Graph y=2 log base 2 of -x+2-2

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

Step 1

Find the asymptotes.

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

$x = 2$

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: $x = 2$

No Horizontal Asymptotes

Vertical Asymptotes: $x = 2$

No Horizontal Asymptotes

Step 2

Find the point at $x = 0$ .

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

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

Simplify the result.

Simplify each term.

Multiply $- 1$ by $0$ .

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

Add $0$ and $2$ .

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

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

$f (0) = 2 \cdot 1 - 2$

Multiply $2$ by $1$ .

$f (0) = 2 - 2$

Subtract $2$ from $2$ .

$f (0) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = 1$ .

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

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

Simplify the result.

Simplify each term.

Multiply $- 1$ by $1$ .

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

Add $- 1$ and $2$ .

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

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

$f (1) = 2 \cdot 0 - 2$

Multiply $2$ by $0$ .

$f (1) = 0 - 2$

Subtract $2$ from $0$ .

$f (1) = - 2$

The final answer is $- 2$ .

$- 2$

Convert $- 2$ to decimal.

$y = - 2$

Step 4

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

Vertical Asymptote: $x = 2$

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

Step 5

Graph y=2 log base 2 of -x+2-2

Related Questions

Support LCMGCF.com by disabling your adblocker.