Shape and features of graph
$$ \text{General equation: } y = a \tan bx $$
Shape and features when a > 0 (click to show):
Note: The graph cannot pass through the vertical asymptote $x = {90^\circ \over b}$
\begin{align*}
\text{Period} & = {180^\circ \over b} = {\pi \text{ radians} \over b}
\end{align*}
Shape and features when a < 0 (click to show):
Note: The graph cannot pass through the vertical asymptote $x = {90^\circ \over b}$
\begin{align*}
\text{Period} & = {180^\circ \over b} = {\pi \text{ radians} \over b}
\end{align*}
Questions
Deduce equation from graph
Q1. The diagram below shows the graph of $y = \tan ax$.
Determine the value of $a$.
Answer: $ a = {3 \over 2} $
Solutions
\begin{align*}
\text{Period} & = {2\pi \over 3} \\
\\
{\pi \over a} & = {2\pi \over 3} \\
3\pi & = 2a \pi \phantom{000000} [\text{Cross-multiply}] \\
{3\pi \over 2\pi} & = a \\
{3 \over 2} & = a
\end{align*}
Sketch question
Q2. Sketch the graph of $y = - \tan 2x$, for $0^\circ \le x \le 360^\circ$.
Solutions
\begin{align}
y & = - \tan 2x \phantom{000000} [\text{Inverted shape since } a < 0] \\
\\
\text{Period} & = {180^\circ \over 2} \\
& = 90^\circ \\
\\
\text{No. of cycles} & = {360^\circ \over 90^\circ} \\
& = 4
\end{align}
Past year O level question: 2011 P2 Question 2
Note: This is the only question on tangent graph (from 2002 to 2023)
Q3. The function $f$ is given by $f(x) = a \tan bx$, where $a$ and $b$ are positive integers and $ -{\pi \over 2} \le x \le {\pi \over 2}$.
(i) Given that $f(x) = 0$ when $x = {\pi \over 2}$, find the smallest possible value of $b$.
[1]
Answer: $ b = 2 $
Solutions
\begin{align*}
f(x) & = a \tan bx \\
f\left({\pi \over 2}\right) & = a \tan \left[b \left({\pi \over 2}\right)\right] \\
0 & = a \tan {{\pi \over 2}b} \\
{0 \over a} & = \tan {{\pi \over 2}b} \\
0 & = \tan {{\pi \over 2}b} \\
\\
\tan x & = 0 \text{ when } x = \pi, 2\pi, 3 \pi, ... \\
\\
\therefore {\pi \over 2}b & = \pi, 2\pi, 3 \pi, ... \\
b & = 2, 4, 6, ... \\
\\
\text{Smallest value of } b & = 2
\end{align*}
(ii) Using the value of $b$ found in part (i) and given that the gradient of the graph of $y = f(x)$ is $12$ at the point where $x = {\pi \over 8}$, find the value of $a$.
(Note: This part is about differentiation)
[3]
Answer: $ a = 3 $
Solutions
\begin{align*}
f(x) & = a \tan bx \\
& = a \tan 2x \\
\\
f'(x) & = a (2)(\sec^2 2x) \phantom{000000} \left[ {d \over dx}[ \tan f(x) ] = f'(x) \sec^2 f(x) \right] \\
& = 2a \sec^2 2x \\
\\
f'\left({\pi \over 8}\right) & = 2a \sec^2 \left[2 \left({\pi \over 8}\right) \right] \\
12 & = 2a \sec^2 {\pi \over 4} \\
12 & = {2a \over \cos^2 {\pi \over 4}} \\
12 & = {2a \over {1 \over 2}} \\
{1 \over 2}(12) & = 2a \\
6 & = 2a \\
{6 \over 2} & = a \\
3 & = a
\end{align*}
(iii) Sketch the graph of $y = f(x)$.
[3]
Solutions
\begin{align*}
y & = f(x) \\
y & = a \tan bx \\
y & = 3 \tan 2x \\
\\
\text{Period} & = {\pi \over 2}
\end{align*}
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