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Algebra

Algebra Practice Test: Practice Test 30

Practice Test 30 for Algebra: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

The function f(x)=x2−6x+5f(x) = x^2 - 6x + 5f(x)=x2−6x+5 represents the height (in feet) of a ball xxx seconds after it is thrown. After factoring to find when the ball hits the ground, what is the axis of symmetry of this parabola?

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Question 1

The function f(x)=x2−6x+5f(x) = x^2 - 6x + 5f(x)=x2−6x+5 represents the height (in feet) of a ball xxx seconds after it is thrown. After factoring to find when the ball hits the ground, what is the axis of symmetry of this parabola?

  1. x=3x = 3x=3 (correct answer)
  2. x=1x = 1x=1
  3. x=5x = 5x=5
  4. x=−3x = -3x=−3

Explanation: First factor: f(x)=x2−6x+5=(x−1)(x−5)f(x) = x^2 - 6x + 5 = (x-1)(x-5)f(x)=x2−6x+5=(x−1)(x−5). The zeros are x=1x = 1x=1 and x=5x = 5x=5. The axis of symmetry is the midpoint between the zeros: x=1+52=3x = \frac{1+5}{2} = 3x=21+5​=3. Choice B gives one zero, choice C gives the other zero, and choice D incorrectly uses the negative of the correct answer.

Question 2

A polynomial is given by P(x)=(x−3)(x+3)(x−1)2.P(x)=(x-3)(x+3)(x-1)^2.P(x)=(x−3)(x+3)(x−1)2. Use zeros, multiplicities, and degree to sketch a rough graph. Which statement is correct?

  1. Zeros: x=−3,1,3x=-3,1,3x=−3,1,3 with x=1x=1x=1 multiplicity 2; touches at x=1x=1x=1; left down, right up
  2. Zeros: x=−3,1,3x=-3,1,3x=−3,1,3 with x=1x=1x=1 multiplicity 2; crosses at all three zeros; both ends up
  3. Zeros: x=−3,−1,3x=-3,-1,3x=−3,−1,3 with x=−1x=-1x=−1 multiplicity 2; touches at x=−1x=-1x=−1; both ends up
  4. Zeros: x=−3,1,3x=-3,1,3x=−3,1,3 with x=1x=1x=1 multiplicity 2; touches at x=1x=1x=1 and crosses at x=−3x=-3x=−3 and x=3x=3x=3; both ends up (correct answer)

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. The multiplicity of a zero (how many times a factor appears) tells you how the graph behaves there: if a zero has odd multiplicity (like just (x−2)(x - 2)(x−2) or (x−2)3(x - 2)^3(x−2)3), the graph crosses the x-axis at that point. If a zero has even multiplicity (like (x−2)2(x - 2)^2(x−2)2 or (x−2)4(x - 2)^4(x−2)4), the graph touches the x-axis but bounces back without crossing—it turns around at that zero! The polynomial P(x)=(x−3)(x+3)(x−1)2P(x) = (x - 3)(x + 3)(x - 1)^2P(x)=(x−3)(x+3)(x−1)2 has a zero at x=1x = 1x=1 with multiplicity 2 because the factor (x−1)(x - 1)(x−1) appears 2 times. Since 2 is even, the graph touches and turns at this zero. This is different from simple zeros where the graph just crosses straight through. Higher multiplicity means the graph 'hugs' the x-axis more at that zero! Zeros at x=−3x = -3x=−3 and 3 have multiplicity 1 (odd), so crosses there. Degree 4 even positive, both ends up. Choice A correctly describes behavior as touches at x=1x=1x=1, crosses at x=−3x=-3x=−3 and x=3x=3x=3 with both ends up by recognizing multiplicity effects and using degree and leading coefficient. Choice C has the end behavior backwards: with degree 4 (even) and leading coefficient positive, the ends should both up, not left down right up. Remember even degree means both ends same direction. The sign of the leading coefficient then determines up or down! Multiplicity matters: Simple zero (appears once) = graph crosses straight through. Even multiplicity (appears 2, 4, 6... times) = graph touches and bounces without crossing, creating a turning point at that zero. Odd multiplicity higher than 1 (appears 3, 5... times) = graph crosses but flattens out at that zero. The more times a factor repeats, the 'flatter' the graph gets at that zero! Quick check: count your zeros (including multiplicities) and it should equal the degree. If P(x)P(x)P(x) is degree 4, you should find 4 zeros total (could be 4 simple zeros, or 1 with multiplicity 2 and 2 simple, etc.). If your count doesn't match the degree, you've either missed a zero or the polynomial isn't completely factored. This check prevents forgetting zeros!

Question 3

Is the expression (34)×π\left(\frac{3}{4}\right)\times\pi(43​)×π rational or irrational? Choose the option with correct reasoning.​

  1. Rational, because multiplying by 34\frac{3}{4}43​ turns any number into a fraction.
  2. Irrational, because π\piπ is irrational and 34\frac{3}{4}43​ is a nonzero rational; if the product were rational, dividing by 34\frac{3}{4}43​ would make π\piπ rational, a contradiction. (correct answer)
  3. Rational, because π≈3.14\pi\approx 3.14π≈3.14 and (34)×3.14=2.355\left(\frac{3}{4}\right)\times 3.14=2.355(43​)×3.14=2.355 is a terminating decimal.
  4. Irrational, because the product of any two real numbers is always irrational unless both are integers.

