Sketching Graphs Revision Notes for VCE SSCE Mathematical Methods

Sketching Graphs

Understanding phase shifts

When we work with trigonometric functions, we often need to sketch graphs that have been shifted horizontally. These horizontal shifts are called phase shifts.

The general forms we work with are:

y = a \sin n(t \pm \varepsilon)

y = a \cos n(t \pm \varepsilon)

The term $\pm \varepsilon$ (epsilon) is called the phase and determines how far the graph shifts along the horizontal axis.

How phase shifts work

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Key principle: The phase shift moves the entire graph horizontally.

For $y = \sin(t - \varepsilon)$ : the graph shifts right by $\varepsilon$ units
For $y = \sin(t + \varepsilon)$ : the graph shifts left by $\varepsilon$ units

This might seem backwards at first! Remember that the sign inside the brackets is opposite to the direction of the shift.

Worked example: sketching with phase shifts

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Worked Example: Sketching Trigonometric Functions with Phase Shifts

Part a: Sketch $y = 3 \sin 2\left(t - \frac{\pi}{4}\right)$ for $\frac{\pi}{4} \leq t \leq \frac{5\pi}{4}$

First, identify the key features:

Range: $[-3, 3]$ (the amplitude is 3)
Period: $\frac{2\pi}{2} = \pi$
Phase shift: The graph of $y = 3\sin(2t)$ is shifted $\frac{\pi}{4}$ units to the right

Part b: Sketch $y = 2 \cos 3\left(t + \frac{\pi}{3}\right)$ for $-\frac{\pi}{3} \leq t \leq \frac{\pi}{3}$

Key features:

Range: $[-2, 2]$ (the amplitude is 2)
Period: $\frac{2\pi}{3}$
Phase shift: The graph of $y = 2\cos(3t)$ is shifted $\frac{\pi}{3}$ units to the left

Finding endpoints and key points

When sketching over a specific interval, it's important to calculate the exact y-values at the endpoints. This ensures your graph is accurate at the boundaries and helps you understand the behavior of the function over the given domain.

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Worked Example: Finding Endpoints and Sketching

For the function $f : [0, 2\pi] \to \mathbb{R}$ , $f(x) = \sin\left(x - \frac{\pi}{3}\right)$ :

Finding the endpoints:

f(0) = \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}

f(2\pi) = \sin\left(2\pi - \frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}

Sketching the graph:

The graph of $y = \sin\left(x - \frac{\pi}{3}\right)$ is the graph of $y = \sin x$ shifted $\frac{\pi}{3}$ units to the right.

Key features:

Period: $2\pi$
Amplitude: $1$
Endpoints: $\left(0, -\frac{\sqrt{3}}{2}\right)$ and $\left(2\pi, -\frac{\sqrt{3}}{2}\right)$

Helpful approach:

Start at $x = \frac{\pi}{3}$ (where the shifted sine curve would normally start at zero)
Each complete cycle has length $\pi$ for a half-loop
Mark key points at intervals of $\frac{\pi}{2}$ from the starting position

Vertical translations

In addition to horizontal shifts, we can also translate graphs vertically. This gives us the general forms:

y = a \sin n(t \pm \varepsilon) \pm b

y = a \cos n(t \pm \varepsilon) \pm b

The term $\pm b$ translates the graph vertically:

$+b$ shifts the graph up by $b$ units
$-b$ shifts the graph down by $b$ units

Effect on range

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When we add a vertical shift $b$ :

The new range becomes $[-a + b, a + b]$
The central axis of the wave moves from $y = 0$ to $y = b$

This means the entire graph shifts vertically, including both the maximum and minimum values.

