Unit 1 · 30 to 40% of the exam

Unit 1 practice test: Polynomial and Rational Functions

twelve questions, every option explained

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value or an expression; sets share one graph or table. Covers: Change in Tandem; Rates of Change; Rates of Change in Linear and Quadratic Functions; Polynomial Functions and Rates of Change; Polynomial Functions and Complex Zeros; Polynomial Functions and End Behavior; Rational Functions and End Behavior; Rational Functions and Zeros; Rational Functions and Vertical Asymptotes; Rational Functions and Holes; Equivalent Representations of Polynomial and Rational Expressions; Transformations of Functions; Function Model Selection and Assumption Articulation; Function Model Construction and Application.

Start at question 1

Graph

Volume of an open box

Squares of side xx inches are cut from the corners of a 1212-inch by 1212-inch sheet, and the sides are folded up to make an open box. The plot shows the volume VV, in cubic inches, for six cut sizes.
0123456020406080100120140x (inches)V (cubic inches)

1. Which type of function model is most appropriate for the box volume VV as a function of the cut size xx?

  1. Quadratic, because the data rise and then fall
  2. Cubic, because volume is length times width times height, each depending on xx
  3. Linear, because the cut size changes by equal steps
  4. Quadratic, because area depends on two lengths
Show the answer

The box is 12−2x12 - 2x by 12−2x12 - 2x by xx, so V(x)=x(12−2x)2V(x) = x(12 - 2x)^2, a product of three linear factors: a cubic. Volume contexts suggest cubic models.

  • Quadratic, because the data rise and then fall: Rising then falling fits a cubic here too; the volume context points to degree 33.
  • Correct: Cubic, because volume is length times width times height, each depending on xx: Three dimensions, each linear in xx, multiply to a cubic.
  • Linear, because the cut size changes by equal steps: Equal steps in the input say nothing about the model type by themselves.
  • Quadratic, because area depends on two lengths: Volume uses three lengths, not two.

2. What domain does the context allow for the volume model?

  1. 0<x<120 < x < 12
  2. x>0x > 0
  3. 0<x<60 < x < 6
  4. 0.5≤x≤5.50.5 \le x \le 5.5
Show the answer

The cut must have positive size, and two cuts of xx must fit along a 1212-inch side: 2x<122x < 12, so x<6x < 6. At x=6x = 6 there is no base left.

  • 0<x<120 < x < 12: Two cuts come out of each side, so 2x<122x < 12.
  • x>0x > 0: A cut of 77 inches is impossible on a 1212-inch sheet.
  • Correct: 0<x<60 < x < 6: Positive cuts that leave a base.
  • 0.5≤x≤5.50.5 \le x \le 5.5: These are only the cut sizes plotted; any size between 00 and 66 makes a box.

3. Which is an assumption built into the model V(x)=x(12−2x)2V(x) = x(12 - 2x)^2?

  1. The sheet has no thickness, so the folded sides add nothing to the size
  2. The volume increases as the cut size increases
  3. The cuts are whole numbers of inches
  4. The box has a lid
Show the answer

The formula uses the full 12−2x12 - 2x for the inside length, which treats the material as having no thickness. Real cardboard would make the box a little smaller inside.

  • Correct: The sheet has no thickness, so the folded sides add nothing to the size: The formula ignores the thickness of the sheet.
  • The volume increases as the cut size increases: The plot shows the volume rising and then falling.
  • The cuts are whole numbers of inches: Any cut size between 00 and 66 works in the formula.
  • The box has a lid: The box is open; the formula has no lid in it.

Graph

Concentration of a medicine

1234567891024681012t (hours)C (mg/L)C

4. The concentration of a medicine in the blood is C(t)=12tt+2C(t) = \frac{12t}{t + 2} milligrams per liter, tt hours after a dose. Using average rates of change on [1,1.1][1, 1.1] and [5,5.1][5, 5.1], at which time is the concentration changing faster?

  1. At t=5t = 5, because C(5)>C(1)C(5) > C(1)
  2. At t=1t = 1: about 2.5812.581 against about 0.4830.483
  3. At both times equally, because the intervals have the same length
  4. At t=5t = 5: about 8.6208.620 compared with about 4.2584.258
Show the answer

C(1.1)−C(1)0.1\frac{C(1.1) - C(1)}{0.1}≈2.581\approx 2.581 and C(5.1)−C(5)0.1\frac{C(5.1) - C(5)}{0.1}≈0.483\approx 0.483. The rate near t=1t = 1 is larger, so the concentration is rising faster then. This matches the graph, which is steeper near t=1t = 1.

