IB Mathematics: Analysis and Approaches Quiz: Rational Functions
20 questions · exam conditions
0:00
Rational FunctionsQuestion 1 of 20

The function f(x)=ax+32x−bf(x) = \frac{ax + 3}{2x - b} has a vertical asymptote at x=4x = 4 and a horizontal asymptote at y=−2y = -2. Find the value of a+ba+b.

-12
-2
4
12
← Back to quizzes

IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Rational Functions

Practice Rational Functions in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rational Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The function f(x)=ax+32x−bf(x) = \frac{ax + 3}{2x - b} has a vertical asymptote at x=4x = 4 and a horizontal asymptote at y=−2y = -2. Find the value of a+ba+b.

  1. -12
  2. -2
  3. 4 (correct answer)
  4. 12
Explanation: A vertical asymptote occurs where the denominator is zero. So, 2x−b=02x - b = 0 at x=4x=4. Substituting x=4x=4 gives 2(4)−b=02(4) - b = 0, which means 8−b=08 - b = 0, so b=8b=8. A horizontal asymptote for a rational function where the degrees of the numerator and denominator are equal is the ratio of their leading coefficients. Here, the HA is y=a2y = \frac{a}{2}. We are given that the HA is at y=−2y=-2. So, a2=−2\frac{a}{2} = -2, which gives a=−4a = -4. The value of a+ba+b is −4+8=4-4 + 8 = 4.

Question 2

The graph of the function f(x)=2x2−2x−12x+2f(x) = \frac{2x^2 - 2x - 12}{x+2} has a removable discontinuity (a hole). What are the coordinates of this hole?

  1. (−2,−10)(-2, -10) (correct answer)
  2. (−2,0)(-2, 0)
  3. (3,−10)(3, -10)
  4. (3,0)(3, 0)
Explanation: A hole occurs when a factor in the denominator cancels with a factor in the numerator. First, factor the numerator: 2x2−2x−12=2(x2−x−6)=2(x−3)(x+2)2x^2 - 2x - 12 = 2(x^2 - x - 6) = 2(x-3)(x+2). The function is f(x)=2(x−3)(x+2)x+2f(x) = \frac{2(x-3)(x+2)}{x+2}. The factor (x+2)(x+2) cancels, which means there is a hole at the value of xx that makes this factor zero, i.e., x=−2x=-2. To find the y-coordinate of the hole, we evaluate the simplified function, g(x)=2(x−3)g(x) = 2(x-3), at x=−2x=-2. g(−2)=2(−2−3)=2(−5)=−10g(-2) = 2(-2-3) = 2(-5) = -10. Thus, the coordinates of the hole are (−2,−10)(-2, -10).

Question 3

Which statement best describes the graph of the function f(x)=x3−8x−2f(x) = \frac{x^3 - 8}{x-2}?

  1. A cubic function with a vertical asymptote at x=2x=2.
  2. A parabola with a vertical asymptote at x=2x=2.
  3. A parabola with a hole at (2,12)(2, 12). (correct answer)
  4. A line with a hole at (2,8)(2, 8).
Explanation: The numerator is a difference of cubes, which can be factored as a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2+ab+b^2). So, x3−8=x3−23=(x−2)(x2+2x+4)x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4). The function can be written as f(x)=(x−2)(x2+2x+4)x−2f(x) = \frac{(x-2)(x^2+2x+4)}{x-2}. For x≠2x \neq 2, the (x−2)(x-2) terms cancel, and the function simplifies to g(x)=x2+2x+4g(x) = x^2 + 2x + 4. This is the equation of a parabola. The cancellation indicates a hole in the graph at x=2x=2. To find the y-coordinate of the hole, substitute x=2x=2 into the simplified function: g(2)=22+2(2)+4=4+4+4=12g(2) = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12. Therefore, the graph is a parabola with a hole at (2,12)(2, 12).

Question 4

The function f(x)=ax+32x−bf(x) = \frac{ax + 3}{2x - b} has a vertical asymptote at x=4x = 4 and a horizontal asymptote at y=−2y = -2. Find the value of a+ba+b.

