AS 91578 · 2025 paper · derivatives, normal, tangents

Question 1(d) Merit

Practises Tangents, normals and curve features

A chain-rule power, an oven cooling written in half-powers of t, a trig normal, a tangent to an exponential-plus-quadratic, and a parametric curve where inflection and stationary have to coincide.

The graph shows f(x) = x² + e^2x and the tangent to the curve when x = 1.

Find the point P, the x-intercept of this tangent.

The 2025 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 A tangent needs a point and a gradient. Which point on the curve is it?
f(1) = 1² + e² = 1 + e² ≈ 8.389, so the point is (1, 1 + e²).
2 What is the gradient there?
f'(x) = 2x + 2e^2x (the chain rule gives the 2 on e^2x), so f'(1) = 2 + 2e² ≈ 16.778.
3 Write the equation of the tangent.
y − (1 + e²) = (2 + 2e²)(x − 1)
which is y ≈ 16.778x − 8.389.
4 P is where the tangent crosses the x-axis. Put y = 0.
x − 1 = −(1 + e²)/(2 + 2e²) = −(1 + e²)/(2(1 + e²)) = −1/2
so x = 1/2. The 1 + e² cancels exactly.

Answer P is the point (0.5, 0).

What the marker wanted

Achieved The derivative f'(x) = 2x + 2e^2x, with its value at x = 1.
Merit P found at x = 0.5, from a correct derivative.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

Finding the gradient correctly and then not turning it into the line and where it meets the axis. NZQA’s 2025 report names this part, with Two (c), as one where many students were unsure how to join differentiation to coordinate geometry.

How did it go?

Got it moves it further out. Review brings it back tomorrow.