Question

The vertical asymptote for \(f(x)=\frac{3x+5}{7-x}\) is given by:

• A. \(x = 7\)
• B. \(x = 3\)
• C. \(x = 5\)
• D. \(x = -1\)

Hint:

To find vertical asymptotes of any rational function:
Simply set the denominator to zero and solve for \(x\), making sure that value doesn't also make the numerator zero (which would indicate a hole instead of an asymptote).
Here, \(7 - x = 0 \implies x = 7\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the vertical asymptote of the rational function \(f(x) = \frac{3x+5}{7-x}\).
A vertical asymptote is a vertical line \(x = c\) where the function values grow without bound (approach \(\pm \infty\)) as \(x\) approaches \(c\).
Key Formula or Approach:
For a rational function \(f(x) = \frac{P(x)}{Q(x)}\) in simplest form, a vertical asymptote occurs at the real roots of the denominator, i.e., where:
\[ Q(x) = 0 \quad \text{and} \quad P(x) \neq 0 \]

Step 2: Detailed Explanation:

• Set the denominator of the function equal to zero:
\[ 7 - x = 0 \]

• Solve for \(x\):
\[ x = 7 \]

• Check if the numerator is non-zero at \(x = 7\):
The numerator is \(P(x) = 3x + 5\).
Substitute \(x = 7\):
\[ P(7) = 3(7) + 5 = 21 + 5 = 26 \neq 0 \]

• Since the denominator is zero and the numerator is non-zero at \(x = 7\), we evaluate the limits to confirm the vertical asymptote:
As \(x \to 7^-\), \(7-x > 0\), so \(f(x) \to \infty\).
As \(x \to 7^+\), \(7-x < 0\), so \(f(x) \to -\infty\).
Therefore, the line \(x = 7\) is indeed a vertical asymptote.

Step 3: Final Answer:
The vertical asymptote of the function is \(x = 7\).