  1. -12
  2. -2
  3. 4 (correct answer)
  4. 12
Explanation: A vertical asymptote occurs where the denominator is zero. So, 2x−b=02x - b = 0 at x=4x=4. Substituting x=4x=4 gives 2(4)−b=02(4) - b = 0, which means 8−b=08 - b = 0, so b=8b=8. A horizontal asymptote for a rational function where the degrees of the numerator and denominator are equal is the ratio of their leading coefficients. Here, the HA is y=a2y = \frac{a}{2}. We are given that the HA is at y=−2y=-2. So, a2=−2\frac{a}{2} = -2, which gives a=−4a = -4. The value of a+ba+b is −4+8=4-4 + 8 = 4.

Question 5

Find the equation of the vertical asymptote of the function f(x)=x2−9x2−2x−3f(x) = \frac{x^2 - 9}{x^2 - 2x - 3}.

  1. x=−3x = -3
  2. x=−1x = -1 (correct answer)
  3. x=1x = 1
  4. x=3x = 3
Explanation: To find vertical asymptotes, we look for values of xx that make the denominator zero. First, we factor the numerator and the denominator: f(x)=(x−3)(x+3)(x−3)(x+1)f(x) = \frac{(x-3)(x+3)}{(x-3)(x+1)}. The factor (x−3)(x-3) appears in both the numerator and the denominator, indicating a hole at x=3x=3, not a vertical asymptote. After simplifying the function for x≠3x \neq 3, we get f(x)=x+3x+1f(x) = \frac{x+3}{x+1}. The denominator is zero when x+1=0x+1=0, which gives x=−1x=-1. This does not make the simplified numerator zero, so there is a vertical asymptote at x=−1x=-1.

Question 6

The graph of a rational function y=f(x)y=f(x) has exactly one vertical asymptote at x=−1x=-1 and a horizontal asymptote at y=2y=2. The graph has an x-intercept at x=3x=3 and a y-intercept at y=−6y=-6. Which of the following could be the equation for f(x)f(x)?

  1. f(x)=x−32(x+1)f(x) = \frac{x-3}{2(x+1)}
  2. f(x)=2(x+1)x−3f(x) = \frac{2(x+1)}{x-3}
  3. f(x)=2x−6x+1f(x) = \frac{2x-6}{x+1} (correct answer)
  4. f(x)=2x+6x−1f(x) = \frac{2x+6}{x-1}
Explanation: From the given information, we can construct the function's form. A vertical asymptote at x=−1x=-1 implies a factor of (x+1)(x+1) in the denominator. An x-intercept at x=3x=3 implies a factor of (x−3)(x-3) in the numerator. A horizontal asymptote at y=2y=2 implies the ratio of the leading coefficients of the numerator and denominator is 2. This gives us the form f(x)=2(x−3)x+1f(x) = \frac{2(x-3)}{x+1}. Let's check if this form satisfies the y-intercept condition. The y-intercept occurs at x=0x=0: f(0)=2(0−3)0+1=−61=−6f(0) = \frac{2(0-3)}{0+1} = \frac{-6}{1} = -6. This matches the given information. Expanding the numerator gives f(x)=2x−6x+1f(x) = \frac{2x-6}{x+1}.

Question 7

A rational function is defined by f(x)=x2−k2x2−5x+6f(x) = \frac{x^2 - k^2}{x^2 - 5x + 6}, where kk is a positive real constant. For which value of kk does the graph of f(x)f(x) have exactly one vertical asymptote?

  1. 1
  2. 5\sqrt{5}
  3. 3 (correct answer)
  4. 6
Explanation: First, find the potential vertical asymptotes by factoring the denominator: x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x-2)(x-3). This gives potential vertical asymptotes at x=2x=2 and x=3x=3. For the graph to have only one vertical asymptote, one of these must be a hole. This means a factor in the denominator must cancel with a factor in the numerator. The numerator is x2−k2=(x−k)(x+k)x^2 - k^2 = (x-k)(x+k). For a factor to cancel, either x−kx-k or x+kx+k must be equal to x−2x-2 or x−3x-3. Since kk is positive, we have two possibilities for cancellation: 1. x−k=x−2  ⟹  k=2x-k = x-2 \implies k=2. In this case, the (x−2)(x-2) factor cancels, leaving a single VA at x=3x=3. 2. x−k=x−3  ⟹  k=3x-k = x-3 \implies k=3. In this case, the (x−3)(x-3) factor cancels, leaving a single VA at x=2x=2. Both k=2k=2 and k=3k=3 result in exactly one vertical asymptote. Of the choices given, k=3k=3 is an option.

