NCEA Level 2 Mathematics 91262 Calculus free practice questions with answers

Question 1 (9 marks) [Gradients, tangents and graphs]

(a) A curve has the equation y = x² − 6x − 3. Find the coordinates of the point on the curve where the gradient is −4.

(a) Answer: (1, −8)

• Evidence level: Achieved (differentiate, set the gradient equal to the given value, solve, then find y).
• Differentiate: dy/dx = 2x − 6.
• Set the gradient equal to −4: 2x − 6 = −4, so x = 1.
• Substitute into the curve: y = −8.
• The point is (1, −8).

Gradient at a point: differentiate to get dy/dx, set it equal to the required gradient to find x, then substitute back into the ORIGINAL equation for y.

(b) For the function f(x) = −x³ + 12x² − 45x, find the values of x for which the function is decreasing.

(b) Answer: x < 3 or x > 5

• Evidence level: Merit (use the sign of the derivative).
• f’(x) = −3x² + 24x − 45 = −3(x − 5)(x − 3).
• f’(x) = 0 at x = 3 and x = 5. This is a downward-opening parabola in f’, so f’(x) is negative outside these values and positive between them.
• The function is decreasing where f’(x) < 0, that is x < 3 or x > 5.

Increasing means f’(x) > 0 and decreasing means f’(x) < 0: find where f’ = 0, then use the shape of f’ to decide the sign on each interval.

(c) The curve y = 2x³ − 3x² − 4x − 1 has a tangent at the point where x = 1. Find the equation of this tangent, in the form y = mx + c.

(c) Answer: y = −4x − 2

• Evidence level: Merit (find the gradient with a derivative, then build the tangent line).
• Differentiate: dy/dx = 6x² − 6x − 4.
• At x = 1 the gradient is m = −4.
• The point on the curve is (1, −6) because y = −6.
• Use y − (−6) = −4(x − 1): y = −4x − 2.
• The tangent is y = −4x − 2.

A tangent has the same gradient as the curve at the point: use the derivative for m, the curve for the point, then y − y₁ = m(x − x₁).