Graph y = log of x+3

$y = \log (x + 3)$

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.

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

Simplify the result.

Add $- 2$ and $3$ .

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

Logarithm base $10$ 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 $x = 7$ .

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

$f (7) = \log ((7) + 3)$

Simplify the result.

Add $7$ and $3$ .

$f (7) = \log (10)$

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

$f (7) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Step 4

Find the point at $x = - 1$ .

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

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

Simplify the result.

Add $- 1$ and $3$ .

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

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

$\log (2)$

Convert $\log (2)$ to decimal.

$y = 0.30102999$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 3$ and the points $(- 2, 0), (7, 1), (- 1, 0.30102999)$ .

Vertical Asymptote: $x = - 3$

$\begin{array}{c} x & y \\ - 2 & 0 \\ - 1 & 0.301 \\ 7 & 1 \end{array}$

Step 6

Graph y = log of x+3

Related Questions

Support LCMGCF.com by disabling your adblocker.