y is inversely proportional to the square of x, $x > 0$.

1. Express the new value of $x$ after a $25\%$ increase and relate it to $y$.

If $x$ is increased by $25\%$, then
$x_{\text{new}}=1.25x=\frac{5}{4}x.$
Since $y$ is inversely proportional to the square of $x$,
$y \propto \frac{1}{x^2}.$

2. Find the percentage change in $y$.

So
$\frac{y_{\text{new}}}{y_{\text{old}}}=\frac{1}{\left(\frac{5}{4}\right)^2}=\frac{1}{\frac{25}{16}}=\frac{16}{25}=0.64.$
Thus the new value of $y$ is $64\%$ of the old value.

Percentage change in $y$:
$64\%-100\%=-36\%.$
So $y$ decreases by $36\%$.

1. Write the inverse proportion relation using a constant of proportionality.

Since $y$ is inversely proportional to the square of $x$,
$y=\frac{k}{x^2}$
where $k$ is a constant.

2. Substitute the given values to find the constant and state the equation.

Given $y=16$ when $x=\frac12$,
$16=\frac{k}{\left(\frac12\right)^2}$
$16=\frac{k}{\frac14}=4k$
$k=4$
Hence,
$y=\frac{4}{x^2}.$

1. Substitute $y=100$ into the equation connecting $y$ and $x$.

From part (b)(i),
$y=\frac{4}{x^2}.$
When $y=100$,
$100=\frac{4}{x^2}.$

2. Solve the equation for $x$, using $x>0$.

$100x^2=4$
$x^2=\frac{4}{100}=\frac{1}{25}$
Since $x>0$,
$x=\frac{1}{5}.$

1. Identify the equation and the permitted region of the graph.

From part (b)(i),
$y=\frac{4}{x^2}.$
Since $x>0$ and $\frac{4}{x^2}>0$, the graph lies only in the first quadrant.

2. Sketch the shape and asymptotic behaviour of the curve.

Sketch a smooth decreasing curve for $x>0$:

• as $x\to 0^+$, $y\to \infty$, so the curve approaches the $y$-axis but does not touch it;
• as $x\to \infty$, $y\to 0$, so the curve approaches the $x$-axis but does not touch it.

The curve should be in the first quadrant only.