Graph Trigonometric Functions 4

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Question 1 of 4

1. Question
Graph the trigonometric function

$y = 4 cos 3 (x + \frac{π}{3})$ $y=4cos 3(x+pi/3)$
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General Form of a Cos Function with Horizontal Shift

$y =$ $y=$ $a$ $a$ $cos$ $\text(cos)$ $b$ $b$ $(x -$ $(x-$ $h$ $h$ $)$ $)$

Period Fomula

$P=\frac{2\pi}{\color{#00880A}{b}}$

First, identify the values of the function

$y$ $y$ $=$ $=$ $a$ $a$ $cos$ $\text(cos)$ $b$ $b$ $(x -$ $(x-$ $h$ $h$ $)$ $)$

$y$ $y$ $=$ $=$ $\color{#004ec4}{4}\;\text{cos}\;\color{#00880A}{3}(x+\color{#9a00c7}{\frac{\pi}{3}})$

$a$ $a$ $=$ $=$ $4$ $4$

$b$ $b$ $=$ $=$ $3$ $3$

$h$ $h$ $=$ $=$ $- \frac{π}{3}$ $-pi/3$

Next, solve for the period of the function

$P$ $=$ $=$ $\frac{2\pi}{\color{#00880A}{b}}$

$=$ $=$ $\frac{2\pi}{\color{#00880A}{3}}$ Substitute known values

To graph the $cos$ $cos$ curve, it is better to divide the value of its period into $4$ $4$ parts and have the curve meet the following conditions

Curve starts at the peak of the amplitude $(a = 4)$ $(a=4)$

Curve intercepts x-axis at 1st quarter

Curve reaches minimum amplitude at 2nd quarter

Curve intercepts x-axis again at 3rd quarter

Curve starts again at the period ( $P = \frac{2 π}{3}$ $P=(2pi)/3$ )

This will be the $cos$ $cos$ curve for $y = 4 cos 3 x$ $y=4\text(cos) 3x$ with a period of $\frac{2 π}{3}$ $(2pi)/3$

Shift the graph horizontally $h = - \frac{π}{3}$ $h=-pi/3$ units to the left. Since the labels are in intervals of $\frac{π}{6}$ $pi/6$ , we move the graph $2$ $2$ labels to the left
Question 2 of 4

2. Question
Graph the trigonometric function within the domain $0 \leq x \leq 2 π$ $0≤x≤2pi$

$y = - cos x + 1$ $y=-cos x+1$
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General Form of a Cos Function with Vertical Shift

$y =$ $y=$ $a$ $a$ $cos$ $\text(cos)$ $b$ $b$ $x +$ $x+$ $k$ $k$

Period Fomula

$P=\frac{2\pi}{\color{#00880A}{b}}$

First, identify the values of the function

$y$ $y$ $=$ $=$ $a$ $a$ $cos$ $\text(cos)$ $b$ $b$ $x +$ $x+$ $k$ $k$

$y$ $y$ $=$ $=$ $\color{#004ec4}{-}\;\text{cos}\;x+\color{#e65021}{1}$

$a$ $a$ $=$ $=$ $- 1$ $-1$

$b$ $b$ $=$ $=$ $1$ $1$

$k$ $k$ $=$ $=$ $1$ $1$

Next, solve for the period of the function

$P$ $=$ $=$ $\frac{2\pi}{\color{#00880A}{b}}$

$=$ $=$ $\frac{2\pi}{\color{#00880A}{1}}$ Substitute known values

$=$ $=$ $2 π$ $2pi$

To graph the $cos$ $cos$ curve, it is better to divide the value of its period into $4$ $4$ parts and have the curve meet the following conditions

Curve starts at the minimum amplitude $(a = - 1)$ $(a=-1)$

Curve intercepts x-axis at 1st quarter

Curve reaches peak of amplitude at 2nd quarter

Curve intercepts x-axis again at 3rd quarter

Curve starts again at the period ( $P = 2 π$ $P=2pi$ )

This will be the $cos$ $cos$ curve for $y = - cos x$ $y=-\text(cos) x$ with a period of $2 π$ $2pi$

Finally, move the curve up to match the recently plotted points
Question 3 of 4

3. Question
Graph the following trigonometric functions within the domain $0 \leq x \leq π$ $0≤x≤pi$

$y = cos 2 x$ $color(red)(y=cos2x)$

$y = \frac{x}{2}$ $color(blue)(y=x/2)$
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General Form of a Cos Function

$y =$ $y=$ $a$ $a$ $cos$ $\text(cos)$ $b$ $b$ $x$ $x$

Period Fomula

$P=\frac{2\pi}{\color{#00880A}{b}}$

Start by graphing the first function

First, identify the values of the function.

