Machine-scored practice Choose or enter each answer. Use the individual solutions for review or mark the whole section at the end.
Questions 1–8 Questions 9–16 Questions 17–24 Questions 25–32
1. Multiple choice The graph of y = f(x) is transformed to the graph of y = −2f(x − 5) + 3.
Which of the following lists the transformations that are applied to the graph of y = f(x)?
Check solution 1 A vertical stretch about the x-axis by a factor of 2, a reflection in the x-axis, then a translation of 5 units right and 3 units up
In y = af[b(x − h)] + k, a = −2 gives a vertical stretch by a factor of |−2| = 2 together with a reflection in the x-axis; h = 5 is 5 units right and k = 3 is 3 units up. Stretches and reflections are applied before translations, so the choice that translates first is out.
2. Multiple choice The point (−6, 4) lies on the graph of y = f(x).
The corresponding point on the graph of y = ½f(3x) − 1 is
Check solution 2 (−2, 1)
The mapping is (x, y) → (x/3, ½y − 1). The x-coordinate is divided by 3, because b = 3 compresses horizontally: −6 ÷ 3 = −2. The y-coordinate is ½(4) − 1 = 1, so the image is (−2, 1). Multiplying by 3 instead of dividing gives the distractor (−18, 1).
3. Multiple choice The domain of y = f(x) is −4 ≤ x ≤ 8.
The domain of y = f(2(x + 1)) is
Check solution 3 −3 ≤ x ≤ 3
Whatever goes inside f must still land between −4 and 8, so −4 ≤ 2(x + 1) ≤ 8. Divide by 2: −2 ≤ x + 1 ≤ 4. Subtract 1: −3 ≤ x ≤ 3. Undo the parameters in the reverse order you apply them.
4. Numerical response Four functions are listed below.
1 y = f(x − 3)
2 y = f(x) − 3
3 y = f(x + 3)
4 y = f(x) + 3
The graph of y = f(x) translated 3 units right is function ___, translated 3 units left is function ___, translated 3 units up is function ___, and translated 3 units down is function ___.
Your response Enter all four digits in the requested order.
Check solution 4 1342
Inside the brackets moves the graph horizontally and does the opposite of what it looks like: f(x − 3) is 3 right, f(x + 3) is 3 left. Outside the brackets moves it vertically and does what it looks like: +3 is up, −3 is down. Right, left, up, down = 1, 3, 4, 2.
5. Multiple choice If f(x) = 2x³ − 5, then f⁻¹(x) =
Check solution 5 ∛((x + 5)/2)
Swap x and y: x = 2y³ − 5. Solve for y: x + 5 = 2y³, so y³ = (x + 5)/2 and y = ∛((x + 5)/2). The inverse is not the reciprocal, which is where 1/(2x³ − 5) comes from.
6. Multiple choice The solution to the equation 4^(x + 1) = 8^(x − 2) is
Check solution 6 x = 8
Write both sides base 2: 2^(2(x + 1)) = 2^(3(x − 2)). Equate exponents: 2x + 2 = 3x − 6, so x = 8. Check: 4⁹ = 262 144 and 8⁶ = 262 144.
7. Multiple choice log_b 2 = m and log_b 5 = n
Written in terms of m and n, log_b 50 =
Check solution 7 m + 2n
50 = 2 × 5², so log_b 50 = log_b 2 + log_b 5² = log_b 2 + 2 log_b 5 = m + 2n. The product law adds; the power law brings the exponent out front as a multiplier, not as an exponent on n.
8. Numerical response The solution to the equation 3^(2x − 1) = 81 is x = ______.
Your response Enter your answer to the nearest tenth.
Check solution 8 2.5
81 = 3⁴, so 2x − 1 = 4, giving 2x = 5 and x = 2.5. A numerical-response answer between 0 and 1 would need the 0 recorded before the decimal point; here the answer is 2.5.
