Graph f(x) = natural log of x+3

$f (x) = \ln (x + 3)$

Step 1

Find the asymptotes.

Set the argument of the logarithm equal to zero.

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The vertical asymptote occurs at $x = - 3$ .

Vertical Asymptote: $x = - 3$

Step 2

Find the point at $x = - 2$ .

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

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

Simplify the result.

Add $- 2$ and $3$ .

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

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

$f (- 2) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = - 1$ .

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

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

Simplify the result.

Add $- 1$ and $3$ .

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

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

$\ln (2)$

Convert $\ln (2)$ to decimal.

$y = 0.69314718$

Step 4

Find the point at $x = 0$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = \ln ((0) + 3)$

Simplify the result.

Add $0$ and $3$ .

$f (0) = \ln (3)$

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

$\ln (3)$

Convert $\ln (3)$ to decimal.

$y = 1.09861228$

Step 5

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

Vertical Asymptote: $x = - 3$

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

Step 6

Graph f(x) = natural log of x+3

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