Graph y = log base 3 of x+2

$y = \log_{3} (x + 2)$

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.

$f (- 1) = \log_{3} ((- 1) + 2)$

Simplify the result.

Add $- 1$ and $2$ .

$f (- 1) = \log_{3} (1)$

Logarithm base $3$ of $1$ is $0$ .

$f (- 1) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Find the point at $x = 1$ .

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

$f (1) = \log_{3} ((1) + 2)$

Simplify the result.

Add $1$ and $2$ .

$f (1) = \log_{3} (3)$

Logarithm base $3$ of $3$ is $1$ .

$f (1) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Find the point at $x = 7$ .

Replace the variable $x$ with $7$ in the expression.

$f (7) = \log_{3} ((7) + 2)$

Simplify the result.

Add $7$ and $2$ .

$f (7) = \log_{3} (9)$

Logarithm base $3$ of $9$ is $2$ .

Rewrite as an equation.

$\log_{3} (9) = x$

Rewrite $\log_{3} (9) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$3^{x} = 9$

Create equivalent expressions in the equation that all have equal bases.

$3^{x} = 3^{2}$

Since the bases are the same, the two expressions are only equal if the exponents are also equal.

$x = 2$

The variable $x$ is equal to $2$ .

$f (7) = 2$

The final answer is $2$ .

$2$

Convert $2$ to decimal.

$y = 2$

The log function can be graphed using the vertical asymptote at $x = - 2$ and the points $(- 1, 0), (1, 1), (7, 2)$ .

Vertical Asymptote: $x = - 2$

$\begin{array}{c} x & y \\ - 1 & 0 \\ 1 & 1 \\ 7 & 2 \end{array}$

Graph y = log base 3 of x+2

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