9. Multiple choice The expression 2 log x − ½ log y + log 3, where x > 0 and y > 0, written as a single logarithm, is
Check solution 9 log(3x²/√y)
Power law first: 2 log x = log x² and ½ log y = log √y. Then the product and quotient laws: log x² − log √y + log 3 = log(3x²/√y). Logarithms of different terms cannot be combined inside one log the way log(2x − ½y + 3) suggests.
10. Multiple choice The graph of y = log₂ x is transformed to the graph of y = log₂(x + 4) − 3.
Which statement about the transformed graph is true?
Check solution 10 Its vertical asymptote is x = −4 and its domain is x > −4
x + 4 replaces x, which translates the graph 4 units left, so the asymptote moves from x = 0 to x = −4 and the domain becomes x > −4. The −3 shifts the graph down and does not affect the domain; a logarithmic function has no horizontal asymptote.
11. Multiple choice When P(x) = 2x³ − 5x² + x − 4 is divided by x + 2, the remainder is
Check solution 11 −42
By the remainder theorem the remainder is P(−2), because x + 2 = x − (−2). P(−2) = 2(−8) − 5(4) + (−2) − 4 = −16 − 20 − 2 − 4 = −42. Using P(2) instead is the classic slip.
12. Numerical response The polynomial function P(x) = x³ + kx² − 4x − 12 has x − 2 as a factor.
The value of k is ______.
Your response Enter your numerical answer.
Check solution 12 3
By the factor theorem, P(2) = 0. So 8 + 4k − 8 − 12 = 0, which gives 4k = 12 and k = 3. Check: P(x) = x³ + 3x² − 4x − 12 = (x − 2)(x + 2)(x + 3).
13. Multiple choice A cubic polynomial function has zeros of −2 with multiplicity 2 and 3 with multiplicity 1, and its graph passes through the point (0, −12).
An equation of this function is
Check solution 13 y = (x + 2)²(x − 3)
The zeros give y = a(x + 2)²(x − 3). Substituting (0, −12): −12 = a(4)(−3) = −12a, so a = 1 and y = (x + 2)²(x − 3). The sign inside each factor is the opposite of the zero.
14. Multiple choice Which of the following best describes the end behaviour of the graph of y = −2x⁴ + 3x² − 7?
Check solution 14 The graph extends down in quadrant 3 and down in quadrant 4
The degree is even and the leading coefficient is negative, so both ends point down: the graph falls in quadrant 3 on the left and in quadrant 4 on the right. Only the leading term matters for end behaviour.
15. Numerical response The graph of y = (2x² − 5x − 3)/(x − 3) has a point of discontinuity at (3, b).
The value of b is ______.
Your response Enter your numerical answer.
Check solution 15 7
Factor the numerator: 2x² − 5x − 3 = (2x + 1)(x − 3). The factor x − 3 cancels, leaving y = 2x + 1 with a hole where x = 3. Substituting: b = 2(3) + 1 = 7. Here the reduced denominator is nonzero at x = 3, so the missing point is a hole.
16. Multiple choice The domain of the function y = √(5 − 2x) + 3 is
Check solution 16 x ≤ 5/2
A square root needs a radicand that is not negative: 5 − 2x ≥ 0, so 5 ≥ 2x and x ≤ 5/2. The + 3 shifts the graph up and affects the range, not the domain.
17. Multiple choice The solution to the equation √(3x + 4) = x is
Check solution 17 x = 4 only
Square both sides: 3x + 4 = x², so x² − 3x − 4 = 0 and (x − 4)(x + 1) = 0, giving x = 4 or x = −1. Check both: √16 = 4 works, but √1 = 1 ≠ −1, so x = −1 is extraneous. Squaring can create roots that do not satisfy the original equation.
18. Multiple choice The exact value of cos(5π/6) is
Check solution 18 −√3/2
5π/6 is in quadrant II with reference angle π/6. Cosine is negative in quadrant II and cos(π/6) = √3/2, so cos(5π/6) = −√3/2. The value −1/2 comes from using π/3 as the reference angle instead of π/6.
19. Multiple choice The terminal arm of an angle θ in standard position passes through the point P(−8, 15).
The value of sin θ is
Check solution 19 15/17
r = √((−8)² + 15²) = √289 = 17. Then sin θ = y/r = 15/17. It is positive because the point is in quadrant II, where sine is positive. The value −8/17 is cos θ and −15/8 is tan θ.