$y$ $y$ $=$ $=$ $a$ $a$ $sin$ $\text(sin)$ $b$ $b$ $x$ $x$

$y$ $y$ $=$ $=$ $\text{cos}\;2x$

$a$ $a$ $=$ $=$ $1$ $1$

$b$ $b$ $=$ $=$ $2$ $2$

Next, solve for the period of the function

$P$ $=$ $=$ $\frac{2\pi}{\color{#00880A}{b}}$

$=$ $=$ $\frac{2\pi}{\color{#00880A}{2}}$ Substitute known values

$=$ $=$ $π$ $pi$

Graph the $1$ $1$ st function, $y = cos 2 x$ $y=cos2x$ . It is better to divide the value of its period into $4$ $4$ parts and have the curve meet the following conditions:

Curve starts at $(0, 1)$ $(0,1)$

Curve intercepts x-axis at 1st quarter

Curve reaches minimum amplitude ( $a = 1$ $a=1$ ) at 2nd quarter

Curve intercepts x-axis at 3rd quarter

Curve starts again at the period ( $π, 1$ $pi,1$ )

Hence, this will be the curve for $y = cos 2 x$ $y=cos2x$ under the domain $0 \leq x \leq π$ $0≤x≤pi$

Now, graph the $2$ $2$ nd function

Convert the current labels in the graph into decimal form and insert whole numbers accordingly

Set up a grid of $x$ $x$ and $y$ $y$ values to help with the graphing

Substitute each $x$ $x$ value to function to solve for the corresponding $y$ $y$ value

$x$ $x$ $0$ $0$ $1$ $1$ $2$ $2$

$y$ $y$

$y$ $y$ $=$ $=$ $\frac{x}{2}$ $x/2$

$=$ $=$ $\frac{0}{2}$ $0/2$ Substitute $x = 0$ $x=0$

$=$ $=$ $0$ $0$

$x$ $x$ $0$ $0$ $1$ $1$ $2$ $2$

$y$ $y$ $0$ $0$

$y$ $y$ $=$ $=$ $\frac{x}{2}$ $x/2$

$=$ $=$ $\frac{1}{2}$ $1/2$ Substitute $x = 1$ $x=1$

$x$ $x$ $0$ $0$ $1$ $1$ $2$ $2$

$y$ $y$ $0$ $0$ $\frac{1}{2}$ $1/2$

$y$ $y$ $=$ $=$ $\frac{x}{2}$ $x/2$

$=$ $2/2$ Substitute $x=2$

$=$ $1$

$x$ $0$ $1$ $2$

$y$ $0$ $1/2$ $1$

Plot the $3$ points on the updated graph

Finally, connect the $3$ dots to form a line
Question 4 of 4

4. Question
Graph the following trigonometric functions within the domain $-2pi≤x≤2pi$

$y=2sinx$

$y=-2sinx$
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General Form of a Sin Function

$y=$ $a$ $\text(sin)$ $b$ $x$

Period Fomula

$P=\frac{2\pi}{\color{#00880A}{b}}$

First, identify the values of the function

$1$ st function:

$y$ $=$ $a$ $\text(sin)$ $b$ $x$

$y$ $=$ $\color{#004ec4}{2}\;\text{sin}\;x$

$a$ $=$ $2$

$b$ $=$ $1$

$2$ nd function:

$y$ $=$ $a$ $\text(sin)$ $b$ $x$

$y$ $=$ $\color{#004ec4}{-2}\;\text{sin}\;x$

$a$ $=$ $-2$

$b$ $=$ $1$

Next, solve for the period of the function

$1$ st Function:

$P_{\text{sin}}$ $=$ $\frac{2\pi}{\color{#00880A}{b}}$

$=$ $\frac{2\pi}{\color{#00880A}{1}}$ Substitute known values

$=$ $2pi$

$2$ st Function:

