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Question 1 of 4
Graph the trigonometric function
y=4cos3(x+π3)
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First, identify the values of the function
y |
= |
a cos b(x-h) |
y |
= |
4cos3(x+π3) |
Next, solve for the period of the function
P |
= |
2πb |
|
|
= |
2π3 |
Substitute known values |
To graph the cos curve, it is better to divide the value of its period into 4 parts and have the curve meet the following conditions
Curve starts at the peak of the amplitude (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=2π3)
This will be the cos curve for y=4cos 3x with a period of 2π3
Shift the graph horizontally h=-π3 units to the left. Since the labels are in intervals of π6, we move the graph 2 labels to the left
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Question 2 of 4
Graph the trigonometric function within the domain 0≤x≤2π
y=-cosx+1
Incorrect
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First, identify the values of the function
y |
= |
a cos bx+k |
y |
= |
−cosx+1 |
Next, solve for the period of the function
P |
= |
2πb |
|
|
= |
2π1 |
Substitute known values |
|
|
= |
2π |
To graph the cos curve, it is better to divide the value of its period into 4 parts and have the curve meet the following conditions
Curve starts at the minimum amplitude (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=2π)
This will be the cos curve for y=-cos x with a period of 2π
Finally, move the curve up to match the recently plotted points
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Question 3 of 4
Graph the following trigonometric functions within the domain 0≤x≤π
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Start by graphing the first function
First, identify the values of the function.
Next, solve for the period of the function
P |
= |
2πb |
|
|
= |
2π2 |
Substitute known values |
|
|
= |
π |
Graph the 1st function, y=cos2x. It is better to divide the value of its period into 4 parts and have the curve meet the following conditions:
Curve intercepts x-axis at 1st quarter
Curve reaches minimum amplitude (a=1) at 2nd quarter
Curve intercepts x-axis at 3rd quarter
Curve starts again at the period (π,1)
Hence, this will be the curve for y=cos2x under the domain 0≤x≤π
Now, graph the 2nd function
Convert the current labels in the graph into decimal form and insert whole numbers accordingly
Set up a grid of x and y values to help with the graphing
Substitute each x value to function to solve for the corresponding y value
y |
= |
x2 |
|
|
= |
02 |
Substitute x=0 |
|
|
= |
0 |
y |
= |
x2 |
|
|
= |
12 |
Substitute x=1 |
y |
= |
x2 |
|
|
= |
22 |
Substitute x=2 |
|
|
= |
1 |
Plot the 3 points on the updated graph
Finally, connect the 3 dots to form a line
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Question 4 of 4
Graph the following trigonometric functions within the domain -2π≤x≤2π
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First, identify the values of the function
Next, solve for the period of the function
Psin |
= |
2πb |
|
|
= |
2π1 |
Substitute known values |
|
|
= |
2π |
Psin |
= |
2πb |
|
|
= |
2π1 |
Substitute known values |
|
|
= |
2π |
The period of both functions is 2π
Graph the 1st 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 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=2π)
Since the domain is -2π≤x≤2π, copy this curve into the left side of the y-axis
Hence, this will be the curve for y=2sin x under the domain -2π≤x≤2π
Lastly, graph the 2nd function, y=-2sinx. Again, divide the value of its period into 4 parts and have the curve meet the following conditions:
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=2π)
Since the domain is -2π≤x≤2π, copy this curve into the left side of the y-axis
Therefore, this will be the curve for the functions y=2sin x and y=-2sin x under the domain -2π≤x≤2π