrac{ 5 }{ 8 }$ .

Watch out!

Practice ★★

Solve $\dfrac{1}{x} + \dfrac{1}{ 11 } = \dfrac{1}{ 5 }$ .
Solve $\dfrac{1}{x} + \dfrac{1}{ 12 } = \dfrac{1}{ 2 }$ .

Solve $\dfrac{x^2 - 4}{x - 2} = 4$ .
1. No solution: simplifying gives x = 2, but x = 2 makes the denominator 0
2. x = 2
3. Every x except 2

Exit ticket

Three questions to check you have got it.

Solve $\dfrac{x}{x + 2} = \dfrac{ 2 }{ 7 }$ .
Solve $\dfrac{x}{x + 2} = \dfrac{ 4 }{ 6 }$ .
Solve $\dfrac{x}{x + 6} = \dfrac{ 5 }{ 8 }$ .

Summary card

Rational equations

Clear denominators → solve → check in the original.

Watch out: reject values that make a denominator 0.

6. Solving radical equations; extraneous solutions

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $\sqrt{x} = -3$ . Squaring gives $x = 9$ . Does 9 work?

No, $\sqrt{9} = 3$
Yes

Key idea

Undo the root, then check.

To solve a radical equation, isolate the radical, raise both sides to the matching power (square for

\sqrt{\ }

, cube for

\sqrt[3]{\ }

), and solve. Squaring can create extraneous solutions: always check in the original equation. Cubing does not.

Worked example

$\sqrt{x + 6} = x$ .

x + 6 = x^2

Square.

x^2 - x - 6 = 0

, so

x = 3

x = -2

Solve.

Only

x = 3

√9 = 3 ✓; √4 = 2 ≠ −2 ✗.

Your turn, guided

Your turn: $\sqrt{x - 1} = 4$ .

x = 17

x − 1 = 16.

Practice ★

Solve $\sqrt{x + 2} = 8$ .
Solve $\sqrt{x + 8} = 7$ .

Solve $\sqrt[3]{x - 8} = 3$ .

Watch out!

Practice ★★

Solve $\sqrt{x + 6} = x$ . (Squaring gives two candidates; enter the one that works.)
Solve $\sqrt{x + 30} = x$ . (Squaring gives two candidates; enter the one that works.)

Exit ticket

Three questions to check you have got it.

Solve $\sqrt{x + 2} = 8$ .
Solve $\sqrt{x + 8} = 7$ .
Solve $\sqrt{x + 4} = 6$ .

Summary card

Radical equations

Isolate → square (or cube) → solve → check.

Watch out: squaring can add extraneous solutions.

Answers: Rational and radical relationships

1. Rational functions: vertical and horizontal asymptotes

$\frac{5}{2}$
$\frac{1}{2}$
$2$

$y = 0$
$y = 0$

$\frac{5}{2}$
$\frac{1}{2}$
$2$

2. Rational functions: intercepts and holes

$4$
$14$

$4$
$14$
$10$

3. Simplifying, multiplying and dividing rational expressions

x + 2, for x ≠ 2
x + 5, for x ≠ 5

$\frac{x + 8}{x + 9}$

$\frac{1}{x}$
$\frac{1}{x}$

$\frac{x + 8}{x + 9}$

x + 2, for x ≠ 2
x + 5, for x ≠ 5
x + 3, for x ≠ 3

4. Adding and subtracting rational expressions

$- \frac{3}{x - 1}$
$\frac{7}{x - 1}$

$x^{2} - 16$

$\frac{2 x + 8}{x \left(x + 8\right)}$
$\frac{2 x + 1}{x \left(x + 1\right)}$

$- \frac{3}{x - 1}$

$\frac{2 x + 8}{x \left(x + 8\right)}$
$\frac{2 x + 1}{x \left(x + 1\right)}$
$\frac{2 x + 4}{x \left(x + 4\right)}$

5. Solving rational equations

$\frac{4}{5}$
$4$
$10$

$\frac{55}{6}$
$\frac{12}{5}$

No solution: simplifying gives x = 2, but x = 2 makes the denominator 0

$\frac{4}{5}$
$4$
$10$

6. Solving radical equations; extraneous solutions

$62$
$41$

$35$

$62$
$41$
$32$