Question 1: Estimate the gradient of the curve y=x^2 at the point where x=2 using the change in y over change in x between x=1.5 and x=2.5.

Answer: Calculate (y2 - y1)/(x2 - x1)

Question 2: On the same curve y=x^2, estimate the gradient at x=3 using points at x=2.5 and x=3.5.

Answer: Calculate (y2 - y1)/(x2 - x1)

Question 3: Construct a tangent line to the curve y=2x+1 at x=4 and estimate its gradient.

Answer: Use the change in y over change in x around x=4

Question 4: Estimate the gradient of y=5x at x=0 by choosing points at x=-1 and x=1.

Answer: Calculate (y2 - y1)/(x2 - x1)

Question 1: The curve y=x^2 is increasing everywhere. Explain how estimating the gradient between two points can help us understand the slope at a specific point.

Answer: Students should describe the concept of secant lines approximating the tangent and how smaller intervals give better estimates.

Question 2: Given the points (2,4) and (3,9) on y=x^2, estimate the gradient at x=2.5. Then, explain why this estimate is close to the actual gradient.

Answer: Calculate (9-4)/(3-2)=5; explanation involves the curve's increasing nature and the proximity of points.

Question 3: Identify and correct the mistake in the following estimate: using points at x=1 and x=3 on y=x^2 to estimate the gradient at x=2.

Answer: Incorrect: (y at 3 - y at 1)/(3 - 1) = (9-1)/2=4; it's an approximation, but the mistake may be assuming it's the exact gradient.

Question 1: A car accelerates along a straight road, and its distance s(t) in meters after t seconds is given by s(t)=4t^2. Estimate the instantaneous rate of change of distance at t=3 seconds using points at t=2.5 and t=3.5.

Answer: Calculate (s(3.5)-s(2.5))/(3.5-2.5)

Question 2: A cyclist rides along a hill with height h(t)=3t+2. Estimate the gradient of the hill at t=4 seconds using points at t=3.5 and t=4.5.

Answer: Calculate (h(4.5)-h(3.5))/(4.5-3.5)

Question 1: Construct a curve y=√x on the grid. Estimate the gradient at x=4 by choosing points at x=3.5 and x=4.5. Comment on the accuracy of your estimate.

Answer: Calculate (√4.5 - √3.5)/(4.5-3.5) and compare with the derivative at x=4

Question 2: If the true gradient of y=ln(x) at x=2 is approximately 0.5, estimate it using points at x=1.8 and x=2.2, then discuss the error made.

Answer: Calculate (ln(2.2)-ln(1.8))/0.4; compare with 0.5 and analyze the error.

Question 1: A student estimates the gradient of y=x^2 at x=2 by using points at x=1 and x=3, and calculates (9-4)/2=2.5. What is the mistake here?

Answer: Incorrect: the change in x is 2, so the division should be by 2, but the calculation was wrong; also, this is an estimate, not the exact gradient.

Question 2: A student claims that for y=x^2, the gradient at x=2 is exactly 4. Explain the mistake.

Answer: Misunderstanding: the gradient at x=2 is actually 4, but this is the exact derivative, not an estimate.