Worked example: combined transformations

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Worked Example: Combining Horizontal and Vertical Shifts

Part a: Sketch $y = 3 \sin 2\left(t - \frac{\pi}{4}\right) + 2$ for $\frac{\pi}{4} \leq t \leq \frac{5\pi}{4}$

The graph of $y = 3\sin(2t)$ is:

Shifted $\frac{\pi}{4}$ units to the right
Shifted $2$ units up

The range is now $[-3 + 2, 3 + 2] = [-1, 5]$

Part b: Sketch $y = 2 \cos 3\left(t + \frac{\pi}{3}\right) - 1$ for $-\frac{\pi}{3} \leq t \leq \frac{\pi}{3}$

The graph of $y = 2\cos(3t)$ is:

Shifted $\frac{\pi}{3}$ units to the left
Shifted $1$ unit down

The range is now $[-2 - 1, 2 - 1] = [-3, 1]$

Finding axis intercepts

When sketching trigonometric graphs, finding where they cross the x-axis (the x-intercepts) helps us draw them accurately. These intercepts represent the points where the function equals zero, which are often key features in understanding the graph's behavior.

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Method: Set $y = 0$ and solve the equation for $x$ .

This gives you the exact x-coordinates where the graph crosses the horizontal axis.

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Worked Example: Determining x-Intercepts

Part a: Sketch $y = \sqrt{2} \sin(x) + 1$ for $x \in [0, 2\pi]$ and find the x-intercepts.

To find x-intercepts, solve $\sqrt{2} \sin(x) + 1 = 0$ :

\sqrt{2} \sin(x) + 1 = 0

\sin x = -\frac{1}{\sqrt{2}}

x = \pi + \frac{\pi}{4} \text{ or } x = 2\pi - \frac{\pi}{4}

x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}

The x-axis intercepts are at $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$ .

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Part b: Sketch $y = 2\cos(2x) - 1$ for $x \in [0, 2\pi]$ and find the x-intercepts.

To find x-intercepts, solve $2\cos(2x) - 1 = 0$ :

2\cos(2x) - 1 = 0

\cos(2x) = \frac{1}{2}

2x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}

x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}

The x-axis intercepts are at $\frac{\pi}{6}$ , $\frac{5\pi}{6}$ , $\frac{7\pi}{6}$ , and $\frac{11\pi}{6}$ .

Part c: Sketch $y = 2\sin 2\left(x - \frac{\pi}{3}\right) - \sqrt{3}$ for $x \in [0, 2\pi]$ and find the x-intercepts.

Set the equation equal to zero and solve:

\sin 2\left(x - \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}

2\left(x - \frac{\pi}{3}\right) = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}

x - \frac{\pi}{3} = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}

x = \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{2}, \frac{5\pi}{3}

The x-axis intercepts are at $\frac{\pi}{2}$ , $\frac{2\pi}{3}$ , $\frac{3\pi}{2}$ , and $\frac{5\pi}{3}$ .

Key formulas and terminology

Amplitude: The value $|a|$ determines how far the graph extends above and below its central axis.

Period: For functions of the form $y = a\sin(nt)$ or $y = a\cos(nt)$ , the period is $\frac{2\pi}{n}$ .

Phase shift: The value $\varepsilon$ in $y = a\sin n(t \pm \varepsilon)$ :

$(t - \varepsilon)$ : shift right by $\varepsilon$ units
$(t + \varepsilon)$ : shift left by $\varepsilon$ units

Vertical shift: The value $b$ in $y = a\sin(nt) \pm b$ shifts the graph up or down by $b$ units.

Range:

For $y = a\sin(nt)$ : the range is $[-|a|, |a|]$
For $y = a\sin(nt) + b$ : the range is $[-|a| + b, |a| + b]$

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Key Points to Remember:

Phase shifts move the graph horizontally: $(t - \varepsilon)$ shifts right, $(t + \varepsilon)$ shifts left
Vertical shifts move the graph up or down by the value of $b$
To find x-intercepts, set $y = 0$ and solve the resulting equation
The period of $y = a\sin(nt)$ or $y = a\cos(nt)$ is $\frac{2\pi}{n}$
Always calculate endpoint values when sketching over a specific interval to ensure accuracy

Sketching Graphs (VCE SSCE Mathematical Methods): Revision Notes