Explanation: This question tests your understanding of how rational and irrational numbers behave under multiplication—specifically, applying the rule that nonzero rational × irrational = irrational. When you multiply a nonzero rational number by an irrational number, the result is ALWAYS irrational: for example, 2√3 is irrational, and (5/2)π is irrational. Again using contradiction: if rational × irrational = rational, then irrational = rational/rational = rational (rationals closed under division), contradicting irrationality. The 'nonzero' qualifier is crucial: 0 · √2 = 0, which IS rational, so we exclude that special case! Let's analyze (3/4) × π: Here 3/4 is a nonzero rational (it's a fraction with integer numerator 3 and denominator 4) and π is irrational. By our rule, nonzero rational × irrational = irrational, so (3/4)π must be irrational. Proof by contradiction: Assume (3/4)π is rational. Then π = [(3/4)π] ÷ (3/4) = [(3/4)π] × (4/3). Since (3/4)π is assumed rational and 4/3 is rational, their product would be rational (closure). So π would be rational, contradicting the fact that π is irrational. Therefore (3/4)π must be irrational. Choice B correctly identifies the product as irrational and provides valid reasoning: π is irrational, 3/4 is a nonzero rational, and if the product were rational, we could divide by 3/4 (multiply by 4/3) to show π is rational, which is a contradiction. Choice C makes a common error: it uses the approximation π ≈ 3.14 and claims the product 2.355 is rational because it's 'terminating.' But π is not equal to 3.14—that's just an approximation! The actual value of π has infinitely many non-repeating decimal places. Using approximations changes the problem entirely. Quick verification: if you claim something is rational, you should be able to write it as p/q with integer p and q. But (3/4)π cannot be written this way—if it could, then π = (4/3) × (p/q) = 4p/3q would be rational, contradiction! The rule 'nonzero rational × irrational = irrational' is universal and reliable.

Question 4

A bathroom scale measures to the nearest 0.1 lb. It reads 156.4 lb. Which reporting shows appropriate precision?

  1. 156 lb
  2. 156.40 lb
  3. 156.4 lb (correct answer)
  4. 156.37 lb

Explanation: This question tests your understanding that the precision of your reported answer should match the precision of your measurements—you can't claim accuracy beyond what your measurement tools or methods allow. The fundamental principle: you cannot report a calculated or measured value more precisely than your least precise measurement. If you measure length to the nearest centimeter (like 24 cm), you can't honestly report area to the nearest millimeter (like 576.00 cm²)—your measurement tool wasn't that precise! Calculations don't magically create precision; they can only preserve (or lose) the precision from your inputs. Reporting excessive decimal places is false precision—claiming accuracy you don't actually have. In the context of a bathroom scale measuring to nearest 0.1 lb, the value 156.4 lb should be reported as 156.4 lb because the scale's precision is to tenths of a pound. Reporting as 156.40 lb would be false precision—the scale can't distinguish between 156.40 lb and 156.44 lb. Context determines meaningful precision level! Choice C correctly reports to tenths of a pound (156.4 lb) which appropriately reflects the scale's capability of measuring to nearest 0.1 lb. Choice B shows false precision: reporting to 156.40 lb when the scale only justifies precision to tenths. Example: if you measure with a tool accurate to 0.1 lb, reporting 156.40 lb claims you can distinguish hundredths of a pound—but you can't with that scale! Report 156.4 lb instead, matching your measurement capability. Trailing zeros matter: 156.4 vs 156.40 represent different claimed precision (tenths vs hundredths). Only include trailing zeros after the decimal if your measurement actually supports that precision! If measured to tenths, write 156.4 (don't add fake zeros). Trailing zeros after decimal are significant—they're precision claims! When using a digital scale or any measuring device, report your measurement to match the device's displayed precision—no more, no less.

Question 5

A student writes Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1}Sn​=a+ar+ar2+⋯+arn−1 and multiplies by rrr to get rSn=ar+ar2+⋯+arnrS_n=ar+ar^2+\cdots+ar^nrSn​=ar+ar2+⋯+arn. When subtracting to make the middle terms cancel, which equation is correct?

(Assume r≠1r\ne 1r=1.)

  1. Sn−rSn=a−arnS_n-rS_n=a-ar^nSn​−rSn​=a−arn (correct answer)
  2. rSn−Sn=a−arnrS_n-S_n=a-ar^nrSn​−Sn​=a−arn
  3. Sn−rSn=a−arn−1S_n-rS_n=a-ar^{n-1}Sn​−rSn​=a−arn−1
  4. Sn+rSn=a+arnS_n+rS_n=a+ar^nSn​+rSn​=a+arn

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. The geometric series formula S_n = a(1 - r^n)/(1 - r) comes from a clever trick: write the sum S_n = a + ar + ar² + ... + ar^(n-1), then multiply by r to get rS_n = ar + ar² + ar³ + ... + ar^n. Subtracting these (S_n - rS_n) makes almost all terms cancel, leaving just S_n(1 - r) = a(1 - r^n), so S_n = a(1 - r^n)/(1 - r). The middle terms canceling is the magic that makes this work! Deriving S_n = a(1 - r^n)/(1 - r): (1) Write the sum: S_n = a + ar + ar² + ... + ar^(n-1). (2) Multiply both sides by r: rS_n = ar + ar² + ar³ + ... + ar^n. (3) Subtract second from first: S_n - rS_n = (a + ar + ar² + ... + ar^(n-1)) - (ar + ar² + ar³ + ... + ar^n). (4) Notice the cancellation: all middle terms cancel, leaving S_n - rS_n = a - ar^n. (5) Factor left side: S_n(1 - r) = a(1 - r^n). (6) Divide by (1 - r): S_n = a(1 - r^n)/(1 - r). This derivation shows why the formula works—it's not just memorization! Choice B correctly derives the formula through the subtraction method by showing S_n - rS_n = a - ar^n after cancellation. Choice A makes an error in the derivation: when subtracting S_n - rS_n, the middle terms should cancel, leaving a - ar^n, but this choice has a - ar^{n-1}, missing the correct exponent on the last term. Derivation memory aid: the trick is writing S_n, then writing rS_n (shifted one term), then subtracting. When you subtract, the middle terms align and cancel: S_n has 'ar' and rS_n has 'ar' (opposite signs, cancel!), S_n has 'ar²' and rS_n has 'ar²' (cancel!), etc. Only a from S_n and ar^n from rS_n don't cancel. This telescoping is the insight! Once you see it, you'll never forget the derivation.

Question 6

For the floor (greatest integer) function f(x)=⌊x⌋f(x)=\lfloor x\rfloorf(x)=⌊x⌋, what is f(2.7)f(2.7)f(2.7)?

