Algebra 2 · Question of the Day

Algebra 2 Question of the Day

A fresh daily question to build accuracy, reinforce recall, and turn practice into a steady habit.
Friday, September 18, 2026

Which rational function has a horizontal asymptote at y=2y = -2 and vertical asymptotes at x=1x = 1 and x=4x = -4?

Keep practicing Algebra 2

Question of the Day

Answer today's Algebra 2 question, reveal the full explanation, then keep the streak going with a new question every day.

Which rational function has a horizontal asymptote at y=2y = -2 and vertical asymptotes at x=1x = 1 and x=4x = -4?

  1. f(x)=2x2+7x6(x1)(x+4)f(x) = \frac{-2x^2 + 7x - 6}{(x-1)(x+4)} because this form clearly shows the desired asymptotic behavior
  2. f(x)=2x2+x+8x2+3x4f(x) = \frac{-2x^2 + x + 8}{x^2 + 3x - 4} because the denominator factors to (x1)(x+4)(x-1)(x+4) giving the vertical asymptotes
  3. f(x)=2x2+5x3x2+3x4f(x) = \frac{-2x^2 + 5x - 3}{x^2 + 3x - 4} because the leading coefficient ratio gives the horizontal asymptote (correct answer)
  4. f(x)=2x3+5x23xx2+3x4f(x) = \frac{-2x^3 + 5x^2 - 3x}{x^2 + 3x - 4} because higher degree numerator creates the horizontal asymptote

Explanation: When analyzing rational functions for asymptotes, you need to examine both the numerator and denominator carefully. Vertical asymptotes occur where the denominator equals zero (and the numerator doesn't), while horizontal asymptotes depend on the degrees and leading coefficients of the numerator and denominator. For vertical asymptotes at x=1x = 1 and x=4x = -4, the denominator must factor as (x1)(x+4)=x2+3x4(x-1)(x+4) = x^2 + 3x - 4. You can verify this by expanding or check that the given expression x2+3x4x^2 + 3x - 4 factors correctly. For the horizontal asymptote at y=2y = -2, you need the numerator and denominator to have the same degree (both quadratic), with the ratio of leading coefficients equal to 2-2. Since the denominator's leading coefficient is 1, the numerator needs a leading coefficient of 2-2. Choice C gives f(x)=2x2+5x3x2+3x4f(x) = \frac{-2x^2 + 5x - 3}{x^2 + 3x - 4}. The denominator factors to (x1)(x+4)(x-1)(x+4) providing the correct vertical asymptotes, and the ratio of leading coefficients is 21=2\frac{-2}{1} = -2, giving the horizontal asymptote y=2y = -2. Choice A has the wrong numerator for the required horizontal asymptote. Choice B's numerator has leading coefficient 2-2, but you'd need to verify the specific form doesn't create unwanted cancellations. Choice D has a cubic numerator over a quadratic denominator, which creates no horizontal asymptote (the function grows without bound). Study tip: Always check both conditions separately—factor the denominator for vertical asymptotes, then compare degrees and leading coefficients for horizontal asymptotes.