20. Numerical response The number of solutions to the equation sin 2θ = −½ on the domain 0 ≤ θ < 2π is ______.
Your response Enter your numerical answer.
Check solution 20 4
In radian mode graph y = sin(2θ) and y = −0.5 from θ = 0 to 2π with vertical limits −1.2 and 1.2. There are two cycles, each intersecting the horizontal line twice. Neither endpoint is an intersection, giving four solutions.
21. Multiple choice A central angle of 2.5 radians in a circle of radius 12 cm subtends an arc of length
Check solution 21 30 cm
a = θr with θ in radians, so a = 2.5 × 12 = 30 cm. Converting to degrees first is unnecessary, and dividing 12 by 2.5 gives the 4.8 cm distractor.
22. Multiple choice The general solution to the equation 2 cos θ + √3 = 0, where n ∈ I, is
Check solution 22 θ = 5π/6 + 2πn and θ = 7π/6 + 2πn
cos θ = −√3/2, so the reference angle is π/6 and cosine is negative in quadrants II and III: θ = π − π/6 = 5π/6 and θ = π + π/6 = 7π/6, each plus 2πn. The pair π/6 and 11π/6 is where cosine is +√3/2, not −√3/2.
23. Multiple choice The expression (1 − cos²θ)/(sin θ cos θ), where sin θ ≠ 0 and cos θ ≠ 0, simplifies to
Check solution 23 tan θ
By the Pythagorean identity, 1 − cos²θ = sin²θ. So the expression is sin²θ/(sin θ cos θ) = sin θ/cos θ = tan θ. Cancelling one sine from the top and bottom is the whole question.
24. Multiple choice sin θ = 3/5, where θ terminates in quadrant II.
The value of sin 2θ is
Check solution 24 −24/25
Using x² + y² = r²: x = −4 since θ is in quadrant II, so cos θ = −4/5. Then sin 2θ = 2 sin θ cos θ = 2(3/5)(−4/5) = −24/25. Forgetting that cosine is negative in quadrant II gives 24/25. −7/25 is sin²θ − cos²θ, the double-angle formula with its terms the wrong way round; cos 2θ itself is +7/25.
25. Multiple choice In the expansion of (2x − 3)⁸, written in descending powers of x, the general term is t_(k+1) = ₈C_k (2x)^(8−k) (−3)^k.
The coefficient of the term containing x⁵ is
Check solution 25 −48 384
The power of x is 8 − k, so 8 − k = 5 and k = 3, which makes it the 4th term. Its coefficient is ₈C₃(2)⁵(−3)³ = 56 × 32 × (−27) = −48 384. Dropping the 2⁵ from (2x)⁵ gives −1 512, and losing the sign on (−3)³ gives 48 384.
26. Multiple choice y = −4 cos[2(x − π/3)] + 7
Which statement about the graph of the function above is true?
Check solution 26 The range is 3 ≤ y ≤ 11 and the period is π
The amplitude is |a| = 4 and the midline is y = 7, so the range is 7 − 4 ≤ y ≤ 7 + 4, or 3 ≤ y ≤ 11. The period is 2π/|b| = 2π/2 = π. Amplitude is never negative — the minus sign is a reflection.
27. Multiple choice The depth of water in a harbour varies sinusoidally. The maximum depth is 8.2 m, the minimum depth is 2.4 m, and one complete cycle takes 12.4 h. The depth is modelled by D(t) = a cos[b(t − c)] + d with a > 0.
The values of a and d, respectively, are
Check solution 27 2.9 and 5.3
The amplitude is half the difference: a = (8.2 − 2.4)/2 = 2.9. The midline is the average: d = (8.2 + 2.4)/2 = 5.3. Using the whole difference instead of half gives 5.8.
28. Numerical response A 100 mg sample of a radioactive isotope decays according to A(t) = 100(½)^(t/5.7), where A is the mass remaining, in milligrams, and t is the time, in days.
