Precalculus: Rational Functions and Asymptotes Worksheet

Question 1

For f(x) = (x + 2)/(x - 3), find the vertical asymptote, horizontal asymptote, x-intercept, and y-intercept.

Accepted Answer

The vertical asymptote is x = 3 because the denominator is zero there. The horizontal asymptote is y = 1 because the numerator and denominator have the same degree and the leading coefficient ratio is 1/1. The x-intercept is (-2, 0), and the y-intercept is (0, -2/3).

Question 2

Simplify g(x) = (x^2 - 9)/(x - 3). Identify any holes or vertical asymptotes.

Accepted Answer

The function simplifies to g(x) = x + 3 for x not equal to 3. There is a hole at x = 3 because the factor x - 3 cancels. The hole is at (3, 6). There is no vertical asymptote.

Question 3

For h(x) = (2x^2 - 5x + 1)/(x^2 - 4), identify the vertical asymptotes and the horizontal asymptote.

Accepted Answer

The denominator factors as (x - 2)(x + 2), so the vertical asymptotes are x = 2 and x = -2. Since the numerator and denominator have the same degree, the horizontal asymptote is y = 2, the ratio of the leading coefficients.

Question 4

Use polynomial division to find the slant asymptote of k(x) = (x^2 + 1)/(x - 1). Also identify the vertical asymptote.

Accepted Answer

Dividing x^2 + 1 by x - 1 gives x + 1 with a remainder of 2, so k(x) = x + 1 + 2/(x - 1). The slant asymptote is y = x + 1, and the vertical asymptote is x = 1.

Question 5

Find the domain of p(x) = (x - 4)/((x + 1)(x - 5)). Write your answer in set notation or interval notation.

Accepted Answer

The denominator is zero at x = -1 and x = 5, so those values are excluded from the domain. The domain is all real numbers except -1 and 5, or (-infinity, -1) union (-1, 5) union (5, infinity).

Question 6

Find the end behavior asymptote of q(x) = (3x^3 - x)/(x^2 + 4).

Accepted Answer

Polynomial division gives q(x) = 3x + (-13x)/(x^2 + 4). Since the remainder term approaches 0 as x becomes very large or very negative, the end behavior asymptote is y = 3x.

Question 7

For r(x) = ((x - 2)(x + 5))/((x - 2)(x - 7)), identify the hole and the vertical asymptote.

Accepted Answer

The factor x - 2 cancels, so there is a hole at x = 2. The simplified function is (x + 5)/(x - 7), so the hole has y-value 7/(-5), which is -7/5. The hole is (2, -7/5), and the vertical asymptote is x = 7.

Question 8

Write one possible rational function that has vertical asymptotes x = -1 and x = 3, a horizontal asymptote y = 0, and an x-intercept at x = 2.

Accepted Answer

One possible function is f(x) = (x - 2)/((x + 1)(x - 3)). The denominator gives vertical asymptotes at x = -1 and x = 3, the numerator gives an x-intercept at x = 2, and the denominator degree is greater than the numerator degree, so the horizontal asymptote is y = 0.

Question 9

Solve the inequality (x + 1)/(x - 2) > 0. Give your answer using interval notation.

Accepted Answer

The critical values are x = -1, where the numerator is zero, and x = 2, where the expression is undefined. A sign chart shows the expression is positive on (-infinity, -1) and (2, infinity). The solution is (-infinity, -1) union (2, infinity).

Question 10

For m(x) = (5x^2 - 3x + 7)/(2x^2 + x - 4), find the horizontal asymptote and explain how you know.

Accepted Answer

The horizontal asymptote is y = 5/2 because the numerator and denominator have the same degree. For equal degrees, the horizontal asymptote is the ratio of the leading coefficients.

Question 11

Find the x-intercepts and y-intercept of s(x) = (x^2 - 4)/(x^2 + x - 6). Be sure to account for any canceled factors.

Accepted Answer

Factoring gives s(x) = ((x - 2)(x + 2))/((x + 3)(x - 2)). The factor x - 2 cancels, so x = 2 is a hole, not an x-intercept. The simplified function has x-intercept (-2, 0). The y-intercept is s(0) = (-4)/(-6) = 2/3, so the y-intercept is (0, 2/3).

Question 12

Evaluate the one-sided behavior of t(x) = (x + 1)/(x - 2)^2 as x approaches 2 from the left and from the right.

Accepted Answer

As x approaches 2 from either side, the numerator approaches 3 and the denominator is a very small positive number because it is squared. Therefore, t(x) approaches positive infinity from both the left and the right.

Question 13

Find the value of a so that f(x) = (x^2 + ax - 12)/(x - 3) has a removable discontinuity at x = 3 instead of a vertical asymptote. Then state the location of the hole.

Accepted Answer

For x = 3 to be a removable discontinuity, the numerator must also be zero at x = 3. Substituting gives 9 + 3a - 12 = 0, so 3a - 3 = 0 and a = 1. Then the numerator is x^2 + x - 12 = (x + 4)(x - 3), so the simplified function is x + 4. The hole is at (3, 7).

Question 14

A company models the average cost per item by C(x) = (500 + 20x)/x, where x is the number of items produced and x > 0. Find the horizontal asymptote and explain its meaning in context.

Accepted Answer

The function can be rewritten as C(x) = 500/x + 20. As x becomes very large, 500/x approaches 0, so the horizontal asymptote is y = 20. In context, the average cost per item approaches 20 dollars as production increases.

Question 15

Construct a rational function with a vertical asymptote at x = -2, a hole at x = 5, and a horizontal asymptote at y = 3. State your function and briefly justify it.

Accepted Answer

One possible function is f(x) = 3(x - 1)(x - 5)/((x + 2)(x - 5)). The factor x - 5 cancels, so there is a hole at x = 5. The remaining denominator factor x + 2 gives a vertical asymptote at x = -2. After cancellation, the numerator and denominator have the same degree with leading coefficient ratio 3, so the horizontal asymptote is y = 3.

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Precalculus: Rational Functions and Asymptotes

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