Question 8

A rational function f(x)f(x) has a numerator and denominator that are both linear polynomials. The graph of f(x)f(x) passes through the origin, has a vertical asymptote at x=−2x=-2, and a horizontal asymptote at y=3y=3. Find the function f(x)f(x).

  1. f(x)=3xx−2f(x) = \frac{3x}{x-2}
  2. f(x)=3x−6x+2f(x) = \frac{3x-6}{x+2}
  3. f(x)=3xx+2f(x) = \frac{3x}{x+2} (correct answer)
  4. f(x)=x+23xf(x) = \frac{x+2}{3x}
Explanation: Let the function be f(x)=ax+bcx+df(x) = \frac{ax+b}{cx+d}. The vertical asymptote at x=−2x=-2 means the denominator is zero at x=−2x=-2, so we can write the denominator as c(x+2)c(x+2). The horizontal asymptote at y=3y=3 means the ratio of leading coefficients ac=3\frac{a}{c} = 3. We can choose c=1c=1 and a=3a=3 for simplicity. So the function has the form f(x)=3x+bx+2f(x) = \frac{3x+b}{x+2}. The graph passes through the origin (0,0)(0,0), which means f(0)=0f(0)=0. Substituting x=0x=0 into the function: f(0)=3(0)+b0+2=b2f(0) = \frac{3(0)+b}{0+2} = \frac{b}{2}. Setting this to 0 gives b2=0\frac{b}{2}=0, so b=0b=0. Therefore, the function is f(x)=3xx+2f(x) = \frac{3x}{x+2}.

Question 9

Let A be the x-intercept and B be the y-intercept of the graph of f(x)=2x−10x−3f(x) = \frac{2x-10}{x-3}. A vertical line L is the asymptote to the graph. Find the product of the horizontal distances from L to A and from L to B.

  1. 5
  2. 6 (correct answer)
  3. 9
  4. 10
Explanation:
  1. Find the x-intercept A by setting f(x)=0f(x)=0: 2x−10=0  ⟹  x=52x-10=0 \implies x=5. So, point A is (5,0)(5,0). 2. Find the y-intercept B by setting x=0x=0: f(0)=−10−3=103f(0) = \frac{-10}{-3} = \frac{10}{3}. So, point B is (0,103)(0, \frac{10}{3}). 3. Find the vertical asymptote L by setting the denominator to 0: x−3=0  ⟹  x=3x-3=0 \implies x=3. The line L is x=3x=3. 4. The horizontal distance from L (x=3x=3) to A (x=5x=5) is ∣5−3∣=2|5-3| = 2. 5. The horizontal distance from L (x=3x=3) to B (x=0x=0) is ∣0−3∣=3|0-3| = 3. 6. The product of these distances is 2×3=62 \times 3 = 6.

Question 10

The graph of a rational function y=f(x)y=f(x) has exactly one vertical asymptote at x=−1x=-1 and a horizontal asymptote at y=2y=2. The graph has an x-intercept at x=3x=3 and a y-intercept at y=−6y=-6. Which of the following could be the equation for f(x)f(x)?

  1. f(x)=x−32(x+1)f(x) = \frac{x-3}{2(x+1)}
  2. f(x)=2(x+1)x−3f(x) = \frac{2(x+1)}{x-3}
  3. f(x)=2x−6x+1f(x) = \frac{2x-6}{x+1} (correct answer)
  4. f(x)=2x+6x−1f(x) = \frac{2x+6}{x-1}
Explanation: From the given information, we can construct the function's form. A vertical asymptote at x=−1x=-1 implies a factor of (x+1)(x+1) in the denominator. An x-intercept at x=3x=3 implies a factor of (x−3)(x-3) in the numerator. A horizontal asymptote at y=2y=2 implies the ratio of the leading coefficients of the numerator and denominator is 2. This gives us the form f(x)=2(x−3)x+1f(x) = \frac{2(x-3)}{x+1}. Let's check if this form satisfies the y-intercept condition. The y-intercept occurs at x=0x=0: f(0)=2(0−3)0+1=−61=−6f(0) = \frac{2(0-3)}{0+1} = \frac{-6}{1} = -6. This matches the given information. Expanding the numerator gives f(x)=2x−6x+1f(x) = \frac{2x-6}{x+1}.