$P_{\text{sin}}$ $=$ $\frac{2\pi}{\color{#00880A}{b}}$

$=$ $\frac{2\pi}{\color{#00880A}{1}}$ Substitute known values

$=$ $2pi$

The period of both functions is $2pi$

Graph the $1$ st function, $y=2sinx$ . It is better to divide the value of its period into $4$ parts and have the curve meet the following conditions:

Curve starts at $(0,0)$

Curve reaches peak of amplitude ( $a=2$ ) at 1st quarter

Curve intercepts x-axis at 2nd quarter

Curve reaches minimum amplitude at 3rd quarter

Curve starts at x-axis again at the period ( $P=2pi$ )

Since the domain is $-2pi≤x≤2pi$ , copy this curve into the left side of the $y$ -axis

Hence, this will be the curve for $y=2\text(sin) x$ under the domain $-2pi≤x≤2pi$

Lastly, graph the $2$ nd function, $y=-2sinx$ . Again, divide the value of its period into $4$ parts and have the curve meet the following conditions:

Curve starts at $(0,0)$

Curve reaches peak of amplitude ( $a=-2$ ) at 1st quarter

Curve intercepts x-axis at 2nd quarter

Curve reaches minimum amplitude at 3rd quarter

Curve starts at x-axis again at the period ( $P=2pi$ )

Since the domain is $-2pi≤x≤2pi$ , copy this curve into the left side of the $y$ -axis

Therefore, this will be the curve for the functions $y=2\text(sin) x$ and $y=-2\text(sin) x$ under the domain $-2pi≤x≤2pi$

Graph Trigonometric Functions 4

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Quiz summary

Information

1. Question

Hint

Great Work!

General Form of a Cos Function with Horizontal Shift

Period Fomula

2. Question

Hint

Nice Job!

General Form of a Cos Function with Vertical Shift

Period Fomula

3. Question

Hint

Fantastic!

General Form of a Cos Function

Period Fomula

4. Question

Hint

Correct!

General Form of a Sin Function

Period Fomula

Quizzes

$y$ $y$	$=$ $=$	$a$ $a$ $cos$ $\text(cos)$ $b$ $b$ $(x -$ $(x-$ $h$ $h$ $)$ $)$
$y$ $y$	$=$ $=$	$\color{#004ec4}{4}\;\text{cos}\;\color{#00880A}{3}(x+\color{#9a00c7}{\frac{\pi}{3}})$

$a$ $a$	$=$ $=$	$4$ $4$
$b$ $b$	$=$ $=$	$3$ $3$

$h$ $h$	$=$ $=$	$- \frac{π}{3}$ $-pi/3$

$P$	$=$ $=$	$\frac{2\pi}{\color{#00880A}{b}}$

	$=$ $=$	$\frac{2\pi}{\color{#00880A}{3}}$	Substitute known values

$a$ $a$	$=$ $=$	$- 1$ $-1$
$b$ $b$	$=$ $=$	$1$ $1$
$k$ $k$	$=$ $=$	$1$ $1$

$y$ $y$	$=$ $=$	$a$ $a$ $sin$ $\text(sin)$ $b$ $b$ $x$ $x$
$y$ $y$	$=$ $=$	$\text{cos}\;2x$

$y$ $y$	$=$ $=$	$\frac{x}{2}$ $x/2$

	$=$ $=$	$\frac{0}{2}$ $0/2$	Substitute $x = 0$ $x=0$

	$=$ $=$	$0$ $0$

$x$ $x$	$0$ $0$	$1$ $1$	$2$ $2$
$y$ $y$	$0$ $0$	$\frac{1}{2}$ $1/2$

$y$	$=$	$a$ $\text(sin)$ $b$ $x$
$y$	$=$	$\color{#004ec4}{2}\;\text{sin}\;x$

$y$	$=$	$a$ $\text(sin)$ $b$ $x$
$y$	$=$	$\color{#004ec4}{-2}\;\text{sin}\;x$

$P_{\text{sin}}$	$=$	$\frac{2\pi}{\color{#00880A}{b}}$

	$=$	$\frac{2\pi}{\color{#00880A}{1}}$	Substitute known values

	$=$	$2pi$