Graph y = log base 6 of x+2

$y = \log_{6} (x + 2)$

Find the asymptotes.

Set the argument of the logarithm equal to zero.

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The vertical asymptote occurs at $x = - 2$ .

Vertical Asymptote: $x = - 2$

Find the point at $x = - 1$ .

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

$f (- 1) = \log_{6} ((- 1) + 2)$

Simplify the result.

Add $- 1$ and $2$ .

$f (- 1) = \log_{6} (1)$

Logarithm base $6$ of $1$ is $0$ .

$f (- 1) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Find the point at $x = 4$ .

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

$f (4) = \log_{6} ((4) + 2)$

Simplify the result.

Add $4$ and $2$ .

$f (4) = \log_{6} (6)$

Logarithm base $6$ of $6$ is $1$ .

$f (4) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Find the point at $x = 0$ .

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

$f (0) = \log_{6} ((0) + 2)$

Simplify the result.

Add $0$ and $2$ .

$f (0) = \log_{6} (2)$

The final answer is $\log_{6} (2)$ .

$\log_{6} (2)$

Convert $\log_{6} (2)$ to decimal.

$y = 0.3868528$

The log function can be graphed using the vertical asymptote at $x = - 2$ and the points $(- 1, 0), (4, 1), (0, 0.3868528)$ .

Vertical Asymptote: $x = - 2$

$\begin{array}{c} x & y \\ - 1 & 0 \\ 0 & 0.387 \\ 4 & 1 \end{array}$