# Given the tangent functions of y = 1– 3 \tan (\frac{2x-\pi}{4}), find the

Given the tangent functions of $$\displaystyle{y}={1}–{3}{\tan{{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}}}$$, find the Equation of all of its vertical asymptotes.

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delilnaT
Step 1
Given- $$\displaystyle{y}={1}–{3}{\tan{{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}}}$$
To find- The equation of vertical asymptotes.
Concept Used- To find the vertical asymptotes of the function $$\displaystyle{y}={\frac{{{p}{\left({x}\right)}}}{{{q}{\left({x}\right)}}}}$$, equate q(x) with 0 and solve accordingly.
Step 2
Explanation- Rewrite the given expression,
$$\displaystyle{y}={1}-{3}{\tan{{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}}}$$
Simplify the above expression, we get,
$$\displaystyle{y}={1}-{\frac{{{3}}}{{{\cot{{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}}}}}}$$
$$\displaystyle{y}={\frac{{{\cot{{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}}}-{3}}}{{{\cot{{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}}}}}}$$
For vertical asymptotes, equate denominator with zero, we get,
$$\displaystyle{\cot{{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}}}={0}$$
$$\displaystyle{\left({\frac{{{2}{x}-\pi}}{{{4}}}}\right)}={0}$$
$$\displaystyle{2}{x}-\pi={0}$$
$$\displaystyle{2}{x}=\pi$$
$$\displaystyle{x}={\frac{{\pi}}{{{2}}}}$$
So, the equation of the vertical asymptotes is $$\displaystyle{x}={\frac{{\pi}}{{{2}}}}$$.
Answer- Hence, the equation of the vertical asymptotes is $$\displaystyle{x}={\frac{{\pi}}{{{2}}}}$$.