  1. 333
  2. 222 (correct answer)
  3. 2.72.72.7
  4. 111

Explanation: This question tests your understanding of how to graph step functions and identify their key features like boundaries. Step functions are constant on intervals but jump to different values at certain points: the floor function f(x) = ⌊x⌋ gives the greatest integer less than or equal to x, creating horizontal segments that jump up by 1 at each integer. So f(2.7) = 2, f(3.0) = 3, f(3.8) = 3—it 'steps up' at whole numbers. These model situations like postage rates or parking fees where cost jumps at thresholds. The greatest integer function f(x) = ⌊x⌋ creates horizontal steps: for any x in the interval [n, n+1), the function value is n (the greatest integer ≤ x). So 0 ≤ x < 1 gives f(x) = 0, 1 ≤ x < 2 gives f(x) = 1, etc. On each interval, draw a horizontal segment at height n with a closed circle on the left endpoint and open circle on the right. The graph looks like stairs going up! Choice B correctly gives f(2.7) = 2 because 2 is the greatest integer less than or equal to 2.7. Choice A has the step function jumping at the wrong places or with wrong values. The floor function ⌊x⌋ equals 2 for all x in [2, 3), jumping to 3 exactly at x = 3. This choice has 3, but for 2.7 it's 2. Step functions need precise boundaries! Step function evaluation is straightforward: ⌊2.7⌋ = 2, ⌊5.1⌋ = 5, ⌊-1.3⌋ = -2. Find the greatest integer that's still less than or equal to your number. For positive decimals, just drop the decimal (2.7 → 2). For negative decimals, go down to next integer (-1.3 → -2, not -1). Graphing: horizontal segment from each integer to the next, jumping at integer values!

Question 7

An investment account is modeled by A(t)=1000(1.05)tA(t)=1000(1.05)^tA(t)=1000(1.05)t, where ttt is the number of years. What is the annual percent growth rate?

  1. 1.05% growth per year
  2. 5% growth per year (correct answer)
  3. 105% growth per year
  4. 0.05% growth per year

Explanation: This question tests your understanding of exponential functions and how to identify whether they represent growth or decay and what the percent rate of change is. To find the percent growth or decay rate from the base, use the formula r = b - 1 and convert to percent: if b = 1.05, then r = 1.05 - 1 = 0.05 = 5% growth. If b = 0.95, then r = 0.95 - 1 = -0.05 = 5% decay (we usually just say '5% decay' and understand it's a decrease). For the function A(t) = 1000(1.05)^t, the base is 1.05. To find the percent rate, we calculate r = 1.05 - 1 = 0.05. Converting to percent: 0.05 × 100% = 5%. Since 1.05 is greater than 1, this is growth, specifically 5% growth per year. Choice B correctly identifies the percent rate as 5% growth per year by showing that 1.05 = 1 + 0.05, which represents 5% growth. Excellent! Choice C makes a common percent mistake: the base 1.05 doesn't mean 105% growth—it means 5% growth! The 1 represents 'what you already have' (100%), and the 0.05 is the additional 5%, for a total of 105% of the previous amount (which is 5% growth). To find the percent rate: (1) Identify the base b = 1.05, (2) Subtract 1: r = 1.05 - 1 = 0.05, (3) Convert to percent: 0.05 × 100 = 5%. Example: base is 1.05, so r = 1.05 - 1 = 0.05 = 5%. Easy!

Question 8

A ball’s height (in feet) is modeled by h(t)=−16t2+32th(t) = -16t^2 + 32th(t)=−16t2+32t. When does it hit the ground (solve h(t)=0h(t)=0h(t)=0)?

  1. t=0, 2t = 0,\ 2t=0, 2 (correct answer)
  2. t=1t = 1t=1
  3. t=−1, 1t = -1,\ 1t=−1, 1
  4. t=−2, 0t = -2,\ 0t=−2, 0

Explanation: This question tests your ability to solve quadratic equations by factoring in a real-world context, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor -16t² + 32t = 0 to get -16t(t - 2) = 0. Using the Zero Product Property, either -16t = 0 or (t - 2) = 0. Solving each: t = 0 and t = 2. These are the two solutions, representing when the ball is thrown (t=0) and when it hits the ground (t=2)! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (t - 2) = 0, we get t = 2 (not t = -2). When you solve t - 2 = 0, you add 2 to both sides, giving the positive value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 9

Solve the system algebraically. (The solution(s) are the intersection point(s) of the line and the circle.)

x^2 + y^2 = 25 \\ y = 4 \end{cases}$$
  1. (3,4)(3,4)(3,4) and (−3,4)(-3,4)(−3,4) (correct answer)
  2. (4,3)(4,3)(4,3) and (−4,3)(-4,3)(−4,3)
  3. (±5,4)(\pm 5,4)(±5,4)
  4. No solution

Explanation: This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually y = mx + b is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system {x² + y² = 25, y = 4} by substitution: (1) The linear equation gives us y = 4. (2) Substitute into the circle equation: x² + 4² = 25. (3) Simplify: x² + 16 = 25, so x² = 9. (4) Solve: x = ±3. (5) The y-value is already given as 4 for both points. Solutions: (3, 4) and (-3, 4). Choice A correctly finds the intersection points as (3, 4) and (-3, 4) through proper substitution and solving. Choice B has the coordinates backwards: the solutions should be (3, 4) and (-3, 4), not (4, 3) and (-4, 3). After solving for x and then finding y, make sure you write them in the correct order: x-coordinate first, y-coordinate second! Circle-line systems are special: when solving x² + y² = r² with a line, you often get solutions with radicals like (√5, √5). Don't be intimidated! These are exact answers. You can estimate decimals if asked, but the radical form is exact and preferred. Also, circle-line systems are symmetric—if (a, b) is a solution and the line passes through the origin or has special symmetry, there's often a matching solution (-a, -b) or similar.

Question 10

What is the completed square form of x2+4x−5=0x^2+4x-5=0x2+4x−5=0 written as (x−p)2=q(x-p)^2=q(x−p)2=q?

