Sketch the Graph of f(x) = sec²x

Question:

Sketch the graph of the following function :

\[ f(x)=\sec^2x \]

Solution:

We know that

\[ \sec^2x=\frac{1}{\cos^2x} \]

Since square of secant is always positive, the graph always lies above the x-axis.

Whenever

\[ \cos x=0 \]

the function becomes undefined.

Thus vertical asymptotes occur at

\[ x=\frac{\pi}{2}+n\pi \]

Important properties:

• Period \(=\pi\)
• Range \(y\ge1\)
• Vertical asymptotes at \(x=\frac{\pi}{2}+n\pi\)
• Minimum value \(=1\)

Now calculate some important points:

\[ \begin{aligned} x=0 &\Rightarrow y=\sec^20=1\\[8pt] x=\frac{\pi}{4} &\Rightarrow y=\sec^2\frac{\pi}{4}=2\\[8pt] x=-\frac{\pi}{4} &\Rightarrow y=\sec^2\left(-\frac{\pi}{4}\right)=2 \end{aligned} \]

Thus the graph passes through the points

\[ (0,1),\quad \left(\frac{\pi}{4},2\right),\quad \left(-\frac{\pi}{4},2\right) \]

Plot these points and draw smooth curves approaching the vertical asymptotes.

Hence, the required graph is shown above.

Graph Features:

• Period \(=\pi\)
• Range \(y\ge1\)
• Vertical asymptotes at \(x=\frac{\pi}{2}+n\pi\)
• The graph always lies above the x-axis
• The curve approaches infinity near the asymptotes