Question 11

Consider the function f(x)=ax2+1bx−2f(x) = \frac{ax^2+1}{bx-2}, where aa and bb are non-zero constants. Which of the following statements about the asymptotes of f(x)f(x) is always true?

  1. The function has a horizontal asymptote at y=0y=0.
  2. The function has a horizontal asymptote at y=aby=\frac{a}{b}.
  3. The function has no vertical or horizontal asymptotes.
  4. The function has a vertical asymptote but no horizontal asymptote. (correct answer)
Explanation: For the vertical asymptote, we set the denominator to zero: bx−2=0bx-2=0. Since bb is non-zero, this gives x=2bx=\frac{2}{b}. As aa is non-zero, the numerator will not be zero at x=2bx=\frac{2}{b} unless a(2b)2+1=0a(\frac{2}{b})^2+1=0, which is only one specific case. Thus, a vertical asymptote generally exists. For the horizontal asymptote, we compare the degree of the numerator (2) with the degree of the denominator (1). Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Therefore, the function always has a vertical asymptote and no horizontal asymptote.

Question 12

The graph of the function f(x)=2x2−2x−12x+2f(x) = \frac{2x^2 - 2x - 12}{x+2} has a removable discontinuity (a hole). What are the coordinates of this hole?

  1. (−2,−10)(-2, -10) (correct answer)
  2. (−2,0)(-2, 0)
  3. (3,−10)(3, -10)
  4. (3,0)(3, 0)
Explanation: A hole occurs when a factor in the denominator cancels with a factor in the numerator. First, factor the numerator: 2x2−2x−12=2(x2−x−6)=2(x−3)(x+2)2x^2 - 2x - 12 = 2(x^2 - x - 6) = 2(x-3)(x+2). The function is f(x)=2(x−3)(x+2)x+2f(x) = \frac{2(x-3)(x+2)}{x+2}. The factor (x+2)(x+2) cancels, which means there is a hole at the value of xx that makes this factor zero, i.e., x=−2x=-2. To find the y-coordinate of the hole, we evaluate the simplified function, g(x)=2(x−3)g(x) = 2(x-3), at x=−2x=-2. g(−2)=2(−2−3)=2(−5)=−10g(-2) = 2(-2-3) = 2(-5) = -10. Thus, the coordinates of the hole are (−2,−10)(-2, -10).

Question 13

Let A be the x-intercept and B be the y-intercept of the graph of f(x)=2x−10x−3f(x) = \frac{2x-10}{x-3}. A vertical line L is the asymptote to the graph. Find the product of the horizontal distances from L to A and from L to B.

  1. 5
  2. 6 (correct answer)
  3. 9
  4. 10
Explanation:
  1. Find the x-intercept A by setting f(x)=0f(x)=0: 2x−10=0  ⟹  x=52x-10=0 \implies x=5. So, point A is (5,0)(5,0). 2. Find the y-intercept B by setting x=0x=0: f(0)=−10−3=103f(0) = \frac{-10}{-3} = \frac{10}{3}. So, point B is (0,103)(0, \frac{10}{3}). 3. Find the vertical asymptote L by setting the denominator to 0: x−3=0  ⟹  x=3x-3=0 \implies x=3. The line L is x=3x=3. 4. The horizontal distance from L (x=3x=3) to A (x=5x=5) is ∣5−3∣=2|5-3| = 2. 5. The horizontal distance from L (x=3x=3) to B (x=0x=0) is ∣0−3∣=3|0-3| = 3. 6. The product of these distances is 2×3=62 \times 3 = 6.

Question 14

A rational function is defined by f(x)=x2−k2x2−5x+6f(x) = \frac{x^2 - k^2}{x^2 - 5x + 6}, where kk is a positive real constant. For which value of kk does the graph of f(x)f(x) have exactly one vertical asymptote?

