Please help solve this

Question

Which of the following are two of the vertical asymptotes for the function f(x)=3tan(4x)-1?

Answer 1

The -1 is a vertical shift (or displacement). It moves the function up and down (down one unit in this case), but does not change the x-value of any asymptotes.

Infinity minus 1 is still infinity. Negative infinity minus 1 is still negative infinity.

The 3 is a vertical stretch. It does not change the x-value of any asymptotes. Three times infinity is still infinity. Three times negative infinity is still negative infinity.

Let g(u) = tan(u). Then g(u) has asymptotes closest to the y-axis at ±(1/2)π.

If we set u=4x, then tan(u) =

tan(4x) has asymptotes closest to the y-axis at ±(1/2)(π)/4 = ±(1/8)π.

Thus, f(x) = 3tan(4x) - 1 has asympyotes closest to the y-axis at the same values.

So the correct answer is (B).

Answer 2

Hi Kay G

You just need to plug your choices into the function. Make sure you are using radians.

Keep in mind that Tangent is (sin/cos) so whenever your 4x yields a value in radians that has a cosine of zero, the function is undefined (because of a domain error) while any of the other choices will return a value. On a calculator you may also see the term domain error.

I hope this helps.