Precalculus Examples

$f (x) = \log (x + 5)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$x + 5 = 0$

Step 1.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 1.3

The vertical asymptote occurs at $x = - 5$ .

Vertical Asymptote: $x = - 5$

Step 2

Find the point at $x = - 4$ .

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Step 2.1

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

$f (- 4) = \log ((- 4) + 5)$

Step 2.2

Simplify the result.

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Step 2.2.1

Add $- 4$ and $5$ .

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

Step 2.2.2

Logarithm base $10$ of $1$ is $0$ .

$f (- 4) = 0$

Step 2.2.3

The final answer is $0$ .

$0$

Step 2.3

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = 5$ .

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Step 3.1

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

$f (5) = \log ((5) + 5)$

Step 3.2

Simplify the result.

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Step 3.2.1

Add $5$ and $5$ .

$f (5) = \log (10)$

Step 3.2.2

Logarithm base $10$ of $10$ is $1$ .

$f (5) = 1$

Step 3.2.3

The final answer is $1$ .

$1$

Step 3.3

Convert $1$ to decimal.

$y = 1$

Step 4

Find the point at $x = - 3$ .

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Step 4.1

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

$f (- 3) = \log ((- 3) + 5)$

Step 4.2

Simplify the result.

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Step 4.2.1

Add $- 3$ and $5$ .

$f (- 3) = \log (2)$

Step 4.2.2

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

$\log (2)$

Step 4.3

Convert $\log (2)$ to decimal.

$y = 0.30102999$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 5$ and the points $(- 4, 0), (5, 1), (- 3, 0.30102999)$ .

Vertical Asymptote: $x = - 5$

$\begin{array}{c} x & y \\ - 4 & 0 \\ - 3 & 0.301 \\ 5 & 1 \end{array}$

Step 6