  1. 1
  2. 5\sqrt{5}
  3. 3 (correct answer)
  4. 6
Explanation: First, find the potential vertical asymptotes by factoring the denominator: x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x-2)(x-3). This gives potential vertical asymptotes at x=2x=2 and x=3x=3. For the graph to have only one vertical asymptote, one of these must be a hole. This means a factor in the denominator must cancel with a factor in the numerator. The numerator is x2−k2=(x−k)(x+k)x^2 - k^2 = (x-k)(x+k). For a factor to cancel, either x−kx-k or x+kx+k must be equal to x−2x-2 or x−3x-3. Since kk is positive, we have two possibilities for cancellation: 1. x−k=x−2  ⟹  k=2x-k = x-2 \implies k=2. In this case, the (x−2)(x-2) factor cancels, leaving a single VA at x=3x=3. 2. x−k=x−3  ⟹  k=3x-k = x-3 \implies k=3. In this case, the (x−3)(x-3) factor cancels, leaving a single VA at x=2x=2. Both k=2k=2 and k=3k=3 result in exactly one vertical asymptote. Of the choices given, k=3k=3 is an option.

Question 15

How many distinct asymptotes (vertical and horizontal) and intercepts (x and y) does the graph of f(x)=2xx2−4f(x) = \frac{2x}{x^2-4} have in total?

  1. 3
  2. 4 (correct answer)
  3. 5
  4. 6
Explanation: Let's find each feature: 1. Vertical Asymptotes: Set the denominator to zero. x2−4=0  ⟹  (x−2)(x+2)=0x^2 - 4 = 0 \implies (x-2)(x+2)=0. The numerator is not zero at x=2x=2 or x=−2x=-2. So, there are two vertical asymptotes: x=2x=2 and x=−2x=-2. 2. Horizontal Asymptote: The degree of the numerator (1) is less than the degree of the denominator (2). So, there is one horizontal asymptote at y=0y=0. 3. x-intercept(s): Set the numerator to zero. 2x=0  ⟹  x=02x=0 \implies x=0. The x-intercept is at (0,0)(0,0). 4. y-intercept: Set x=0x=0. f(0)=2(0)02−4=0f(0) = \frac{2(0)}{0^2-4} = 0. The y-intercept is at (0,0)(0,0). The x- and y-intercepts are the same point. So, the distinct features are: VA at x=2x=2, VA at x=−2x=-2, HA at y=0y=0, and an intercept at (0,0)(0,0). This is a total of 4 distinct features.

Question 16

A rational function f(x)f(x) has a numerator and denominator that are both linear polynomials. The graph of f(x)f(x) passes through the origin, has a vertical asymptote at x=−2x=-2, and a horizontal asymptote at y=3y=3. Find the function f(x)f(x).

  1. f(x)=3xx−2f(x) = \frac{3x}{x-2}
  2. f(x)=3x−6x+2f(x) = \frac{3x-6}{x+2}
  3. f(x)=3xx+2f(x) = \frac{3x}{x+2} (correct answer)
  4. f(x)=x+23xf(x) = \frac{x+2}{3x}
Explanation: Let the function be f(x)=ax+bcx+df(x) = \frac{ax+b}{cx+d}. The vertical asymptote at x=−2x=-2 means the denominator is zero at x=−2x=-2, so we can write the denominator as c(x+2)c(x+2). The horizontal asymptote at y=3y=3 means the ratio of leading coefficients ac=3\frac{a}{c} = 3. We can choose c=1c=1 and a=3a=3 for simplicity. So the function has the form f(x)=3x+bx+2f(x) = \frac{3x+b}{x+2}. The graph passes through the origin (0,0)(0,0), which means f(0)=0f(0)=0. Substituting x=0x=0 into the function: f(0)=3(0)+b0+2=b2f(0) = \frac{3(0)+b}{0+2} = \frac{b}{2}. Setting this to 0 gives b2=0\frac{b}{2}=0, so b=0b=0. Therefore, the function is f(x)=3xx+2f(x) = \frac{3x}{x+2}.

Question 17

The graph of y=x2+ax+bx−2y = \frac{x^2 + ax + b}{x-2} has an x-intercept at x=1x=1 and a y-intercept at y=−2.5y=-2.5. Find the value of aa.