  1. (x+2)2=1(x+2)^2=1(x+2)2=1
  2. (x+2)2=9(x+2)^2=9(x+2)2=9 (correct answer)
  3. (x−2)2=−9(x-2)^2=-9(x−2)2=−9
  4. (x−2)2=9(x-2)^2=9(x−2)2=9

Explanation: This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The form (x - p)² = q tells us immediately: if q > 0, there are two real solutions (x = p ± √q); if q = 0, there's one solution (x = p); if q < 0, there are no real solutions (we'd need to take square root of negative). So completing the square not only solves the equation, it also reveals the nature of the solutions before we calculate them! Solving x² + 4x - 5 = 0 by completing the square: (1) Move constant to right: x² + 4x = 5. (2) Take half of 4 to get 2, square it to get 4, add to both sides: x² + 4x + 4 = 5 + 4. (3) Left side is perfect square (x + 2)², right side simplifies to 9: (x + 2)² = 9. (4) Take square roots: x + 2 = ±3. (5) Solve: x = -2 ± 3. Done! Choice C correctly completes the square to get (x + 2)² = 9 with accurate arithmetic and proper form. Choice D doesn't add (b/2)² to both sides—only adds to one side or miscalculates, leading to negative q incorrectly! Completing the square requires adding the perfect square term to BOTH sides to maintain equality. If you only add to the left, you've changed the equation and will get wrong solutions. Both sides, always! The ± is crucial: from (x - 3)² = 16, taking square roots gives x - 3 = ±4 (both +4 and -4), so x = 3 + 4 = 7 OR x = 3 - 4 = -1. Two solutions! Don't forget the ± and don't forget to split it into two separate solutions. Check both: 7 and -1 both satisfy the original equation? Yes!

Question 11

Factor x2−49x^2-49x2−49 by recognizing it as a difference of squares.​

  1. (x+7)(x−7)(x+7)(x-7)(x+7)(x−7) (correct answer)
  2. (x+49)(x−1)(x+49)(x-1)(x+49)(x−1)
  3. (x+7)2(x+7)^2(x+7)2
  4. (x−7)2\left(x-7\right)^2(x−7)2

Explanation: This question tests your ability to recognize algebraic patterns and structures—like difference of squares or perfect square trinomials—that let you rewrite expressions more efficiently. The difference of squares pattern a² - b² = (a + b)(a - b) is super useful: whenever you see two perfect squares being subtracted (with no middle term), you can factor it as the sum and difference of what's being squared. For example, x² - 9 = (x)² - (3)² = (x + 3)(x - 3). Here, x² - 49 is x² - 7², so it factors directly to (x + 7)(x - 7) using the pattern. Choice B correctly recognizes the pattern as difference of squares and applies the formula to get (x + 7)(x - 7). Excellent pattern recognition! Choice A might come from confusing it with a perfect square trinomial, but remember, difference of squares has no middle term—keep practicing to spot the subtraction between two squares! To spot difference of squares: (1) Are there exactly two terms? (2) Are they being subtracted? (3) Is each term a perfect square (like x², 9, 4x², 25)? If yes to all three, you've got a² - b², which factors as (a + b)(a - b). Try it with x² - 16: yes two terms, yes subtraction, yes both perfect squares → (x + 4)(x - 4). Boom!

Question 12

Which equation correctly demonstrates why we define 271/327^{1/3}271/3 to equal 3?

  1. Because (271/3)3=27(1/3)⋅3=271=27(27^{1/3})^3 = 27^{(1/3) \cdot 3} = 27^1 = 27(271/3)3=27(1/3)⋅3=271=27, which preserves the power property from integer exponents (correct answer)
  2. Because 271/3=271273=271968327^{1/3} = \frac{27^1}{27^3} = \frac{27}{19683}271/3=273271​=1968327​, which follows the quotient rule for exponents directly
  3. Because 27÷3=927 \div 3 = 927÷3=9, so the fractional exponent represents division of the base by the denominator
  4. Because 271/3=132727^{1/3} = \frac{1}{3^{27}}271/3=3271​, which follows from the negative exponent rule applied to fractions

Explanation: The correct answer is A. We define 271/3=327^{1/3} = 3271/3=3 precisely because we want the power property (am)n=amn(a^m)^n = a^{mn}(am)n=amn to continue working for rational exponents. If 271/3=327^{1/3} = 3271/3=3, then (271/3)3=33=27(27^{1/3})^3 = 3^3 = 27(271/3)3=33=27, which equals 27(1/3)⋅3=271=2727^{(1/3) \cdot 3} = 27^1 = 2727(1/3)⋅3=271=27, confirming the property holds. B incorrectly applies the quotient rule. C incorrectly suggests that fractional exponents represent division. D incorrectly applies negative exponent rules to positive fractional exponents.

Question 13

Prove or disprove the claim: “If rrr is a nonzero rational number and iii is an irrational number, then r×ir\times ir×i is irrational.” Which reasoning is correct?

  1. Assume r×ir\times ir×i is rational. Then i=r×iri=\frac{r\times i}{r}i=rr×i​ would be rational (rational divided by nonzero rational), contradicting that iii is irrational. Therefore r×ir\times ir×i is irrational. (correct answer)
  2. The claim is false because 0×i=00\times i=00×i=0 is rational, so r×ir\times ir×i can be rational even when rrr is nonzero.
  3. The claim is true because multiplication always makes decimals longer, so the result cannot be a fraction.
  4. The claim is true because r×ir\times ir×i is irrational by definition whenever iii is irrational.

Explanation: This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. When you multiply a nonzero rational number by an irrational number, the result is ALWAYS irrational: for example, 2√3 is irrational, and (5/2)π is irrational. Again using contradiction: if rational × irrational = rational, then irrational = rational/rational = rational (rationals closed under division), contradicting irrationality. The 'nonzero' qualifier is crucial: 0 · √2 = 0, which IS rational, so we exclude that special case! Proving nonzero rational × irrational = irrational by contradiction: Let r be a nonzero rational and i be irrational. Assume r · i = q for some rational q. Rearranging: i = q/r. Since q is rational, r is rational and nonzero, and rationals are closed under division (by nonzero), q/r is rational. So i is rational. Contradiction with i being irrational! Therefore r · i must be irrational when r ≠ 0. (Note: 0 · irrational = 0, which IS rational, so we need r ≠ 0.) Choice A correctly uses proof by contradiction, showing that if r × i were rational, then i = (r × i)/r would be rational divided by nonzero rational = rational, contradicting that i is irrational. Choice B has the conclusion backwards: it claims the statement is false because 0 × i = 0 is rational, but the claim specifically states 'nonzero rational,' which excludes the case r = 0. The statement is true as written with the nonzero condition! Check the logical flow: does the reasoning actually support the conclusion? The three key facts to remember: (1) Rational + rational = rational, always (closure property—proven by showing p/q + r/s = (ps+qr)/(qs), still a fraction). (2) Rational + irrational = irrational, always (proven by contradiction—if sum were rational, we could rearrange to show the irrational is rational, contradiction!). (3) Nonzero rational × irrational = irrational, always (same contradiction structure as addition). These are universal rules you can rely on! Quick verification: if you claim something is rational, you should (in principle) be able to write it as p/q with integer p and q. If you claim it's irrational, you should explain why it CAN'T be written that way (often via contradiction). Don't just assert—provide reasoning! That's what mathematical understanding means: knowing not just WHAT is true, but WHY it's true.

