I need to state the period and 2 consecutive asymptotes on the graph for the following questions.

1: y = -3 tan pi*x
period: pi (?)
asymptotes: ?

2: y = 2 sec 4x
period: ?
asymptotes: ?

3: y = csc (x/3)
period: ?
asymptotes: ?

4: y = 3 cot (pi*x/2)
period: ?
asymptotes: ?

Hey,

I ca help you with the first one as this is the only one I understand so far. Since you have pi*x in the tan, you have to take the pi and divide by the k of the function, which in this case is pi. This gets you a perio of one, meaning that there is a point at every single interval. The aymptotes are in between each interval. Since you have to get half way between each quarter of pi, there is an asymptote at pi/8 and then add pi/4 to find the other asymptotes there of. A few I found are pi/8, which is the first one, and (3*pi)/8. Hope it helps. If I find out how to do the other three problems, I'll help! :D

F(x)=sec(4x)-1

To determine the period and asymptotes for each of the given equations, let's analyze each equation individually:

1: y = -3 tan pi*x
The general equation for tangent function is y = tan(x), but in this case, the function is multiplied by -3 and the x-value is multiplied by pi.
Period: The period of the tangent function is pi, so the period of this function is also pi.
Asymptotes: The asymptotes of the tangent function occur whenever the x-value makes the tangent function undefined. In this case, the asymptotes occur when x is equal to (2n + 1)/2, where n is an integer. So, the two consecutive asymptotes for this function would be x = 1/2 and x = 3/2.

2: y = 2 sec 4x
The general equation for secant function is y = sec(x), but in this case, the function is multiplied by 2 and the x-value is multiplied by 4.
Period: The period of the secant function is 2pi, but because the x-value is multiplied by 4, the period of this function is (2pi)/4 = pi/2.
Asymptotes: The asymptotes of the secant function occur whenever the x-value makes the secant function undefined. In this case, the asymptotes occur when x is equal to (2n + 1)/2, where n is an integer. So, the two consecutive asymptotes for this function would be x = 1/2 and x = 3/2.

3: y = csc (x/3)
The general equation for cosecant function is y = csc(x), but in this case, the x-value is divided by 3.
Period: The period of the cosecant function is 2pi, but because the x-value is divided by 3, the period of this function is 2pi * 3 = 6pi.
Asymptotes: The asymptotes of the cosecant function occur whenever the x-value makes the cosecant function undefined. In this case, the asymptotes occur when x is equal to n*pi, where n is an integer. So, the two consecutive asymptotes for this function would be x = 0 and x = pi.

4: y = 3 cot (pi*x/2)
The general equation for cotangent function is y = cot(x), but in this case, the function is multiplied by 3 and the x-value is multiplied by pi/2.
Period: The period of the cotangent function is pi, but because the x-value is multiplied by pi/2, the period of this function is pi * 2/pi = 2.
Asymptotes: The asymptotes of the cotangent function occur whenever the x-value makes the cotangent function undefined. In this case, the asymptotes occur when x is equal to n*pi, where n is an integer. So, the two consecutive asymptotes for this function would be x = 0 and x = pi.

To determine the period and asymptotes of the given trigonometric functions, let's analyze each function one by one.

1: y = -3 tan π*x
To find the period of the tangent function, you need to identify the coefficient in front of x, which in this case is π. The period (T) of the tangent function is given by T = π/|b|, where b is the coefficient of x. Therefore, the period of this function is π/π = 1.

The tangent function has asymptotes at every multiple of π/2. Since the coefficient in front of x is 1, the asymptotes occur at x = (n * π/2), where n is an integer.

So, for y = -3 tan π*x,
Period = 1
Asymptotes occur at x = π/2, x = -π/2, x = 3π/2, x = -3π/2, and so on.

2: y = 2 sec 4x
To find the period of the secant function, you divide the coefficient of x in the argument by 4. In this case, the coefficient is 4. The period (T) is given by T = 2π/|b|. Therefore, the period is 2π/4 = π/2.

The secant function has asymptotes at every multiple of π. Since the coefficient in front of x is 4, the asymptotes occur at x = (n * π/4), where n is an integer.

So, for y = 2 sec 4x,
Period = π/2
Asymptotes occur at x = π/4, x = -π/4, x = 3π/4, x = -3π/4, and so on.

3: y = csc (x/3)
The cosecant function is the reciprocal of the sine function. The period of the sine function is 2π, but here we have (x/3) as the argument, which means the period is multiplied by 3. So, the period of the csc function is (2π) * 3 = 6π.

The cosecant function has asymptotes at every multiple of π. Since (x/3) is the argument, the asymptotes occur at x = (n * π/3), where n is an integer.

So, for y = csc (x/3),
Period = 6π
Asymptotes occur at x = π/3, x = -π/3, x = 2π/3, x = -2π/3, and so on.

4: y = 3 cot (π*x/2)
The cotangent function has the same period as the tangent function, which is π/|b|. Here, the coefficient in front of x is π/2. So, the period is π/(π/2) = 2.

The cotangent function has asymptotes at every multiple of π. Since we have (π*x/2) as the argument, the asymptotes occur at x = (2n + 1) * π/2, where n is an integer.

So, for y = 3 cot (π*x/2),
Period = 2
Asymptotes occur at x = π/2, x = -π/2, x = 3π/2, x = -3π/2, and so on.

I hope this explanation helps you understand how to determine the period and asymptotes of trigonometric functions.