Let r(x)=1x+8
. This function has a vertical asymptote at x=−8
. What is the equation of the vertical asymptote of r(−8x)
?


The given function is:

\[ r(x) = \frac{1}{x+8} \]

The vertical asymptote occurs where the denominator is zero. For \( r(x) \), this is at:

\[ x + 8 = 0 \]
\[ x = -8 \]

Now we need to find the vertical asymptote of \( r(-8x) \). The function \( r(-8x) \) is:

\[ r(-8x) = \frac{1}{-8x + 8} \]

We find the vertical asymptote by setting the denominator equal to zero:

\[ -8x + 8 = 0 \]

Solving for \( x \):

\[ -8x = -8 \]
\[ x = 1 \]

So, the equation of the vertical asymptote of \( r(-8x) \) is:

\[ x = 1 \]

Would you like more details or have any questions? Here are some related problems to consider:

1. What is the horizontal asymptote of \( r(x) \)?
2. How does the transformation \( r(-8x) \) affect the original function \( r(x) \)?
3. What happens to the asymptote if the function is \( r(-x+8) \)?
4. How would the vertical asymptote change if the function was \( r(2x) \)?
5. What are the domain and range of \( r(-8x) \)?

**Tip:** The vertical asymptote of a rational function \( \frac{1}{f(x)} \) occurs where \( f(x) = 0 \).

Let r(x)=1/(x+8). This function has a vertical asymptote at x=−8. What is the equation of the vertical asymptote of r(−8x)?

This problem explores the vertical asymptote of the transformed rational function r(-8x) = 1/(-8x + 8). Learn how to solve for the vertical asymptote by setting the denominator equal to zero and analyzing function transformations. Suitable for high school students in Grades 9-12.

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