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what is the horizontal asymptote of \\(f(x) = \\frac{-2x}{x+1}\\)? \\(y…

Question

what is the horizontal asymptote of \\(f(x) = \frac{-2x}{x+1}\\)?

\\(y = -2\\)
\\(y = -1\\)
\\(y = 0\\)
\\(y = 1\\)

Explanation:

Identify the degrees of the numerator and denominator

The given rational function is:

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

The numerator is \(p(x) = -2x\), which has degree \(n = 1\).
The denominator is \(q(x) = x + 1\), which has degree \(m = 1\).

Compare degrees and find the ratio of leading coefficients

Since the degrees of the numerator and denominator are equal (\(n = m = 1\)), the horizontal asymptote is determined by the ratio of their leading coefficients:

$$y = \frac{\text{leading coefficient of } p(x)}{\text{leading coefficient of } q(x)}$$

The leading coefficient of the numerator is \(-2\), and the leading coefficient of the denominator is \(1\).

Calculate the horizontal asymptote

$$y = \frac{-2}{1} = -2$$

Thus, the horizontal asymptote is \(y = -2\).

Answer:

  • (A) \(y = -2\) (Correct answer)
  • (B) \(y = -1\)
  • (C) \(y = 0\)
  • (D) \(y = 1\)