Pre-Algebra Examples

$\frac{1}{\log_{4} (p^{2})}$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $\frac{1}{\log_{4} (x^{2})}$ is undefined.

$x = - 1, x = 0, x = 1$

Step 1.2

Since $\frac{1}{\log_{4} (x^{2})}$ $\to$ $\infty$ as $x$ $\to$ $- 1$ from the left and $\frac{1}{\log_{4} (x^{2})}$ $\to$ $- \infty$ as $x$ $\to$ $- 1$ from the right, then $x = - 1$ is a vertical asymptote.

$x = - 1$

Step 1.3

Since $\frac{1}{\log_{4} (x^{2})}$ $\to$ $- \infty$ as $x$ $\to$ $1$ from the left and $\frac{1}{\log_{4} (x^{2})}$ $\to$ $\infty$ as $x$ $\to$ $1$ from the right, then $x = 1$ is a vertical asymptote.

$x = 1$

Step 1.4

List all of the vertical asymptotes:

$x = - 1, 1$

Step 1.5

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{\log_{4} (x^{2})}$ approaches $0$ .

$0$

Step 1.6

List the horizontal asymptotes:

$y = 0$

Step 1.7

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.8

This is the set of all asymptotes.

Vertical Asymptotes: $x = - 1, 1$

Horizontal Asymptotes: $y = 0$

Vertical Asymptotes: $x = - 1, 1$

Horizontal Asymptotes: $y = 0$

Step 2

Find the point at $x = - 1$ .

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Step 2.1

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

$f (- 1) = \frac{1}{\log_{4} (p^{2})}$

Step 2.2

The final answer is $\frac{1}{\log_{4} (p^{2})}$ .

$\frac{1}{\log_{4} (p^{2})}$

Step 2.3

Convert $\frac{1}{\log_{4} (p^{2})}$ to decimal.

$= \frac{1}{\log_{4} (p^{2})}$

Step 3

The log function can be graphed using the vertical asymptote at $x = - 1, 1$ and the points $(- 1, \frac{1}{\log_{4} (p^{2})}), (- 2, \frac{1}{\log_{4} (p^{2})}), (- 3, \frac{1}{\log_{4} (p^{2})})$ .

Vertical Asymptote: $x = - 1, 1$

$\begin{array}{c} x & y \\ - 3 & \frac{1}{\log_{4} (p^{2})} \\ - 2 & \frac{1}{\log_{4} (p^{2})} \\ - 1 & \frac{1}{\log_{4} (p^{2})} \end{array}$

Step 4