The time it takes for the sample to decay to 12 mg is ______ days.
Your response Enter your answer to the nearest tenth.
Check solution 28 17.4
12 = 100(½)^(t/5.7), so (½)^(t/5.7) = 0.12. Taking logs: (t/5.7) log 0.5 = log 0.12, so t = 5.7 × log 0.12/log 0.5 = 17.435… ≈ 17.4 days. Keep every decimal until the final rounding.
29. Multiple choice A sensor reports f(t) = 3t + 2 for 0 ≤ t ≤ 4. A calibration function is g(x) = (x − 2)/3.
What are the rule and domain of the calibrated reading g(f(t))?
Check solution 29 t for 0 ≤ t ≤ 4
Use the measured reading as the input of the calibration. Then g(f(t)) = ((3t + 2) − 2)/3 = t. The original sensor supplies readings only for 0 ≤ t ≤ 4, so simplifying the rule does not extend its domain.
30. Multiple choice A stage crew spends $240 on lighting and $6 per ticket on other costs. For n paid tickets, the average cost per ticket is C(n) = (240 + 6n)/n.
Which statement describes the average cost in the context of the model?
Check solution 30 It approaches $6 from above as the positive integer n increases.
Write C(n) = 6 + 240/n. The fixed cost per ticket stays positive and gets smaller as more people buy tickets. The horizontal asymptote is C = 6, and the context permits only positive whole ticket counts.
31. Numerical response Four expressions are listed below.
1 ₆C₂
2 ₆P₃
3 5!
4 ₁₀C₃
The three expressions above that have a value of 120 are numbered ______, ______, and ______.
Your response Enter all three digits in any order.
Check solution 31 234
₆C₂ = 15, ₆P₃ = 6 × 5 × 4 = 120, 5! = 120, and ₁₀C₃ = (10 × 9 × 8)/(3 × 2 × 1) = 120. So expressions 2, 3 and 4. Any order of those three digits is scored correct.
32. Numerical response The number of different arrangements of all the letters of the word BANANA is ______.
Your response Enter your numerical answer.
Check solution 32 60
There are 6 letters with 3 A's and 2 N's repeated, so the count is 6!/(3!2!) = 720/12 = 60. Divide by the factorial of each repeat, or you count identical arrangements more than once.
Mark these 32 questions
Written responses Show the setup, justify the steps and interpret your answer. Use the model and marking notes to assess each part yourself. Each response has a two-mark part and a three-mark part. Use half marks for partial reasoning. To combine your scores, multiply the machine score out of 32 by 75% and the written score out of 15 by 25%, then add.
Written Response 1 · Relations and Functions The polynomial function f(x) = x³ − 4x² + x + 6 is transformed to g(x) = −2f(x − 1).
Part 1 · 2 practice marks Algebraically determine the zeros of f(x).
Your working Compare with the model response
Test factors of the constant term 6: f(−1) = −1 − 4 − 1 + 6 = 0, so x + 1 is a factor.
Divide: (x³ − 4x² + x + 6) ÷ (x + 1) = x² − 5x + 6.
Factor the quotient: x² − 5x + 6 = (x − 2)(x − 3).
So f(x) = (x + 1)(x − 2)(x − 3), and setting each factor equal to zero gives the zeros x = −1, x = 2 and x = 3 .
Check your work. Show an algebraic factorization and set each factor equal to zero. Any justified algebraic route to all three zeros earns credit. Reading the zeros off a graphing calculator answers a question that was not asked — algebraically rules that out.
Part 2 · 3 practice marks Describe the transformations that map the graph of f(x) onto the graph of g(x), and determine the zeros of g(x).
Your working Compare with the model response
g(x) = −2f(x − 1), so a = −2 and h = 1: a vertical stretch about the x-axis by a factor of 2, a reflection in the x-axis, and a translation of 1 unit right.
Zeros are where the graph meets the x-axis, and neither a vertical stretch nor a reflection in the x-axis moves a point that is already at y = 0. Only the horizontal translation moves them, so each zero shifts 1 unit right.
