Transforming Reciprocal Trig Graphs | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Find where the whole input reaches a key angle

In $y=1+2\sec(2x)$ , the input to secant is $2x$ . Its period is halved because increasing $x$ by $180^\circ$ increases the input by one full turn. Multiplication by 2 doubles each output; adding 1 moves the result up by 1.

Worked example

Sketch y = 1 + 2 sec(2x) for 0° ≤ x ≤ 180°.

Breaks occur when the denominator $\cos(2x)$ is zero.

\begin{gathered}2x=90^\circ,\ 270^\circ\\x=45^\circ,\ 135^\circ\end{gathered}

At $x=0^\circ,90^\circ,180^\circ$ , cosine is respectively 1, −1, 1, giving $y=3,-1,3$ .

Red dashed lines are excluded inputs. The curve continues beyond the window.

Transform both parts of the secant range: $\sec(2x)\ge1$ gives $y\ge3$ , while $\sec(2x)\le-1$ gives $y\le-1$ . The function never takes an output between −1 and 3.

Your turnPractice

For $y=2\sec(x-30^\circ)$ on $0^\circ\le x\le360^\circ$ , find the asymptotes, period and range.

Show worked answer

x-30^\circ=90^\circ,\ 270^\circ

Answer

\begin{gathered}x=120^\circ,\ 300^\circ\\ \text{period }360^\circ\\y\le-2\ \text{or}\ y\ge2\end{gathered}

Transform the zeros as well as the breaks

For $y=\cot(2x)$ , a break needs $\sin(2x)=0$ ; a zero needs $\cos(2x)=0$ . On $0^\circ\le x\le180^\circ$ , the breaks are $0^\circ,90^\circ,180^\circ$ and the zeros are $45^\circ,135^\circ$ .

Its period is $180^\circ/2=90^\circ$ . Each branch decreases from large positive to large negative outputs. A negative multiplier would reflect those outputs in the horizontal axis.

Your turnPractice

Find the period, breaks and zeros of $y=1-\sec x$ on $0\le x\le2\pi$ . Does the subtraction move its vertical asymptotes?

Show worked answer

The period is $2\pi$ ; breaks stay at $\pi/2,3\pi/2$ . Zeros require $\sec x=1$ , giving $x=0,2\pi$ . Changing outputs does not change where the original denominator is zero.

Find where the whole input reaches a key angle

Worked example

Sketch y = 1 + 2 sec(2x) for 0° ≤ x ≤ 180°.

Breaks occur when the denominator $\cos(2x)$ is zero.

\begin{gathered}2x=90^\circ,\ 270^\circ\\x=45^\circ,\ 135^\circ\end{gathered}

At $x=0^\circ,90^\circ,180^\circ$ , cosine is respectively 1, −1, 1, giving $y=3,-1,3$ .

Red dashed lines are excluded inputs. The curve continues beyond the window.

Transform both parts of the secant range: $\sec(2x)\ge1$ gives $y\ge3$ , while $\sec(2x)\le-1$ gives $y\le-1$ . The function never takes an output between −1 and 3.

Your turnPractice

For $y=2\sec(x-30^\circ)$ on $0^\circ\le x\le360^\circ$ , find the asymptotes, period and range.

Show worked answer

x-30^\circ=90^\circ,\ 270^\circ

Answer

\begin{gathered}x=120^\circ,\ 300^\circ\\ \text{period }360^\circ\\y\le-2\ \text{or}\ y\ge2\end{gathered}

Transform the zeros as well as the breaks

For $y=\cot(2x)$ , a break needs $\sin(2x)=0$ ; a zero needs $\cos(2x)=0$ . On $0^\circ\le x\le180^\circ$ , the breaks are $0^\circ,90^\circ,180^\circ$ and the zeros are $45^\circ,135^\circ$ .

Its period is $180^\circ/2=90^\circ$ . Each branch decreases from large positive to large negative outputs. A negative multiplier would reflect those outputs in the horizontal axis.

Your turnPractice

Find the period, breaks and zeros of $y=1-\sec x$ on $0\le x\le2\pi$ . Does the subtraction move its vertical asymptotes?

Show worked answer

The period is $2\pi$ ; breaks stay at $\pi/2,3\pi/2$ . Zeros require $\sec x=1$ , giving $x=0,2\pi$ . Changing outputs does not change where the original denominator is zero.