345 exam-style questions on Functions with answers and explanations. Six real samples below — the full set is free with an account.
A population follows the logistic model P(t) = 8000 / (1 + 15e^(-0.3t)). As t becomes very large, P(t) approaches
Answer: 8000
As t → ∞, e^(−0.3t) → 0, so P → 8000/1 = 8000. The numerator L is the carrying capacity; 500 = 8000/16 is the initial value, not the limit.
Data is linearised so that ln y = 2.3 + 0.5x. Writing the model as y = a e^(0.5x), the value of a is closest to
Answer: 9.97
From ln y = ln a + 0.5x, we get ln a = 2.3, so a = e^2.3 ≈ 9.97. The intercept of the linearised graph is ln a, not a itself.
The graph of y = f(x − 3) is obtained from the graph of y = f(x) by a translation of
Answer: 3 units right
Changes inside the bracket act horizontally and in the opposite direction to the sign: x − 3 shifts the graph 3 units right.
The vertex of the parabola y = x² − 6x + 11 is at
Answer: (3, 2)
Complete the square: y = (x − 3)² + 2, so the vertex is (3, 2). Check: x = −b/(2a) = 3 and y(3) = 9 − 18 + 11 = 2.
The horizontal asymptote of the graph of y = 5e⁻ˣ + 2 is
Answer: y = 2
As x → ∞, e⁻ˣ → 0, so y → 2. The “+2” shifts the natural asymptote y = 0 up to y = 2.
The gradient of the straight line through the points (1, 2) and (3, 10) is
Answer: 4
m = (10 − 2) / (3 − 1) = 8 / 2 = 4.
That was 6 of 345 Functions questions.
Practise all 345 freeA student calculates the gradient of the line through A(−1, 6) and B(3, −2) and writes m = (−2 − 6)/(−1 − 3) = −8/−4 = 2.
See the correction →FindAO2 · 2–6 marks
Obtain the answer and show the method; if the calculator does it, say what was entered.
StateAO1 · 1–2 marks
One word or one short sentence. No explanation, no working: the mark is for the fact itself.
ExplainAO3 · 1–3 marks
Give reasons or causes: say what happens AND why, linking each step with 'because' or 'so that'.
345 Functions questions — free with an account. Spaced repetition, streaks and full exam simulations included.
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