Describe the graph of the function in terms of a basic trigonometric function. Locate the vertical asymptotes and graph two periods of the function. ​h(x)=sec12x

Question

Describe the graph of the function in terms of a basic trigonometric function. Locate the vertical asymptotes and graph two periods of the function. ​h(x)=sec12x

Use transformations to describe how the graph of the given function is obtained from the graph of a basic trigonometric function. Select the correct choice below and fill in the answer box to complete your choice. (Type integers or simplified​ fractions.)

A.To obtain the graph of the given​ function, shrink the graph of the basic function horizontally by a factor of

__.

B. To obtain the graph of the given​ function, stretch the graph of the basic function vertically by a factor of __.

C. To obtain the graph of the given​ function, stretch the graph of the basic function horizontally by a factor of __

D. To obtain the graph of the given​ function, translate the graph of the basic function to the right by __ units.

E. To obtain the graph of the given​ function, translate the graph of the basic function to the left by __units.

F.To obtain the graph of the given​ function, shrink the graph of the basic function vertically by a factor of __

Answer 1

The normal period of sec x is 2*pi. The 12 in front of the x will make the function run through its values 12 times faster, so the new period will be 1/12th of 2*pi. More generally, a horizontal scaling by a factor of k is accomplished by replacing x with x/k. To get from sec x to sec 12x we must replace x with x / (1/12) = 12x, so the horizontal scale factor is 1/12. (Some texts may say, we shrunk it by a factor of 12, but this can be a little misleading, or ambiguous.)

Note that sec x = 1 / cos x has a vertical asymptote wherever cos x = 0, or at every odd multiple of pi / 2. Thus, applying the scale factor of k = 1 / 12, the graph of sec 12x will have vertical asymptotes at every odd multiple of

pi / 2 * 1 / 12 = pi / 24.