Question 14

A food truck operates under the following conditions: They can prepare at most 100 meals per day due to kitchen capacity. Each regular meal costs 3tomakeandsellsfor3 to make and sells for 3tomakeandsellsfor8. Each premium meal costs 5tomakeandsellsfor5 to make and sells for 5tomakeandsellsfor12. They need at least $200 in profit daily to stay viable.

If rrr represents regular meals and ppp represents premium meals, which system correctly models all constraints for viable daily operation?

  1. r+p≤100r + p \leq 100r+p≤100, 8r+12p−3r−5p≥2008r + 12p - 3r - 5p \geq 2008r+12p−3r−5p≥200, r≥0r \geq 0r≥0, p≥0p \geq 0p≥0
  2. r+p≤100r + p \leq 100r+p≤100, 8r+12p≥2008r + 12p \geq 2008r+12p≥200, r≥0r \geq 0r≥0, p≥0p \geq 0p≥0
  3. r+p≤100r + p \leq 100r+p≤100, (8−3)r+(12−5)p≥200(8-3)r + (12-5)p \geq 200(8−3)r+(12−5)p≥200, r≥0r \geq 0r≥0, p≥0p \geq 0p≥0 (correct answer)
  4. r+p=100r + p = 100r+p=100, 5r+7p≥2005r + 7p \geq 2005r+7p≥200, r>0r > 0r>0, p>0p > 0p>0

Explanation: The constraints are: (1) At most 100 meals: r+p≤100r + p \leq 100r+p≤100, (2) At least $200 profit: profit per regular meal is 8−3=58 - 3 = 58−3=5, profit per premium meal is 12−5=712 - 5 = 712−5=7, so 5r+7p≥2005r + 7p \geq 2005r+7p≥200, (3) Non-negative production: r≥0,p≥0r \geq 0, p \geq 0r≥0,p≥0. Choice C shows the profit calculation explicitly. Choice A shows the calculation but doesn't simplify. Choice B uses revenue instead of profit. Choice D requires exactly 100 meals and positive production, which aren't required.

Question 15

View the expression x+6x−2\frac{x+6}{x-2}x−2x+6​ by identifying what’s being done to what. Which statement best describes its structure?

  1. It is the difference of the two units (x+6)(x+6)(x+6) and (x−2)(x-2)(x−2)
  2. It is the square of the unit (x+6)(x+6)(x+6) divided by x−2x-2x−2
  3. It is the product of (x+6)(x+6)(x+6) and (x−2)(x-2)(x−2)
  4. It is the quotient of the two units (x+6)(x+6)(x+6) and (x−2)(x-2)(x−2) (correct answer)

Explanation: This question tests your ability to look at a complicated expression and understand its overall structure by seeing certain parts as single 'chunks' rather than getting lost in all the details. When an expression looks overwhelming, we can make sense of it by identifying the main parts and temporarily treating complex subexpressions as single units—like viewing P(1+r)nP(1 + r)^nP(1+r)n as 'P times [some factor]' where we don't worry about what's inside that factor yet. This 'chunking' helps us see the big picture structure: is it a product? A sum? Something raised to a power? In (x+6)/(x−2)(x+6)/(x-2)(x+6)/(x−2), treating (x+6) and (x-2) as units shows the structure as one unit divided by the other, making it a quotient. Choice B correctly views the expression as the quotient of the two units (x+6) and (x-2), recognizing that division is the main operation here. It's tempting to see it as a product like in choice A, but by chunking the numerator and denominator separately, we clarify it's division—keep that in mind for fractions! A helpful trick: circle or box the parts you want to treat as units. For example, in P(1+r)nP(1 + r)^nP(1+r)n, box the (1+r)n(1 + r)^n(1+r)n part and think 'P times [box].' This visual chunking helps your brain organize the structure. In applied formulas, chunking helps you understand what each factor means: in A=πr2A = \pi r^2A=πr2, we see 'π times [radius squared],' which helps us remember that doubling the radius quadruples the area (because we're squaring the radius). Structure reveals relationships!

Question 16

Solve using the quadratic formula: x2+6x+25=0x^2+6x+25=0x2+6x+25=0. The discriminant is negative, so the solutions are complex conjugates. Use i2=−1i^2=-1i2=−1 and write solutions in a+bia+bia+bi form.​

  1. x=−3±4ix=-3\pm 4ix=−3±4i (correct answer)
  2. x=3±4ix=3\pm 4ix=3±4i
  3. x=−3±2ix=-3\pm 2ix=−3±2i
  4. x=−6±4ix=-6\pm 4ix=−6±4i

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. Complex solutions appear in conjugate pairs for quadratics with real coefficients: if a + bi is a solution, then a - bi is automatically also a solution (same real part, opposite imaginary part). This pairing is guaranteed by the ± in the quadratic formula and has important implications: the sum is real, and the product is real, which is why quadratics with complex roots can still have real coefficients! Solving x² + 6x + 25 = 0 using the quadratic formula: (1) Identify a = 1, b = 6, c = 25. (2) Calculate discriminant: b² - 4ac = (6)² - 4(1)(25) = 36 - 100 = -64. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-6 ± √(-64))/2 = (-6 ± 8i)/2. (5) Simplify: x = -6/2 ± 8i/2 = -3 ± 4i. Solutions: x = -3 + 4i and x = -3 - 4i. Choice A correctly solves to get x = -3 ± 4i with accurate calculation using i² = -1. Choice B has a sign error in the real part: from the quadratic formula x = (-b ± i√|discriminant|)/(2a), the real part is -b/(2a) = -6/2 = -3, not 3. With the ± and the i both appearing in solutions, it's easy to lose track of signs. Write out each solution separately: x = -3 + 4i AND x = -3 - 4i to make sure both signs are correct. Conjugate pair shortcut: if one solution is a + bi, immediately write down a - bi as the other without recalculating! For real-coefficient quadratics, complex solutions ALWAYS come in conjugate pairs. If the quadratic formula gives you -3 + 4i from the + version, the - version automatically gives -3 - 4i. Same real part, flip the imaginary sign. Done!

