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<ol class="i-list"><li data-list-item-id="item-1769585930368-duptmo2q6" data-list-item-path="item-1769585930368-duptmo2q6"><div class="list-item-paragraph">The diagram shows the graph of $y=\mathrm{f}(x)$ which has a minimum point at $(0, 2)$ and asymptotes $x=2$ and $y=5.$</div><div class="list-item-paragraph">On separate diagrams, sketch the graphs of $y\ =\ \ \frac{1}{\mathrm{f}(x)}$.</div></li></ol>

The diagram shows the graph of $y=\mathrm{f}(x)$ which has a minimum point at $(0, 2)$ and asymptotes $x=2$ and $y=5.$

On separate diagrams, sketch the graphs of $y\ =\ \ \frac{1}{\mathrm{f}(x)}$.

Correction to the retained handwritten sketch: $f$ is undefined at $x=2$, so $1/f$ is also undefined there. The point $(2,0)$ is a hole, not an $x$-intercept. The original image is retained for reference; the typed sketch shows the exclusion.

Default solution

The horizontal asymptote $y=5$ of $y=\mathrm f(x)$ maps to the horizontal asymptote $y=\frac15$ of $y=\frac{1}{\mathrm f(x)}$.

The minimum point $(0,2)$ becomes the maximum point $\left(0,\frac12\right)$.

As $x\to2$, the original graph becomes unbounded at its vertical asymptote, so the reciprocal tends to $0$. The original zero at $x=3$ becomes the vertical asymptote $x=3$ of the reciprocal graph.

The reciprocal has the same sign as $\mathrm f(x)$ on every interval of its domain. These features determine the required sketch.

Reciprocation gives maximum $(0,1/2)$, vertical asymptote $x=3$, horizontal asymptote $y=1/5$, and an excluded point $(2,0)$. Since $f$ is undefined at $2$, this point is a hole, not an intercept.

2013 VJC P1 Q6

2013 VJC P1 Q6

Video Solution:

Default solution