  • At t=5t = 5, because C(5)>C(1)C(5) > C(1): A larger output does not mean a faster change. Compare the rates: about 0.4830.483 at t=5t = 5.
  • Correct: At t=1t = 1: about 2.5812.581 against about 0.4830.483: Right: the average rate on [1,1.1][1, 1.1] is about 2.5812.581 and on [5,5.1][5, 5.1] about 0.4830.483.
  • At both times equally, because the intervals have the same length: Equal lengths make the rates fair to compare, but the rates themselves differ.
  • At t=5t = 5: about 8.6208.620 compared with about 4.2584.258: These are values of CC at 5.15.1 and 1.11.1, not rates of change.

5. The concentration of a medicine in the blood is C(t)=12tt+2C(t) = \frac{12t}{t + 2} milligrams per liter, tt hours after a dose. The average rate of change of CC on [0,2][0, 2] is 33. Which sentence interprets it?

  1. From t=0t = 0 to t=2t = 2, CC rose an average of 33 milligrams per liter per hour.
  2. After 22 hours, the concentration is 33 milligrams per liter.
  3. From t=0t = 0 on, CC rises by exactly 33 milligrams per liter in each hour.
  4. The concentration falls by 33 milligrams per liter each hour.
Show the answer

A positive average rate of change means the output went up over the interval; its value is the average change in output per unit of input.

  • Correct: From t=0t = 0 to t=2t = 2, CC rose an average of 33 milligrams per liter per hour.: Right: C(2)−C(0)2−0\frac{C(2) - C(0)}{2 - 0}=62= \frac{6}{2}=3= 3, in milligrams per liter per hour.
  • After 22 hours, the concentration is 33 milligrams per liter.: C(2)=6C(2) = 6; 33 is a rate, not a concentration.
  • From t=0t = 0 on, CC rises by exactly 33 milligrams per liter in each hour.: An average on [0,2][0, 2] says nothing about later hours, and the rate is not constant.
  • The concentration falls by 33 milligrams per liter each hour.: A positive rate means the output rises as time goes on.

6. The concentration of a medicine in the blood is C(t)=12tt+2C(t) = \frac{12t}{t + 2} milligrams per liter, tt hours after a dose. Which interval's average rate of change gives the best estimate of how fast CC changes at t=4t = 4?

  1. [0,8][0, 8]
  2. [4,10][4, 10]
  3. [3.9,4.1][3.9, 4.1]
  4. [2,6][2, 6]
Show the answer

The shorter the interval around a point, the closer the average rate of change is to how fast the function changes at that point.

  • [0,8][0, 8]: A long interval mixes the fast early change with the slow later change.
  • [4,10][4, 10]: This interval is long and starts at 44, so it reflects mostly later, slower change.
  • Correct: [3.9,4.1][3.9, 4.1]: Right: a short interval around t=4t = 4 uses outputs very close to the one at t=4t = 4.
  • [2,6][2, 6]: Better than a long interval, but [3.9,4.1][3.9, 4.1] is much shorter around t=4t = 4.

7. For y=20−3xy = 20 - 3x, how do xx and yy vary together?

  1. As xx increases, yy increases.
  2. As xx increases, yy decreases, by 33 for each 11 that xx goes up.
  3. As xx increases, yy decreases, by 2020 for each 11 that xx goes up.
  4. yy stays at 2020 as xx changes.
Show the answer

Compare outputs: y(0)=20y(0) = 20, y(1)=17y(1) = 17, y(2)=14y(2) = 14. The input and output move in opposite directions, and the output changes by −3-3 for every 11-unit step in the input.

  • As xx increases, yy increases.: Try two inputs: x=0x = 0 gives 2020 and x=1x = 1 gives 1717. The output went down.
  • Correct: As xx increases, yy decreases, by 33 for each 11 that xx goes up.: Right: each 11-unit increase in xx subtracts another 33 from yy.
  • As xx increases, yy decreases, by 2020 for each 11 that xx goes up.: 2020 is the output at x=0x = 0; the change per unit is the −3-3.
  • yy stays at 2020 as xx changes.: The −3x-3x term changes yy whenever xx changes.