The zeros of g(x) are x = 0, x = 3 and x = 4 .
Check: g(0) = −2f(−1) = −2(0) = 0. ✓
Check your work. Name the transformations in full — "vertical stretch about the x-axis by a factor of 2", not "stretched by 2" — since markers look for correct vocabulary. Saying why the stretch and reflection leave the zeros in place is what separates a complete response from a partial one.
Written Response 2 · Trigonometry A Ferris wheel has a diameter of 16 m. Its lowest point is 1.5 m above the ground, and it completes one rotation every 40 s. A rider boards at the lowest point at t = 0 and the wheel turns at a constant rate. The rider's height above the ground, h, in metres, after t seconds is modelled by a sinusoidal function.
Part 1 · 2 practice marks Determine an equation for h as a function of t.
Your working Compare with the model response
Amplitude: |a| = 16/2 = 8. Midline: the lowest point is 1.5 m, so d = 1.5 + 8 = 9.5.
Period 40 s, so b = 2π/40 = π/20.
At t = 0 the rider is at the minimum, which is a cosine curve reflected in its midline, so a is negative and c = 0.
h(t) = −8 cos(πt/20) + 9.5 , where h is in metres and t is in seconds.
Check: h(0) = −8(1) + 9.5 = 1.5 m ✓ and h(20) = −8(−1) + 9.5 = 17.5 m, which is 1.5 + 16. ✓
Check your work. Every parameter needs a reason, and the answer must be an equation with units identified — not a list of values. A sine version with a phase shift, such as h(t) = 8 sin[(π/20)(t − 10)] + 9.5, is equally correct.
Part 2 · 3 practice marks Algebraically determine the first time the rider is 12 m above the ground, to the nearest tenth of a second, and interpret your answer.
Your working Compare with the model response
Set h = 12: 12 = −8 cos(πt/20) + 9.5.
2.5 = −8 cos(πt/20), so cos(πt/20) = −0.3125.
πt/20 = cos⁻¹(−0.3125) = 1.8886… (radians, second quadrant).
t = 20(1.8886…)/π = 12.023… ≈ 12.0 s .
The rider first passes 12 m above the ground about 12.0 s after boarding, while still rising toward the top at t = 20 s.
Check your work. Set the calculator to radians, keep the decimals until the last step, and finish with a sentence that puts the number back in context — interpret asks for that. The second solution of the equation, at about t = 27.98 s, is the rider coming back down, so it is not the first time.
Written Response 3 · Permutations, Combinations, and the Binomial Theorem A student council has 7 women and 6 men.
Part 1 · 2 practice marks Determine the number of different 5-person committees that can be formed with at least 3 women.
Your working Compare with the model response
Order does not matter, so use combinations, and split "at least 3" into cases.
3 women, 2 men: ₇C₃ × ₆C₂ = 35 × 15 = 525
4 women, 1 man: ₇C₄ × ₆C₁ = 35 × 6 = 210
5 women, 0 men: ₇C₅ = 21
Total: 525 + 210 + 21 = 756 committees .
Check your work. Show all three cases and add them. A response that gives only the largest case, or that multiplies the cases instead of adding, shows partial understanding at best.
Part 2 · 3 practice marks The council spells out the word ALBERTA with 7 cards. Determine the number of arrangements of all 7 letters in which the two A's are not beside each other, and explain your strategy.
Your working Compare with the model response
All arrangements: there are 7 letters with the A repeated twice, so 7!/2! = 5040/2 = 2520.
Arrangements with the A's together: glue them into one block, leaving 6 objects to arrange: 6! = 720. The two A's are identical, so the block has only one internal order.
Not together = all − together = 2520 − 720 = 1800 arrangements .
Strategy: counting the arrangements you do not want and subtracting is easier than building the arrangements you do want, because "not beside each other" would otherwise need several cases.
Check your work. Account for the identical letters. Another valid method arranges the five distinct letters in 5! ways and chooses two of their six gaps for the A cards, giving 5! × ₆C₂ = 1800. The question also asks you to explain , so the complement strategy has to be stated, not just used.
maddyhelps.