Graph y = log base 2 of x+4

$y = \log_{2} (x + 4)$

Step 1

Find the asymptotes.

Set the argument of the logarithm equal to zero.

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

Vertical Asymptote: $x = - 4$

Step 2

Find the point at $x = - 3$ .

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

$f (- 3) = \log_{2} ((- 3) + 4)$

Simplify the result.

Add $- 3$ and $4$ .

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

Logarithm base $2$ of $1$ is $0$ .

$f (- 3) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = - 2$ .

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

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

Simplify the result.

Add $- 2$ and $4$ .

$f (- 2) = \log_{2} (2)$

Logarithm base $2$ of $2$ is $1$ .

$f (- 2) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Step 4

Find the point at $x = 0$ .

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

$f (0) = \log_{2} ((0) + 4)$

Simplify the result.

Add $0$ and $4$ .

$f (0) = \log_{2} (4)$

Logarithm base $2$ of $4$ is $2$ .

Rewrite as an equation.

$\log_{2} (4) = x$

Rewrite $\log_{2} (4) = x$ 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 $b^{y} = x$ .

$2^{x} = 4$

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

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

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 $2$ .

$f (0) = 2$

The final answer is $2$ .

$2$

Convert $2$ to decimal.

$y = 2$

Step 5

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

Vertical Asymptote: $x = - 4$

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

Step 6

Graph y = log base 2 of x+4

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