Algebra Examples

$y = \log_{2} (1 - x)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$1 - x = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 1.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 1.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 1.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 1.3

The vertical asymptote occurs at $x = 1$ .

Vertical Asymptote: $x = 1$

Step 2

Find the point at $x = 0$ .

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Step 2.1

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

$f (0) = \log_{2} (1 - (0))$

Step 2.2

Simplify the result.

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Step 2.2.1

Subtract $0$ from $1$ .

$f (0) = \log_{2} (1)$

Step 2.2.2

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

$f (0) = 0$

Step 2.2.3

The final answer is $0$ .

$0$

Step 2.3

Convert $0$ to decimal.

$= 0$

Step 3

Find the point at $x = - 1$ .

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2.2

Add $1$ and $1$ .

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

Step 3.2.3

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

$f (- 1) = 1$

Step 3.2.4

The final answer is $1$ .

$1$

Step 3.3

Convert $1$ to decimal.

$= 1$

Step 4

Find the point at $x = - 2$ .

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Step 4.1

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

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

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $- 1$ by $- 2$ .

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

Step 4.2.2

Add $1$ and $2$ .

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

Step 4.2.3

The final answer is $\log_{2} (3)$ .

$\log_{2} (3)$

Step 4.3

Convert $\log_{2} (3)$ to decimal.

$= 1.5849625$

Step 5

The log function can be graphed using the vertical asymptote at $x = 1$ and the points $(0, 0), (- 1, 1), (- 2, 1.5849625)$ .

Vertical Asymptote: $x = 1$

$\begin{array}{c} x & y \\ - 2 & 1.585 \\ - 1 & 1 \\ 0 & 0 \end{array}$

Step 6