Written response
Three sample questions in the exam's format — one per topic, each with a 2-mark part and a 3-mark part, for 5 marks. Work the question first, then open the model response and compare, paying attention to the parts that earn the marks.
Written by us in the diploma format, not copied from an exam. Alberta Education publishes real examples with scoring rationales in the documents linked on the diploma prep page.
How these are scored
The three written-response questions together are 25% of your exam mark. Each has a 2-mark part and a 3-mark part that are scored separately and added, for 5 marks a question. Markers are judging four things: that you understood the problem, that you applied the right mathematics, that your strategy and reasoning are visible, and that you communicated the solution clearly. Half marks (0.5, 1.5, 2.5) exist for responses between the benchmarks — a complete method with one slip still earns most of the marks, and a bare answer with no work earns very few.
Written Response 1
Relations and Functions · 5 marks
The polynomial function f(x) = x³ − 4x² + x + 6 is transformed to g(x) = −2f(x − 1).
a) Algebraically determine the zeros of f(x). 2 marks
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.
What earns the marks: Full marks need the algebra: a zero found by the factor theorem, the division or an equivalent factoring step, the factors set equal to zero, and all three zeros stated. Reading the zeros off a graphing calculator answers a question that was not asked — algebraically rules that out.
b) Describe the transformations that map the graph of f(x) onto the graph of g(x), and determine the zeros of g(x). 3 marks
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. ✓
What earns the marks: 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 · 5 marks
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.
a) Determine an equation for h as a function of t. 2 marks
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. ✓
What earns the marks: 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.
b) Algebraically determine the first time the rider is 12 m above the ground, to the nearest tenth of a second, and interpret your answer. 3 marks
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) = 1.8882… (radians, second quadrant).
t = 20(1.8882…)/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.
What earns the marks: 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 · 5 marks
A student council has 7 women and 6 men.
a) Determine the number of different 5-person committees that can be formed with at least 3 women. 2 marks
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.
What earns the marks: 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.
b) 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. 3 marks
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.
What earns the marks: Dividing by 2! for the repeated A in the total, and not multiplying the block by 2! because the A's are identical, is the part markers see missed most often. The question also asks you to explain, so the complement strategy has to be stated, not just used.
Directing words
The bolded word in a question tells you what kind of answer is being marked. These are Alberta Education's definitions, in short form.
| Word | What it asks for |
|---|---|
| Algebraically | Work it out with variables and symbols — not by reading a graph or guessing and checking. |
| Determine | Find the answer and show the formula, procedure or calculation that got you there, to the accuracy asked for. |
| Describe | Write an account of the concept in words — for transformations, name each one properly. |
| Explain | Say why. Give the reason or cause, not just the result. |
| Justify | Give the reasons or evidence that support a conclusion you have already stated. |
| Verify | Show a statement holds for one particular case, usually by substituting. |
| Prove | Show a statement is true in general, by argument — not by trying values. |
| Sketch | Draw the key features: labelled axes with a scale, intercepts, vertices, endpoints, asymptotes, correct shape and end behaviour. |
| Interpret | Say what the number means back in the situation the question described. |
Watch out: the most common lost marks have nothing to do with hard mathematics — missing units, a transformation described loosely instead of named, an expression written where an equation belongs, brackets left out, a solution never related back to the situation, or an answer rounded partway through.