Question 17

The sequence 7,5,3,1,…7, 5, 3, 1, \dots7,5,3,1,… is a function whose input is n=1,2,3,…n = 1,2,3,\dotsn=1,2,3,… and output is ana_nan​. Which recursive rule generates this sequence?

  1. a1=5, an+1=an−2a_1=5,\ a_{n+1}=a_n-2a1​=5, an+1​=an​−2
  2. a1=7, an+1=an+2a_1=7,\ a_{n+1}=a_n+2a1​=7, an+1​=an​+2
  3. a1=7, an+1=an−2a_1=7,\ a_{n+1}=a_n-2a1​=7, an+1​=an​−2 (correct answer)
  4. a1=7, an+1=2ana_1=7,\ a_{n+1}=2a_na1​=7, an+1​=2an​

Explanation: This question tests your understanding that sequences are special functions with integer domains, and how to work with recursively defined sequences. A sequence is just a function where the inputs are whole numbers (like 1, 2, 3, ...) and each output is called a term: the first term a₁, the second term a₂, and so on. We can think of a sequence as a list of numbers where each number's position (1st, 2nd, 3rd) is its input! Looking at the sequence 7, 5, 3, 1, ..., let's find the pattern between consecutive terms: from 7 to 5, we subtract 2. From 5 to 3, we subtract 2. This pattern holds for all consecutive terms, so the recursive rule is aₙ₊₁ = aₙ - 2, with starting value a₁ = 7. Choice A is correct because it properly identifies the pattern as subtracting 2 each time with the correct starting value. Choice B applies the wrong operation in the recursive rule, adding 2 when the pattern shows we should subtract 2. Look carefully at how consecutive terms relate: are we adding the same amount? Multiplying? Doing something else? To write a recursive rule from a sequence, compare consecutive terms: What's happening from term to term? Are we adding the same number (arithmetic)? Multiplying by the same number (geometric)? Adding the previous two (Fibonacci-style)? Once you spot the pattern, write it as aₙ₊₁ = [rule using aₙ], and don't forget to state your starting value! Remember: sequences ARE functions, just special ones! The input is always a positive integer (the term number), and the output is the term value.

Question 18

A bacteria culture starts with 1000 cells and increases by 20% each hour. Which statement correctly identifies the growth factor bbb and the percent rate rrr per hour in the exponential form N(t)=a⋅btN(t)=a\cdot b^tN(t)=a⋅bt?

  1. b=1.02b=1.02b=1.02 and r=0.02r=0.02r=0.02 (2% growth)
  2. b=0.80b=0.80b=0.80 and r=−0.20r=-0.20r=−0.20 (20% decay)
  3. b=1.20b=1.20b=1.20 and r=0.20r=0.20r=0.20 (20% growth) (correct answer)
  4. b=20b=20b=20 and r=20r=20r=20 (20% growth)

Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent growth means the quantity is multiplied by the same factor greater than 1 each time period: if a population grows by 5% per year, it's multiplied by 1.05 each year (since 105% = 1 + 0.05 = 1.05 means 'keep all of what you had plus gain 5% more'). This creates exponential growth where the amount added each period gets larger because you're taking a percent of an increasing base! The context describes 'a bacteria culture starts with 1000 cells and increases by 20% each hour.' Key phrase: 'increases by 20% each hour' directly tells us this is exponential growth with a constant percent rate. Each hour, the quantity is multiplied by 1 + 0.20 = 1.20, making this exponential rather than linear. In one hour, you have 120% of what you started with (original 100% plus 20%). Choice B correctly identifies this as b=1.20 and r=0.20 (20% growth) because the growth factor includes the original 100% plus the 20% increase. Choice A says growth when it's actually decay (or vice versa): looking at the context, since it describes increase, this is growth, not decay. When the base is greater than 1, or when the context says 'increases,' that's exponential growth! To find percent rate from a growth/decay factor: (1) Identify b (the base or factor), (2) Subtract 1: r = b - 1, (3) Convert to percent: multiply by 100. Example: b = 1.12 → r = 0.12 → 12% growth. For decay: b = 0.95 → r = -0.05 → 5% decay (we usually state as positive '5% decay' rather than 'negative 5%'). The subtraction of 1 is the crucial step! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!

Question 19

A water tank is being filled at a rate of 15 gallons per minute for the first 20 minutes, then at 8 gallons per minute for the next 30 minutes. Which statement best describes the rate of change of water volume with respect to time?

  1. The water volume changes at a constant rate of 11.5 gallons per minute throughout the entire process
  2. The water volume changes at a constant rate during each interval, but the overall process does not have a constant rate (correct answer)
  3. The water volume changes at a constant rate of 23 gallons per minute when considering the combined flow rates
  4. The water volume does not change at a constant rate during any part of the filling process due to varying conditions

Explanation: The rate of change is constant within each time interval (15 gal/min for 0-20 min, 8 gal/min for 20-50 min), but the overall process has two different constant rates, making the entire process non-linear. A is incorrect because it averages the rates incorrectly. C is wrong because you don't add the rates. D is incorrect because each individual interval does have a constant rate.

Question 20

If y=x34y = x^{\frac{3}{4}}y=x43​, then y43y^{\frac{4}{3}}y34​ equals which expression?

