Problem

Find the Asymptotes 5csc(1/2pix+1/6pi)

The problem at hand is asking for the identification of the asymptotes of the trigonometric function given, which is 5 times the cosecant of (1/2 pi x + 1/6 pi). An asymptote of a function is a line that the graph of the function approaches as the independent variable (in this case, 'x') either goes to infinity or negative infinity, or at certain finite points, but the function never actually reaches the asymptote. For trigonometric functions like cosecant, which is the reciprocal of the sine function, asymptotes typically occur at values of 'x' where the sine function is zero (since dividing by zero is undefined). The problem requires finding those values of 'x' and determining the vertical lines that the function approaches but does not intersect.

5csc(12πx+16π)

Answer

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Solution:

Step 1:

Eliminate the parentheses in the expression: y=5csc(12πx+16π).

Step 2:

Repeat the elimination of parentheses: y=5csc(12πx+16π).

Step 3:

Simplify each component in the expression.

Step 3.1:

Multiply 12 by πx to simplify.

Step 3.1.1:

Combine π and 12 to get y=5csc(π2x+16π).

Step 3.1.2:

Merge π2 with x to obtain y=5csc(πx2+16π).

Step 3.2:

Combine 16 and π to form y=5csc(πx2+π6).

Step 4:

Identify vertical asymptotes for the function y=csc(x), which occur at x=nπ, where n is an integer. To find the vertical asymptotes for y=5csc(πx2+π6), set the argument of the cosecant function to zero: πx2+π6=0.

Step 5:

Solve for x.

Step 5.1:

Subtract π6 from both sides: πx2=π6.

Step 5.2:

Multiply both sides by 2π: x=13.

Step 6:

Set the argument of the cosecant function equal to 2π: πx2+π6=2π.

Step 7:

Solve for x.

Step 7.1:

Isolate terms containing x: πx2=2ππ6.

Step 7.2:

Multiply both sides by 2π to find x=113.

Step 8:

The basic period for y=5csc(πx2+π6) is between the vertical asymptotes at x=13 and x=113.

Step 9:

Determine the period of the function to locate all vertical asymptotes, which occur every half period: 2π|π2|=4.

Step 10:

The vertical asymptotes for y=5csc(πx2+π6) occur at x=13+4n, where n is an integer.

Step 11:

The function y=5csc(πx2+π6) has no horizontal or oblique asymptotes, only vertical asymptotes at x=13+4n.

Step 12:

There is no further action required in this step.

Knowledge Notes:

The problem involves finding the asymptotes of the function y=5csc(12πx+16π). An asymptote is a line that the graph of a function approaches but never touches.

  • Vertical Asymptotes: These occur in trigonometric functions like cosecant (csc) and secant (sec) where the function is undefined. For the cosecant function, vertical asymptotes occur at values of x where sin(x)=0, since csc(x)=1sin(x).

  • Horizontal Asymptotes: These occur when the values of y approach a constant as x goes to infinity or negative infinity. The cosecant function does not have horizontal asymptotes because it has an infinite range.

  • Oblique Asymptotes: These are diagonal lines that the function approaches as x goes to infinity or negative infinity. The cosecant function does not have oblique asymptotes.

  • Period of Trigonometric Functions: The period of a function is the length of the smallest interval over which the function repeats itself. For y=csc(x), the period is 2π. For a transformed function like y=acsc(bx+c), the period is 2π|b|.

  • Solving Equations: To find the specific values of x for the vertical asymptotes, we set the argument of the cosecant function equal to the values where sin(x)=0 and solve for x. This involves algebraic manipulation, such as isolating terms and multiplying by reciprocals.

In this problem, we use the properties of the cosecant function and algebraic techniques to determine the locations of the vertical asymptotes. The function y=5csc(πx2+π6) has vertical asymptotes at x=13+4n, where n is an integer, and no horizontal or oblique asymptotes.

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