Trigonometry Examples

$\log_{4} (x - 1) + 2$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $\log_{4} (x - 1) + 2$ is undefined.

$x \leq 1$

Step 1.2

Since $\log_{4} (x - 1) + 2$ $\to$ $\infty$ as $x$ $\to$ $1$ from the left and $\log_{4} (x - 1) + 2$ $\to$ $- \infty$ as $x$ $\to$ $1$ from the right, then $x = 1$ is a vertical asymptote.

$x = 1$

Step 1.3

Ignoring the logarithm, consider the rational function $R (x) = \frac{a x^{n}}{b x^{m}}$ where $n$ is the degree of the numerator and $m$ is the degree of the denominator.

1. If $n < m$ , then the x-axis, $y = 0$ , is the horizontal asymptote.

2. If $n = m$ , then the horizontal asymptote is the line $y = \frac{a}{b}$ .

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

Step 1.4

There are no horizontal asymptotes because $Q (x)$ is $1$ .

No Horizontal Asymptotes

Step 1.5

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.6

This is the set of all asymptotes.

Vertical Asymptotes: $x = 1$

No Horizontal Asymptotes

Vertical Asymptotes: $x = 1$

No Horizontal Asymptotes

Step 2

Find the point at $x = 2$ .

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Step 2.1

Replace the variable $x$ with $2$ in the expression.

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

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 $1$ from $2$ .

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

Step 2.2.1.2

Logarithm base $4$ of $1$ is $0$ .

$f (2) = 0 + 2$

Step 2.2.2

Add $0$ and $2$ .

$f (2) = 2$

Step 2.2.3

The final answer is $2$ .

$2$

Step 2.3

Convert $2$ to decimal.

$y = 2$

Step 3

Find the point at $x = 5$ .

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Step 3.1

Replace the variable $x$ with $5$ in the expression.

$f (5) = \log_{4} ((5) - 1) + 2$

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 $1$ from $5$ .

$f (5) = \log_{4} (4) + 2$

Step 3.2.1.2

Logarithm base $4$ of $4$ is $1$ .

$f (5) = 1 + 2$

Step 3.2.2

Add $1$ and $2$ .

$f (5) = 3$

Step 3.2.3

The final answer is $3$ .

$3$

Step 3.3

Convert $3$ to decimal.

$y = 3$

Step 4

Find the point at $x = 3$ .

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Step 4.1

Replace the variable $x$ with $3$ in the expression.

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

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 $1$ from $3$ .

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

Step 4.2.1.2

Logarithm base $4$ of $2$ is $\frac{1}{2}$ .

$f (3) = \frac{1}{2} + 2$

Step 4.2.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (3) = \frac{1}{2} + 2 \cdot \frac{2}{2}$

Step 4.2.3

Combine $2$ and $\frac{2}{2}$ .

$f (3) = \frac{1}{2} + \frac{2 \cdot 2}{2}$

Step 4.2.4

Combine the numerators over the common denominator.

$f (3) = \frac{1 + 2 \cdot 2}{2}$

Step 4.2.5

Simplify the numerator.

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Step 4.2.5.1

Multiply $2$ by $2$ .

$f (3) = \frac{1 + 4}{2}$

Step 4.2.5.2

Add $1$ and $4$ .

$f (3) = \frac{5}{2}$

Step 4.2.6

The final answer is $\frac{5}{2}$ .

$\frac{5}{2}$

Step 4.3

Convert $\frac{5}{2}$ to decimal.

$y = 2.5$

Step 5

The log function can be graphed using the vertical asymptote at $x = 1$ and the points $(2, 2), (5, 3), (3, 2.5)$ .

Vertical Asymptote: $x = 1$

$\begin{array}{c} x & y \\ 2 & 2 \\ 3 & 2.5 \\ 5 & 3 \end{array}$

Step 6