  1. x43x^{\frac{4}{3}}x34​
  2. xxx (correct answer)
  3. x916x^{\frac{9}{16}}x169​
  4. x124x^{\frac{12}{4}}x412​

Explanation: When you encounter problems involving exponents raised to other exponents, you're working with the power rule for exponents. This rule states that (am)n=am⋅n(a^m)^n = a^{m \cdot n}(am)n=am⋅n - you multiply the exponents together. Given that y=x34y = x^{\frac{3}{4}}y=x43​, you need to find what y43y^{\frac{4}{3}}y34​ equals. Substitute the expression for yyy into the problem: y43=(x34)43y^{\frac{4}{3}} = (x^{\frac{3}{4}})^{\frac{4}{3}}y34​=(x43​)34​. Now apply the power rule by multiplying the exponents: 34×43=3×44×3=1212=1\frac{3}{4} \times \frac{4}{3} = \frac{3 \times 4}{4 \times 3} = \frac{12}{12} = 143​×34​=4×33×4​=1212​=1. Therefore, y43=x1=xy^{\frac{4}{3}} = x^1 = xy34​=x1=x, which is choice B. Let's examine why the other answers are incorrect. Choice A gives x43x^{\frac{4}{3}}x34​, which would result from incorrectly thinking you add exponents instead of multiplying them (34+43=2512\frac{3}{4} + \frac{4}{3} = \frac{25}{12}43​+34​=1225​, but this still doesn't equal 43\frac{4}{3}34​). Choice C shows x916x^{\frac{9}{16}}x169​, which comes from the error of squaring each part of the original fraction: (34)2=916(\frac{3}{4})^2 = \frac{9}{16}(43​)2=169​. Choice D presents x124x^{\frac{12}{4}}x412​, which equals x3x^3x3. This might come from multiplying the numerators and denominators separately without simplifying: 3×44\frac{3 \times 4}{4}43×4​. Remember this key pattern: when you see fractional exponents that are reciprocals of each other (like 34\frac{3}{4}43​ and 43\frac{4}{3}34​), they'll always multiply to equal 1, giving you the base variable without any exponent.

Question 21

The quadratic equation x2−4x+k=0x^2 - 4x + k = 0x2−4x+k=0 is transformed by completing the square to (x−2)2=7(x - 2)^2 = 7(x−2)2=7. What is the value of kkk, and what are the solutions to the original equation?

  1. k=11k = 11k=11; solutions are x=2±7x = 2 \pm \sqrt{7}x=2±7​
  2. k=3k = 3k=3; solutions are x=2±7x = 2 \pm \sqrt{7}x=2±7​
  3. k=−3k = -3k=−3; solutions are x=−2±7x = -2 \pm \sqrt{7}x=−2±7​
  4. k=−3k = -3k=−3; solutions are x=2±7x = 2 \pm \sqrt{7}x=2±7​ (correct answer)

Explanation: When you encounter a quadratic equation that's been transformed by completing the square, you need to work backwards to find the original coefficient and then solve using the completed square form. Starting with x2−4x+k=0x^2 - 4x + k = 0x2−4x+k=0, let's complete the square ourselves. Take half of the coefficient of xxx (which is −4-4−4), square it: (−4/2)2=(−2)2=4(-4/2)^2 = (-2)^2 = 4(−4/2)2=(−2)2=4. So we rewrite as x2−4x+4+k−4=0x^2 - 4x + 4 + k - 4 = 0x2−4x+4+k−4=0, which becomes (x−2)2+k−4=0(x - 2)^2 + k - 4 = 0(x−2)2+k−4=0, or (x−2)2=4−k(x - 2)^2 = 4 - k(x−2)2=4−k. Since we're told this equals (x−2)2=7(x - 2)^2 = 7(x−2)2=7, we have 4−k=74 - k = 74−k=7, so k=−3k = -3k=−3. To find the solutions, we solve (x−2)2=7(x - 2)^2 = 7(x−2)2=7 by taking the square root of both sides: x−2=±7x - 2 = \pm\sqrt{7}x−2=±7​, giving us x=2±7x = 2 \pm \sqrt{7}x=2±7​. Looking at the wrong answers: Choice A incorrectly calculates k=11k = 11k=11 (likely from 4+k=74 + k = 74+k=7). Choice B gets k=3k = 3k=3 (perhaps from k−4=7k - 4 = 7k−4=7) but has the right solutions. Choice C has the correct k=−3k = -3k=−3 but wrong solutions x=−2±7x = -2 \pm \sqrt{7}x=−2±7​, missing that the vertex form shows (x−2)(x - 2)(x−2), not (x+2)(x + 2)(x+2). Study tip: When completing the square, always double-check your work by expanding the completed square form back to standard form. This catches sign errors and helps you verify both the constant term and the solutions.

Question 22

An item’s value is modeled by V(t)=1200(1+r)tV(t)=1200(1+r)^tV(t)=1200(1+r)t. If the value decreases by 15% each year, which value of rrr should be used?

  1. r=0.15r=0.15r=0.15
  2. r=−0.15r=-0.15r=−0.15 (correct answer)
  3. r=0.85r=0.85r=0.85
  4. r=−0.85r=-0.85r=−0.85

Explanation: This question tests your understanding of exponential functions and how to identify whether they represent growth or decay and what the percent rate of change is. The key difference between growth factor and growth rate: the factor b is what you multiply by each time, while the rate r is how much it's changing by percent. They're related by b = 1 + r, so knowing one gives you the other! The context tells us the value decreases by 15% each year. For a decrease, the rate r is negative: r = -15% = -0.15. We can verify: if r = -0.15, then b = 1 + r = 1 + (-0.15) = 0.85, which means we keep 85% of the value each year (losing 15%). Choice B correctly identifies r = -0.15 because a 15% decrease means r = -0.15. Excellent! Choice A forgets that decreases require negative rates: if r = 0.15, that would mean 15% growth, not decay! For decay, r must be negative. The form y = a(1 + r)^x makes the rate super obvious: if the value decreases by 15%, then r = -0.15. But remember, we often say '15% depreciation' and understand the negative is implied. The mathematical form needs r = -0.15 to show decay!

