Precalculus Examples

$- \ln (x - 1) + 3$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $- \ln (x - 1) + 3$ is undefined.

$x \leq 1$

Step 1.2

Since $- \ln (x - 1) + 3$ $\to$ $\infty$ as $x$ $\to$ $1$ from the left and $- \ln (x - 1) + 3$ $\to$ $\infty$ 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.

$f (2) = - \ln ((2) - 1) + 3$

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 $1$ from $2$ .

$f (2) = - \ln (1) + 3$

Step 2.2.1.2

The natural logarithm of $1$ is $0$ .

$f (2) = - 0 + 3$

Step 2.2.1.3

Multiply $- 1$ by $0$ .

$f (2) = 0 + 3$

Step 2.2.2

Add $0$ and $3$ .

$f (2) = 3$

Step 2.2.3

The final answer is $3$ .

$3$

Step 2.3

Convert $3$ to decimal.

$y = 3$

Step 3

Find the point at $x = 3$ .

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Step 3.1

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

$f (3) = - \ln ((3) - 1) + 3$

Step 3.2

Simplify the result.

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Step 3.2.1

Subtract $1$ from $3$ .

$f (3) = - \ln (2) + 3$

Step 3.2.2

The final answer is $- \ln (2) + 3$ .

$- \ln (2) + 3$

Step 3.3

Convert $- \ln (2) + 3$ to decimal.

$y = 2.30685281$

Step 4

Find the point at $x = 4$ .

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Step 4.1

Replace the variable $x$ with $4$ in the expression.

$f (4) = - \ln ((4) - 1) + 3$

Step 4.2

Simplify the result.

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Step 4.2.1

Subtract $1$ from $4$ .

$f (4) = - \ln (3) + 3$

Step 4.2.2

The final answer is $- \ln (3) + 3$ .

$- \ln (3) + 3$

Step 4.3

Convert $- \ln (3) + 3$ to decimal.

$y = 1.90138771$

Step 5

The log function can be graphed using the vertical asymptote at $x = 1$ and the points $(2, 3), (3, 2.30685281), (4, 1.90138771)$ .

Vertical Asymptote: $x = 1$

$\begin{array}{c} x & y \\ 2 & 3 \\ 3 & 2.307 \\ 4 & 1.901 \end{array}$

Step 6