  1. -6 (correct answer)
  2. -4
  3. 4
  4. 5
Explanation: The y-intercept occurs when x=0x=0. We are given that this is y=−2.5y=-2.5. So, 02+a(0)+b0−2=−2.5\frac{0^2 + a(0) + b}{0-2} = -2.5, which simplifies to b−2=−2.5\frac{b}{-2} = -2.5. Solving for bb gives b=5b = 5. The x-intercept occurs when y=0y=0, which means the numerator is zero. We are given an x-intercept at x=1x=1. So, 12+a(1)+b=01^2 + a(1) + b = 0, which simplifies to 1+a+b=01 + a + b = 0. Substituting the value b=5b=5 we found, we get 1+a+5=01 + a + 5 = 0, which means a+6=0a+6=0. Therefore, a=−6a=-6.

Question 18

The graph of f(x)=2x2+3x−1x2+2f(x) = \frac{2x^2 + 3x - 1}{x^2 + 2} intersects its horizontal asymptote. Find the x-coordinate of the point of intersection.

  1. -1
  2. 35\frac{3}{5}
  3. 1
  4. 53\frac{5}{3} (correct answer)
Explanation: First, find the equation of the horizontal asymptote. Since the degrees of the numerator and denominator are both 2, the horizontal asymptote is the ratio of the leading coefficients, which is y=21=2y = \frac{2}{1} = 2. To find the point of intersection, we set the function equal to the value of the asymptote: 2x2+3x−1x2+2=2\frac{2x^2 + 3x - 1}{x^2 + 2} = 2. Now, we solve for xx: 2x2+3x−1=2(x2+2)  ⟹  2x2+3x−1=2x2+42x^2 + 3x - 1 = 2(x^2 + 2) \implies 2x^2 + 3x - 1 = 2x^2 + 4. The 2x22x^2 terms cancel out, leaving 3x−1=43x - 1 = 4. Solving for xx gives 3x=53x = 5, so x=53x = \frac{5}{3}.

Question 19

The graph of the function f(x)=x−ax−bf(x) = \frac{x-a}{x-b} has a vertical asymptote at x=3x=3 and a positive y-intercept. Which of the following must be true?

  1. a<0a < 0 and b=3b=3
  2. a>0a > 0 and b=3b=3 (correct answer)
  3. a<0a < 0 and b=−3b=-3
  4. a>0a > 0 and b=−3b=-3
Explanation: The vertical asymptote of a rational function is found where the denominator is zero. Given the VA is at x=3x=3, we must have x−b=0x-b=0 when x=3x=3, which implies b=3b=3. The y-intercept is the value of the function when x=0x=0. So, f(0)=0−a0−b=−a−b=abf(0) = \frac{0-a}{0-b} = \frac{-a}{-b} = \frac{a}{b}. We know b=3b=3, so the y-intercept is a3\frac{a}{3}. We are given that the y-intercept is positive, so a3>0\frac{a}{3} > 0. This implies that aa must be positive, i.e., a>0a > 0. Therefore, we must have a>0a > 0 and b=3b=3.

Question 20

Which statement best describes the graph of the function f(x)=x3−8x−2f(x) = \frac{x^3 - 8}{x-2}?

  1. A cubic function with a vertical asymptote at x=2x=2.
  2. A parabola with a vertical asymptote at x=2x=2.
  3. A parabola with a hole at (2,12)(2, 12). (correct answer)
  4. A line with a hole at (2,8)(2, 8).
Explanation: The numerator is a difference of cubes, which can be factored as a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2+ab+b^2). So, x3−8=x3−23=(x−2)(x2+2x+4)x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4). The function can be written as f(x)=(x−2)(x2+2x+4)x−2f(x) = \frac{(x-2)(x^2+2x+4)}{x-2}. For x≠2x \neq 2, the (x−2)(x-2) terms cancel, and the function simplifies to g(x)=x2+2x+4g(x) = x^2 + 2x + 4. This is the equation of a parabola. The cancellation indicates a hole in the graph at x=2x=2. To find the y-coordinate of the hole, substitute x=2x=2 into the simplified function: g(2)=22+2(2)+4=4+4+4=12g(2) = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12. Therefore, the graph is a parabola with a hole at (2,12)(2, 12).