AS 91578 · 2024 paper · sec, graph, cone, triangle area

Question 3(c) Merit

Practises Tangents, normals and curve features

A product with sec, three readings off a drawn graph (continuity, stationary points, a limit), stationary points of a rational function, a related-rates cone, and an optimisation that has to be proved to a given exact value.

Find the x-value(s) of any stationary point(s) on the graph of f(x) = (x² − 5x + 4)/(x² + 5x + 4). You do not need to determine the nature of any stationary point(s).

The 2024 paper, page 11 ↗ Marking schedule, page 5 ↗

Work it out

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

1 Quotient rule. What are u′ and v′?
u = x² − 5x + 4, u' = 2x − 5
v = x² + 5x + 4, v' = 2x + 5
2 Write out the numerator, u'v − uv', and expand it.
(2x − 5)(x² + 5x + 4) − (x² − 5x + 4)(2x + 5)
= (2x³ + 5x² − 17x − 20) − (2x³ − 5x² − 17x + 20) = 10x² − 40
so f'(x) = (10x² − 40)/(x² + 5x + 4)².
3 Set it to zero. How many solutions are there?
10x² − 40 = 0, so x² = 4 and x = 2 or x = −2. Neither makes the denominator zero (that happens at −1 and −4).

Answer x = 2 and x = −2.

What the marker wanted

Achieved A correct f′(x).
Merit Both x-values, worked from the calculus.

In our words, from NZQA’s 2024 marking schedule.

Common mistake

Giving only x = 2. NZQA’s 2024 report lists taking only one of the two square roots among the reasons for Not Achieved, and here both are valid.

How did it go?

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