Precalculus Examples

f (x) = 2 \log_{3} (x - 1) - 3

$f (x) = 2 \log_{3} (x - 1) - 3$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression

\log_{3} ({(x - 1)}^{2}) - 3

$\log_{3} ({(x - 1)}^{2}) - 3$ is undefined.

x = 1

$x = 1$

Step 1.2

Ignoring the logarithm, consider the rational function

R (x) = \frac{a x^{n}}{b x^{m}}

$R (x) = \frac{a x^{n}}{b x^{m}}$ where

n

$n$ is the degree of the numerator and

m

$m$ is the degree of the denominator.

1. If

n < m

$n < m$ , then the x-axis,

y = 0

$y = 0$ , is the horizontal asymptote.

2. If

n = m

$n = m$ , then the horizontal asymptote is the line

y = \frac{a}{b}

$y = \frac{a}{b}$ .

3. If

n > m

$n > m$ , then there is no horizontal asymptote (there is an oblique asymptote).

Step 1.3

There are no horizontal asymptotes because

Q (x)

$Q (x)$ is

1

$1$ .

No Horizontal Asymptotes

Step 1.4

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.5

This is the set of all asymptotes.

Vertical Asymptotes:

x = 1

$x = 1$

No Horizontal Asymptotes

Vertical Asymptotes:

x = 1

$x = 1$

No Horizontal Asymptotes

Step 2

Find the point at

x = 2

$x = 2$ .

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Step 2.1

Replace the variable

x

$x$ with

2

$2$ in the expression.

f (2) = 2 \log_{3} ((2) - 1) - 3

$f (2) = 2 \log_{3} ((2) - 1) - 3$

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Subtract

1

$1$ from

2

$2$ .

f (2) = 2 \log_{3} (1) - 3

$f (2) = 2 \log_{3} (1) - 3$

Step 2.2.1.2

Logarithm base

3

$3$ of

1

$1$ is

0

$0$ .

f (2) = 2 \cdot 0 - 3

$f (2) = 2 \cdot 0 - 3$

Step 2.2.1.3

Multiply

2

$2$ by

0

$0$ .

f (2) = 0 - 3

$f (2) = 0 - 3$

f (2) = 0 - 3

$f (2) = 0 - 3$

Step 2.2.2

Subtract

3

$3$ from

0

$0$ .

f (2) = - 3

$f (2) = - 3$

Step 2.2.3

The final answer is

- 3

$- 3$ .

- 3

$- 3$

- 3

$- 3$

Step 2.3

Convert

- 3

$- 3$ to decimal.

y = - 3

$y = - 3$

y = - 3

$y = - 3$

Step 3

Find the point at

x = 4

$x = 4$ .

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Step 3.1

Replace the variable

x

$x$ with

4

$4$ in the expression.

f (4) = 2 \log_{3} ((4) - 1) - 3

$f (4) = 2 \log_{3} ((4) - 1) - 3$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Subtract

1

$1$ from

4

$4$ .

f (4) = 2 \log_{3} (3) - 3

$f (4) = 2 \log_{3} (3) - 3$

Step 3.2.1.2

Logarithm base

3

$3$ of

3

$3$ is

1

$1$ .

f (4) = 2 \cdot 1 - 3

$f (4) = 2 \cdot 1 - 3$

Step 3.2.1.3

Multiply

2

$2$ by

1

$1$ .

f (4) = 2 - 3

$f (4) = 2 - 3$

f (4) = 2 - 3

$f (4) = 2 - 3$

Step 3.2.2

Subtract

3

$3$ from

2

$2$ .

f (4) = - 1

$f (4) = - 1$

Step 3.2.3

The final answer is

- 1

$- 1$ .

- 1

$- 1$

- 1

$- 1$

Step 3.3

Convert

- 1

$- 1$ to decimal.

y = - 1

$y = - 1$

y = - 1

$y = - 1$

Step 4

Find the point at

x = 10

$x = 10$ .

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Step 4.1

Replace the variable

x

$x$ with

10

$10$ in the expression.

f (10) = 2 \log_{3} ((10) - 1) - 3

$f (10) = 2 \log_{3} ((10) - 1) - 3$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Subtract

1

$1$ from

10

$10$ .

f (10) = 2 \log_{3} (9) - 3

$f (10) = 2 \log_{3} (9) - 3$

Step 4.2.1.2

Logarithm base

3

$3$ of

9

$9$ is

2

$2$ .

f (10) = 2 \cdot 2 - 3

$f (10) = 2 \cdot 2 - 3$

Step 4.2.1.3

Multiply

2

$2$ by

2

$2$ .

f (10) = 4 - 3

$f (10) = 4 - 3$

f (10) = 4 - 3

$f (10) = 4 - 3$

Step 4.2.2

Subtract

3

$3$ from

4

$4$ .

f (10) = 1

$f (10) = 1$

Step 4.2.3

The final answer is

1

$1$ .

1

$1$

1

$1$

Step 4.3

Convert

1

$1$ to decimal.

y = 1

$y = 1$

y = 1

$y = 1$

Step 5

The log function can be graphed using the vertical asymptote at

x = 1

$x = 1$ and the points

(2, - 3), (4, - 1), (10, 1)

$(2, - 3), (4, - 1), (10, 1)$ .

Vertical Asymptote:

x = 1

$x = 1$

\begin{array}{c} x & y \\ 2 & - 3 \\ 4 & - 1 \\ 10 & 1 \end{array}

$\begin{array}{c} x & y \\ 2 & - 3 \\ 4 & - 1 \\ 10 & 1 \end{array}$

Step 6