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

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.

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.

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.

Make a table of values for ordered pairs of the form (x, 2cot(x2)).

Plot the points and connect them with a smooth curve. Draw dashed vertical lines at the points where the function y = 2cot(x2) is not defined.

Notice that the function repeats at intervals of 2π units . Therefore, the period of y = 2cot(x2) is 2π. The function increases without bound over each interval, so the amplitude is not defined.

The graph of y = acosec bθ takes values above a and below - a.

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

Set bθ = 0 and set bθ = 2π

We obtain θ = 0 and θ = 2π.

Make a table for θ = 0 to 2π.

The above table matches with graph 3.

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

y = 2 cos 3(x + π3) + 5

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

From the argument (x + π3), the graph attains highest point at - π3.

The phase angle of the cosine function 2 cos 3(x + π3) + 5 equals the angle nearest the origin where the curve reaches the highest point above the centerline. This eliminates Graph 1.

The curve reaches the highest point at every half period. The period of this function is 2π3 and its half period is π3. This eliminates Graph 3.

As the half period of the function is - π3, label the other points left and right of - π3

Indicate the location of the y - axis between - π3 and π3 to complete the graph.

Among the given graphs, Graph 4 represents the function y = 2 cos 3(x + π3) + 5.

y = sin 4(x - π2) + 5.

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

From the argument (x - π2) has a phase angle of π2, thus one crossing point is phase angle. This eliminates Graph 1.

The curve will cross its centerline at every half period. The period for this function is π2. Half period is π4. so label the other crossing points π4 apart to the left and right of + π2.

Indicate the location of y - axis between - π2 and π2 to complete the graph.

Among the given graphs, Graph 3 represents the function y = sin 4(x - π2) + 5.

y = 5 - sin 2(x - π2)

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

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

From the argument (x - π2) the phase angle is + π2. Thus one crossing point is the phase angle. This eliminates Graph 2.

The curve will cross its centerline every half period. The period for this function is π. Half period is π2. So label the other crossing points π2 apart to the left and right of + π2.

Since 0° will lie between - π2 and + π2, so indicate the location of y-axis in between - π2 and + π2 to complete the graph.

Thus Graph 4 represents the function y = 5 - sin 2(x - π2).

y = 4 + sin 2(x - 5π12)

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

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

From the argument (x - 5π12) the phase angle is + 5π12. Thus one crossing point is the phase angle. This eliminates Graph 3.

The curve will cross its centerline every half period. The period for this function is π. Half period is π2. So label the other crossing points π2 apart to the left and right of + 5π12.

Since 0° will lie between - π12 and + 5π12, so indicate the location of y-axis in between - π12 and + 5π12 to complete the graph.

Thus Graph 2 represents the function y = 4 + sin 2(x - 5π12).

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

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

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

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

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

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

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

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

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