How to solve inverse proportion questions and graph asymptotes

Expert intro: These questions mix inverse proportion with reading a rational graph. The key is to keep the constant of proportionality fixed, then use that same idea to reason about the graph’s shape and asymptotes.

Detailed walkthrough

(a) Inverse proportion

If $V$ is inversely proportional to $t$, then

$V\propto \frac{1}{t} \quad \text{so} \quad Vt=k$

Using $t=3.6$ and $V=10$:

$k=3.6\times 10=36$

So the relationship is

$V=\frac{36}{t}$

(i) How does $V$ change as $t$ increases?

As $t$ increases, $V$ decreases.

That is the defining feature of inverse proportion: one variable goes up while the other comes down.

(ii) Determine when $V=3$

Substitute $V=3$ into $Vt=36$:

$3t=36$

$t=12$

So $V=3$ when $t=12$.

(b) Graph of $y=\frac{a}{x+2}$

For a graph of the form

$y=\frac{a}{x+2}$

the vertical asymptote is

$x=-2$

and the horizontal asymptote is

$y=0$

If one point on the graph is shown, use it to substitute into the equation and find $a$. After that, the graph should be sketched as a rectangular hyperbola approaching both asymptotes.

If the curve is above the $x$-axis for values to the right of $x=-2$, then $a>0$. If it lies below, then $a<0$.

💡 Pitfall guide

A common mistake is to write that $V$ is directly proportional to $t$ just because one value gets smaller. In inverse proportion, you must state $Vt=k$ or $V=\frac{k}{t}$.

For the graph, don’t forget both asymptotes. Students often draw only the curve and miss the dashed lines at $x=-2$ and $y=0$.

🔄 Real-world variant

If the given pair were different, the method stays the same: find $k$ using $Vt=k$, then solve for the missing variable.

For the graph, if a point such as $(0,4)$ were given, then

$4=\frac{a}{0+2}$

so $a=8$.

If the graph had been $y=\frac{a}{x-h}$ instead, the vertical asymptote would shift to $x=h$ rather than $x=-2$.

🔍 Related terms

inverse proportion, constant of proportionality, asymptote