Precalculus Examples

$y = \ln (4 + x)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$4 + x = 0$

Step 1.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 1.3

The vertical asymptote occurs at $x = - 4$ .

Vertical Asymptote: $x = - 4$

Step 2

Find the point at $x = - 3$ .

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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

Remove parentheses.

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

Step 2.2.2

Subtract $3$ from $4$ .

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

Step 2.2.3

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

$f (- 3) = 0$

Step 2.2.4

The final answer is $0$ .

$0$

Step 2.3

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = - 2$ .

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Step 3.1

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

$f (- 2) = \ln (4 - 2)$

Step 3.2

Simplify the result.

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Step 3.2.1

Remove parentheses.

$f (- 2) = \ln (4 - 2)$

Step 3.2.2

Subtract $2$ from $4$ .

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

Step 3.2.3

The final answer is $\ln (2)$ .

$\ln (2)$

Step 3.3

Convert $\ln (2)$ to decimal.

$y = 0.69314718$

Step 4

Find the point at $x = - 1$ .

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Step 4.1

Replace the variable $x$ with $- 1$ in the expression.

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

Step 4.2

Simplify the result.

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Step 4.2.1

Remove parentheses.

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

Step 4.2.2

Subtract $1$ from $4$ .

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

Step 4.2.3

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

$\ln (3)$

Step 4.3

Convert $\ln (3)$ to decimal.

$y = 1.09861228$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 4$ and the points $(- 3, 0), (- 2, 0.69314718), (- 1, 1.09861228)$ .

Vertical Asymptote: $x = - 4$

$\begin{array}{c} x & y \\ - 3 & 0 \\ - 2 & 0.693 \\ - 1 & 1.099 \end{array}$

Step 6