Determine the equation of the asymptotes of

[Baron]

a) y = 2 tan (3mx) + 1

My work:

Let 3mx = a

The equation of the asymptotes of y = tan (a) + 1 is pi/2 + n*pi [where n is an integer]

so a = pi/2 + n*pi
a = 3mx

3mx = pi/2 + n*pi
x = (pi/2 + n*pi) / 3m
x = pi/(6m) + n*pi/ (3m) [where n is an integer]

Is this correct?

b) y = -4 sec(1/3x + pi/12) - 1

My work:

Let 1/3x + pi/12 = a

The equation of the asymptote of y = -4sec(a) - 1 is pi/2 +n*pi [where n is an integer]

so a = pi/2 + n*pi
a = 1/3x + pi/12

1/3x + pi/12 = pi/2 + n*pi
1/3(x+pi/4) = pi/2 + n*pi
x+pi/4 = 3(pi/2 + n*pi)
x = 3pi/2 + 3n*pi - pi/4
x = 6pi/4 + 3n*pi - pi/4
x = 5pi/4 + 3n*pi [ where n is an integer]

Is my answer correct?

 

[Deleted member 4993]

a) y = 2 tan (3mx) + 1

My work:

Let 3mx = a

The equation of the asymptotes of y = tan (a) + 1 is pi/2 + n*pi [where n is an integer]

so a = pi/2 + n*pi
a = 3mx

3mx = pi/2 + n*pi
x = (pi/2 + n*pi) / 3m
x = pi/(6m) + n*pi/ (3m) [where n is an integer] <<<< Correct

Since there is no bound on the domain, "more correct" answer would be

x = pi/(6m) ± n*pi/ (3m)

Is this correct?

b) y = -4 sec(1/3x + pi/12) - 1

My work:

Let 1/3x + pi/12 = a

The equation of the asymptote of y = -4sec(a) - 1 is pi/2 +n*pi [where n is an integer]

so a = pi/2 ± n*pi
a = 1/3x ± pi/12

1/3x + pi/12 = pi/2 + n*pi ? x/3 = 5?/12 ± n*? ? x = 5?/4 ± 3n*?
1/3(x+pi/4) = pi/2 + n*pi
x+pi/4 = 3(pi/2 + n*pi)
x = 3pi/2 + 3n*pi - pi/4
x = 6pi/4 + 3n*pi - pi/4
x = 5pi/4 ± 3n*pi [ where n is an integer]

Is my answer correct?

<<< Yes

 

[Baron]

Thanks for the reply.

With the plus-minus sign, is it necessary to write "n is an integer"?

 

[Deleted member 4993]

Thanks for the reply.

With the plus-minus sign, is it necessary to write "n is an integer"?

Yes - 'n' cannot have a fractional value.

 

[Baron]

If "n" is an integer, that means it can be positive or negative. So what is the point of writing the plus-minus sign?

 

[Deleted member 4993]

So what is the point of writing the plus-minus sign?

Sometimes it is better to emphasize that the values can become negative.