Direct and inverse variation and reciprocal functions: modelling y = kx^n and y = k/x^n, finding the constant k, graphing y = k/x with its asymptotes and end behaviour, and solving variation problems in context (the power models and end behaviour are new in the 2024 Mathematics Advanced syllabus)

$F = \frac{k}{d^2}$. From $12 = \frac{k}{25}$, $k = 300$. Then $F = \frac{300}{9} = \frac{100}{3} \approx 33.3$ N.

Markers give one mark for the model, one for $k$ and one for the final value. Check direction: a smaller distance must give a larger force, which it does.

With $k = -4 < 0$ the branches lie in the second and fourth quadrants. The asymptotes are $x = 0$ and $y = 0$. At $x = 2$, $y = -2$, so the point $(2, -2)$ is on the lower right branch. At $x = -2$, $y = 2$, so $(-2, 2)$ is on the upper left branch.

Markers award one mark for the correct quadrants, one for the asymptotes and one for the two plotted points.

Write the model

Direct variation with the square gives $y = kx^2$.

Find $k$

Substituting $x = 4$ and $y = 48$: $48 = k \times 16$, so $k = 3$.

Use it

$y = 3x^2$, so when $x = 5$, $y = 3 \times 25 = 75$.

Marker's note: one mark for the model, one for $k$, one for the answer. Always state the equation with its constant before you substitute again.

Write the model

Inverse variation gives $y = \dfrac{k}{x}$.

Find $k$

$6 = \dfrac{k}{4}$, so $k = 24$.

Use it

$10 = \dfrac{24}{x}$, so $x = \dfrac{24}{10} = 2.4$.

Marker's note: one mark for the model, one for $k = 24$, one for solving. Check the answer makes sense: a larger $y$ must correspond to a smaller $x$.

(a) $y = \dfrac{k}{x^2}$. Substituting, $5 = \dfrac{k}{4}$, so $k = 20$ and $y = \dfrac{20}{x^2}$.

(b) $y = \dfrac{20}{100} = 0.2$.

(c) Replace $x$ by $2x$: $y = \dfrac{20}{(2x)^2} = \dfrac{20}{4x^2} = \dfrac{1}{4} \times \dfrac{20}{x^2}$, so $y$ is divided by $4$.

Marker's note: one mark for each part. Part (c) is a standard reasoning question: show it algebraically rather than by one example.

(a) $6 = \dfrac{k}{2}$, so $k = 12$ and $f(x) = \dfrac{12}{x}$.

(b) The asymptotes are the vertical line $x = 0$ (the $y$-axis, since $f$ is undefined there) and the horizontal line $y = 0$ (the $x$-axis). As $x \to \infty$, $f(x) \to 0$ from above (positive values). As $x \to -\infty$, $f(x) \to 0$ from below (negative values).

(c) $g$ is $\frac{3}{x}$ translated up by $2$, so the asymptotes are $x = 0$ and $y = 2$. For the $x$-intercept, set $\dfrac{3}{x} + 2 = 0$, so $\dfrac{3}{x} = -2$ and $x = -\dfrac{3}{2}$. There is no $y$-intercept because $x = 0$ is not in the domain.

Marker's note: one mark for $k$, one for both asymptotes, one for the end behaviour and one for part (c).

Model. $I = \dfrac{k}{d^2}$. Using $d = 3$ and $I = 80$: $80 = \dfrac{k}{9}$, so $k = 720$ and $I = \dfrac{720}{d^2}$.

(a) At $d = 6$, $I = \dfrac{720}{36} = 20$ units. Doubling the distance divided the intensity by $4$, as expected.

(b) $5 = \dfrac{720}{d^2}$, so $d^2 = 144$ and $d = 12$ m (the negative root is not a distance).

Marker's note: one mark for $k$, one for $I = \frac{720}{d^2}$, one for each part. Reject $d = -12$ and say why.

Model. $T = \dfrac{k}{n}$. With $n = 4$ and $T = 9$: $9 = \dfrac{k}{4}$, so $k = 36$ and $T = \dfrac{36}{n}$.

(a) $T = \dfrac{36}{6} = 6$ hours.

(b) $4 = \dfrac{36}{n}$, so $n = 9$ pumps.

Marker's note: one mark for the model, one for $k = 36$, one for each part. The constant $36$ is the total pump-hours of work, a useful interpretation.

Find $k$. $2 = k\sqrt{1}$, so $k = 2$ and $T = 2\sqrt{L}$.

Solve. $3 = 2\sqrt{L}$, so $\sqrt{L} = \dfrac{3}{2}$ and $L = \dfrac{9}{4} = 2.25$ m.

Marker's note: one mark for $k$, one for isolating $\sqrt{L}$, one for squaring to get $L$. Direct variation covers fractional powers too, as $y = kx^{1/2}$.