Graph f(x)=-2 log base 2 of -x-6-5

$f (x) = - 2 \log_{2} (- x - 6) - 5$

Step 1

Find the asymptotes.

Find where the expression $- \log_{2} ({(- x - 6)}^{2}) - 5$ is undefined.

$x = - 6$

Ignoring the logarithm, consider the rational function $R (x) = \frac{a x^{n}}{b x^{m}}$ where $n$ is the degree of the numerator and $m$ is the degree of the denominator.

1. If $n < m$ , then the x-axis, $y = 0$ , is the horizontal asymptote.

2. If $n = m$ , then the horizontal asymptote is the line $y = \frac{a}{b}$ .

3. If $n > m$ , then there is no horizontal asymptote (there is an oblique asymptote).

There are no horizontal asymptotes because $Q (x)$ is $1$ .

No Horizontal Asymptotes

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

This is the set of all asymptotes.

Vertical Asymptotes: $x = - 6$

No Horizontal Asymptotes

Vertical Asymptotes: $x = - 6$

No Horizontal Asymptotes

Step 2

Find the point at $x = - 8$ .

Replace the variable $x$ with $- 8$ in the expression.

$f (- 8) = - 2 \log_{2} (- (- 8) - 6) - 5$

Simplify the result.

Simplify each term.

Multiply $- 1$ by $- 8$ .

$f (- 8) = - 2 \log_{2} (8 - 6) - 5$

Subtract $6$ from $8$ .

$f (- 8) = - 2 \log_{2} (2) - 5$

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

$f (- 8) = - 2 \cdot 1 - 5$

Multiply $- 2$ by $1$ .

$f (- 8) = - 2 - 5$

Subtract $5$ from $- 2$ .

$f (- 8) = - 7$

The final answer is $- 7$ .

$- 7$

Convert $- 7$ to decimal.

$y = - 7$

Step 3

Find the point at $x = - 7$ .

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

$f (- 7) = - 2 \log_{2} (- (- 7) - 6) - 5$

Simplify the result.

Simplify each term.

Multiply $- 1$ by $- 7$ .

$f (- 7) = - 2 \log_{2} (7 - 6) - 5$

Subtract $6$ from $7$ .

$f (- 7) = - 2 \log_{2} (1) - 5$

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

$f (- 7) = - 2 \cdot 0 - 5$

Multiply $- 2$ by $0$ .

$f (- 7) = 0 - 5$

Subtract $5$ from $0$ .

$f (- 7) = - 5$

The final answer is $- 5$ .

$- 5$

Convert $- 5$ to decimal.

$y = - 5$

Step 4

The log function can be graphed using the vertical asymptote at $x = - 6$ and the points $(- 8, - 7), (- 7, - 5)$ .

Vertical Asymptote: $x = - 6$

$\begin{array}{c} x & y \\ - 8 & - 7 \\ - 7 & - 5 \end{array}$

Step 5

Graph f(x)=-2 log base 2 of -x-6-5

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