Question 23

A movie theater has 200 seats. The function R(n)=12nR(n)=12nR(n)=12n gives the revenue (in dollars) from selling nnn tickets, where nnn is the number of tickets sold. What is an appropriate realistic domain for RRR?​

  1. All real numbers nnn
  2. All integers nnn such that 0≤n≤2000\le n\le 2000≤n≤200 (correct answer)
  3. All real numbers nnn such that 0≤n≤2000\le n\le 2000≤n≤200
  4. All integers nnn such that n≥200n\ge 200n≥200

Explanation: This question tests your understanding of how the domain of a function relates to both the context (what makes sense in real life) and the graph (how the function is represented visually). The domain is the set of all possible input values, and in real-world contexts, we need to think about what values actually make sense: if n represents the number of tickets sold, we can't have n = -5 (negative tickets) or n = 2.7 (partial tickets), so the appropriate domain is non-negative integers, and we're also limited by the theater's capacity of 200 seats. Since you can sell 0 tickets (nobody shows up) up to 200 tickets (sold out), the domain includes all whole numbers from 0 to 200. Choice B correctly identifies the domain as all integers n such that 0 ≤ n ≤ 200 because you can only sell whole tickets and you're limited by the 200-seat capacity. Choice C is incorrect because it allows fractional tickets (like 150.5 tickets), which doesn't make sense in this context. When determining domain from context, ask yourself: (1) What does the variable represent? (2) Can it be negative? (usually not for quantities, time, etc.) (3) Are there upper limits? (capacity, time limits) (4) Must it be whole numbers? (counting people/items) or can it be any number? (measuring distance/time). Answer these questions and the domain becomes clear!

Question 24

A sequence is defined by the explicit formula an=12−2na_n=12-2nan​=12−2n for n=1,2,3,…n=1,2,3,\dotsn=1,2,3,…. What is a4a_4a4​?

  1. 666
  2. 222
  3. 444 (correct answer)
  4. 888

Explanation: This question tests your understanding that sequences are special functions with integer domains, and how to work with explicitly defined sequences. A sequence is just a function where the inputs are whole numbers (like 1, 2, 3, ...) and each output is called a term: the first term a₁, the second term a₂, and so on. We can think of a sequence as a list of numbers where each number's position (1st, 2nd, 3rd) is its input! Unlike recursive definitions, an explicit formula like aₙ = 12 - 2n lets us find any term directly by plugging in the value of n. To find a₄, we simply substitute n = 4 into the formula: a₄ = 12 - 2(4) = 12 - 8 = 4. So the 4th term is 4. Choice C is correct because it properly evaluates the explicit formula with n = 4 to get 4. Great work following the pattern! Choice D gives 8, which might come from calculating 12 - 4 = 8 instead of 12 - 2(4) = 4. Remember to multiply 2 times n before subtracting from 12! The beauty of explicit formulas is that you can jump directly to any term—no need to find all the terms before it! Notice that this sequence decreases by 2 each time: a₁ = 10, a₂ = 8, a₃ = 6, a₄ = 4, and so on.

Question 25

Use the zeros and degree to sketch a rough graph of P(x)=(x+2)2(x−3)P(x)=(x+2)^2(x-3)P(x)=(x+2)2(x−3). Which statement is correct about x-intercepts and end behavior?

  1. x-intercepts at x=−2x=-2x=−2 (touches) and x=3x=3x=3 (crosses); as x→−∞x\to -\inftyx→−∞, P(x)→−∞P(x)\to -\inftyP(x)→−∞ and as x→∞x\to \inftyx→∞, P(x)→∞P(x)\to \inftyP(x)→∞ (correct answer)
  2. x-intercepts at x=−2x=-2x=−2 (crosses) and x=3x=3x=3 (touches); as x→−∞x\to -\inftyx→−∞, P(x)→−∞P(x)\to -\inftyP(x)→−∞ and as x→∞x\to \inftyx→∞, P(x)→∞P(x)\to \inftyP(x)→∞
  3. x-intercepts at x=−2x=-2x=−2 (touches) and x=3x=3x=3 (crosses); as x→−∞x\to -\inftyx→−∞, P(x)→∞P(x)\to \inftyP(x)→∞ and as x→∞x\to \inftyx→∞, P(x)→∞P(x)\to \inftyP(x)→∞
  4. x-intercepts at x=−2x=-2x=−2 (crosses) and x=3x=3x=3 (crosses); as x→−∞x\to -\inftyx→−∞, P(x)→∞P(x)\to \inftyP(x)→∞ and as x→∞x\to \inftyx→∞, P(x)→−∞P(x)\to -\inftyP(x)→−∞

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. To construct a rough sketch from zeros: (1) Mark the zeros on the x-axis, (2) Determine end behavior from degree and leading coefficient sign, (3) Connect the zeros with a smooth curve that crosses/touches appropriately and has the right end behavior. You don't need exact heights—just show the general shape, where it crosses the x-axis, and which direction the ends go! To sketch P(x) = (x+2)²(x-3): (1) Zeros are at x = -2 (multiplicity 2) and x = 3 (multiplicity 1), so mark these on the x-axis. (2) This polynomial has degree 3 (add exponents: 2 + 1) with leading coefficient +1 (positive), so end behavior is: left end down, right end up. (3) At x = -2, multiplicity 2 (even) means the graph touches and turns around. At x = 3, multiplicity 1 (odd) means the graph crosses. (4) Connect with smooth curve showing these behaviors. Choice A correctly states x-intercepts at x = -2 (touches) and x = 3 (crosses) with end behavior showing left end to -∞ and right end to ∞ by properly analyzing multiplicities and using degree 3 with positive leading coefficient. Choice B incorrectly swaps the crossing/touching behavior: at x = -2 with even multiplicity 2, the graph should touch and bounce, not cross. At x = 3 with odd multiplicity 1, the graph should cross, not touch. Even multiplicities create turnarounds at zeros, while odd multiplicities allow crossings. Check the power on each factor! The four-step polynomial sketching recipe: (1) Find and mark all zeros on the x-axis, (2) Determine end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive = eventually up, negative = eventually down), (3) For each zero, check if it crosses (odd multiplicity) or touches (even multiplicity), (4) Connect with smooth curve that goes through/touches all zeros with correct end behavior. Don't worry about exact curve—just the general shape!