8. What is the average rate of change of g(x)=−3x+7g(x) = -3x + 7 on [−2,10][-2, 10]?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

A linear function has the same average rate of change on every interval: its slope, −3-3. Check: g(10)−g(−2)12\frac{g(10) - g(-2)}{12}=−23−1312= \frac{-23 - 13}{12}=−3= -3.

  • Answer: -3

9. What are the degree and the leading coefficient of f(x)=−2(x−1)(x+3)2f(x) = -2(x - 1)(x + 3)^2?

  1. Degree 33, leading coefficient −2-2
  2. Degree 22, leading coefficient −2-2
  3. Degree 33, leading coefficient −18-18
  4. Degree 33, leading coefficient 22
Show the answer

Multiplying the xx from each factor gives the leading term −2⋅x⋅x2=−2x3-2 \cdot x \cdot x^2 = -2x^3: degree 33, leading coefficient −2-2.

  • Correct: Degree 33, leading coefficient −2-2: Right: the highest power is x⋅x2=x3x \cdot x^2 = x^3, with coefficient −2⋅1⋅1=−2-2 \cdot 1 \cdot 1 = -2.
  • Degree 22, leading coefficient −2-2: Count every factor of xx: (x−1)(x+3)2(x - 1)(x + 3)^2 has three.
  • Degree 33, leading coefficient −18-18: −18-18 is the constant term, −2(−1)(3)2-2(-1)(3)^2; the leading coefficient multiplies the xx in each factor.
  • Degree 33, leading coefficient 22: Keep the sign: the factor in front is −2-2.

10. A polynomial function pp of degree 55 has real coefficients. Its zeros include 33, with multiplicity 22, and 1+i1 + i. How many more zeros does pp have, counting multiplicity?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Degree 55 means 55 zeros counting multiplicity. Known: 33, 33, 1+i1 + i and its conjugate 1−i1 - i. That is 44, so 11 more remains (and it must be real).

  • Answer: 1

11. Which describes g(x)=−3x5+100x4g(x) = -3x^5 + 100x^4 as x→−∞x \to -\infty?

  1. limx→−∞g(x)=−∞\lim_{x \to -\infty} g(x) = -\infty
  2. limx→−∞g(x)=−3\lim_{x \to -\infty} g(x) = -3
  3. limx→−∞g(x)\lim_{x \to -\infty} g(x)=100= 100
  4. limx→−∞g(x)=∞\lim_{x \to -\infty} g(x) = \infty
Show the answer

For large |x||x| the leading term −3x5-3x^5 dominates. If xx is negative, x5x^5 is negative and −3x5-3x^5 is positive: limx→−∞g(x)=∞\lim_{x \to -\infty} g(x) = \infty.

  • limx→−∞g(x)=−∞\lim_{x \to -\infty} g(x) = -\infty: −3x5-3x^5 is positive for negative xx: a negative times a negative.
  • limx→−∞g(x)=−3\lim_{x \to -\infty} g(x) = -3: −3-3 is the leading coefficient, not a value the outputs approach.
  • limx→−∞g(x)\lim_{x \to -\infty} g(x)=100= 100: The coefficients do not give the limit; the leading term does.
  • Correct: limx→−∞g(x)=∞\lim_{x \to -\infty} g(x) = \infty: Right: the leading term −3x5-3x^5 is positive and huge for large negative xx.

12. What is limx→∞4x3−x2x3+5\lim_{x \to \infty} \frac{4x^3 - x}{2x^3 + 5}?

  1. ∞\infty
  2. −15-\frac{1}{5}
  3. 44
  4. 22
Show the answer

Equal degrees: the limit is the ratio of the leading coefficients, 42=2\frac{4}{2} = 2.

  • ∞\infty: Equal degrees: the outputs level off instead of growing.
  • −15-\frac{1}{5}: That compares non-leading terms; use the leading terms.
  • 44: Divide by the denominator's leading coefficient too: 42\frac{4}{2}.
  • Correct: 22: Right: 4x32x3=2\frac{4x^3}{2x^3} = 2.