Step: 1

Make a table of values for ordered pairs of the form (x , - 2cotx ).

Step: 2

Plot the points and connect them with a smooth curve. Draw dashed vertical lines at the points where the function y = - 2cotx is not defined.

Step: 3

The dashed vertical lines are the asymptotes. That is the graph of y = - 2cotx gets closer and closer to the lines, but never meets them.

Step: 4

Notice that the - 2 cot x function repeats at intervals of π units or 180^{o} . Therefore, the period of y = - 2 cot x is π . The function increases without bound over each interval, so the amplitude is not defined.

Correct Answer is : Graph 2

Step: 1

Make a table of values for ordered pairs of the form (x , 2cot(x 2 )).

Step: 2

Plot the points and connect them with a smooth curve. Draw dashed vertical lines at the points where the function y = 2cot(x 2 ) is not defined.

Step: 3

Notice that the function repeats at intervals of 2π units . Therefore, the period of y = 2cot(x 2 ) is 2π . The function increases without bound over each interval, so the amplitude is not defined.

Correct Answer is : Graph 1

Step: 1

The graph of y = a cosec b θ takes values above a and below - a .

Step: 2

From the given function, we can identify the values of a and b as -5 and 1.

Step: 3

Set b θ = 0 and set b θ = 2π

Step: 4

We obtain θ = 0 and θ = 2π .

[b = 1.]

Step: 5

Make a table for θ = 0 to 2π .

Step: 6

The above table matches with graph 3.

Step: 7

So, the graph of y = - 5cosec θ matches with graph 3.

Correct Answer is : Graph 3

Step: 1

[Original function.]

Step: 2

It is clear that the centerline is 5 and amplitude is 2, this eliminates Graph 2.

[Since the centerline of Graph 2 is 6.]

Step: 3

From the argument (x + π 3 ), the graph attains highest point at - π 3 .

Step: 4

The phase angle of the cosine function 2 cos 3(x + π 3 ) + 5 equals the angle nearest the origin where the curve reaches the highest point above the centerline. This eliminates Graph 1.

Step: 5

The curve reaches the highest point at every half period. The period of this function is 2 π 3 and its half period is π 3 . This eliminates Graph 3.

Step: 6

As the half period of the function is - π 3 , label the other points left and right of - π 3

Step: 7

Indicate the location of the y - axis between - π 3 and π 3 to complete the graph.

Step: 8

Among the given graphs, Graph 4 represents the function y = 2 cos 3(x + π 3 ) + 5.

Correct Answer is : Graph 4

Step: 1

[Original function.]

Step: 2

It is clear that centerline is 5 and amplitude is 1. This eliminates Graph 2 and Graph 4.

[Centerline of Graph 2 is 4, and amplitude of Graph 4 is 2.]

Step: 3

From the argument (x - π 2 ) has a phase angle of π 2 , thus one crossing point is phase angle. This eliminates Graph 1.

[Phase angle of Graph 1 is taken as - π 2 .]

Step: 4

The curve will cross its centerline at every half period. The period for this function is π 2 . Half period is π 4 . so label the other crossing points π 4 apart to the left and right of + π 2 .

Step: 5

Indicate the location of y - axis between - π 2 and π 2 to complete the graph.

Step: 6

Among the given graphs, Graph 3 represents the function y = sin 4(x - π 2 ) + 5.

Correct Answer is : Graph 3

Step: 1

[Original function.]

Step: 2

It is clear that the centerline is + 5 and the amplitude is 1. This eliminates Graph 1.

[Since the center line of Graph 1 is + 6.]

Step: 3

Since the given function has negative sine function, so we can eliminate Graph 3.

Step: 4

From the argument (x - π 2 ) the phase angle is + π 2 . Thus one crossing point is the phase angle. This eliminates Graph 2.

Step: 5

The curve will cross its centerline every half period. The period for this function is π. Half period is π 2 . So label the other crossing points π 2 apart to the left and right of + π 2 .

Step: 6

Since 0° will lie between - π 2 and + π 2 , so indicate the location of y -axis in between - π 2 and + π 2 to complete the graph.

Step: 7

Thus Graph 4 represents the function y = 5 - sin 2(x - π 2 ) .

Correct Answer is : Graph 4

Step: 1

[Original function.]

Step: 2

It is clear that the centerline is + 4 and the amplitude is 1. This eliminates Graph 4.

[Since the center line of Graph 4 is + 3.5.]

Step: 3

Since the given function has positive sine function, so we can eliminate Graph 1.

Step: 4

From the argument (x - 5 π 1 2 ) the phase angle is + 5 π 1 2 . Thus one crossing point is the phase angle. This eliminates Graph 3.

Step: 5

The curve will cross its centerline every half period. The period for this function is π. Half period is π 2 . So label the other crossing points π 2 apart to the left and right of + 5 π 1 2 .

Step: 6

Since 0° will lie between - π 1 2 and + 5 π 1 2 , so indicate the location of y -axis in between - π 1 2 and + 5 π 1 2 to complete the graph.

Step: 7

Thus Graph 2 represents the function y = 4 + sin 2(x - 5 π 1 2 ) .

Correct Answer is : Graph 2

Step: 1

To get the equation of the graph first draw and identify the sinusoidal.

Step: 2

Step: 3

The centerline is at 2. The sinusoidal is y = 2 + 0.5 sin x .

Step: 4

So the graph is y = 2 + 0.5 csc x .

Correct Answer is : y = 2 + 0.5 csc x

Step: 1

To get the equation of the graph first draw and identify the sinusoidal.

Step: 2

Step: 3

The centerline is at 2. The sinusoidal is y = 2 + 0.5 cos x .

Step: 4

So the graph is y = 2 + 0.5 sec x .

Correct Answer is : y = 2 + 0.5 sec x

Step: 1

To get the equation of the graph first draw and identify the sinusoidal.

Step: 2

Step: 3

The centerline is at 2. The sinusoidal is y = 2 + 2 sin x .

Step: 4

So the graph is y = 2 + 2 csc x .

Correct Answer is : y = 2 + 2 csc x

- Determining a given Relation is a Function-Algebra1-Solved Examples
- Finding Values of a Function for a given Domain-Algebra1-Solved Examples
- Graphing Linear Functions for a Given Domain-Algebra1-Solved Examples
- Graphing Linear Functions-Algebra1-Solved Examples
- Finding the Rate of Change-Algebra1-Solved Examples
- Finding Intercepts of a Linear Equation-Algebra1-Solved Examples
- Graphing Piecewise and Step Functions-Algebra1-Solved Examples
- Graphing Absolute Value Functions-Algebra1-Solved Examples
- Finding Intercepts of Quadratic Functions-Algebra1-Solved Examples
- Graphing and Analyzing Quadratic Functions-Algebra1-Solved Examples
- Identifying and Graphing Square Root Functions-Algebra1-Solved Examples
- Comparing Properties of Two Functions-Algebra1-Solved Examples
- Finding Intervals for Increasing and Decreasing Functions-Algebra1-Solved Examples
- Representing Exponential Functions Using Tables or Graphs-Algebra1-Solved Examples
- Graphing Logarithmic Functions-Algebra1-Solved Examples
- Identifying and Using Recursive Formula-Algebra1-Solved Examples

- Surface Area