what are the x - intercepts of the graph of the function (y = 6\tan(\frac{x}{2})-3)? use your graphing…

Answer:

A. $(0.9273 + n\pi,0)$, where $n$ is any integer

Explanation:

Step1: Set $y = 0$

$6\tan(\frac{x}{2})-3 = 0$

Step2: Isolate the tangent - function

$6\tan(\frac{x}{2})=3$, then $\tan(\frac{x}{2})=\frac{3}{6}=\frac{1}{2}$

Step3: Solve for $\frac{x}{2}$

$\frac{x}{2}=\arctan(\frac{1}{2})+n\pi$, where $n\in\mathbb{Z}$ (since the period of the tangent function $y = \tan t$ is $\pi$)

Step4: Solve for $x$

$x = 2\arctan(\frac{1}{2})+2n\pi$. Using a calculator, $\arctan(\frac{1}{2})\approx0.4636$, so $x\approx0.9273 + 2n\pi$. But the general form of the $x$ - intercepts of the tangent - type function $y = A\tan(Bx - C)+D$ with period $\frac{\pi}{|B|}$ (here $B=\frac{1}{2}$, period $= 2\pi$) can also be written as $x=x_0 + nT$ where $x_0$ is a particular solution and $T$ is the period. The particular solution $x_0 = 2\arctan(\frac{1}{2})\approx0.9273$ and the period $T = 2\pi$, so the $x$ - intercepts are of the form $(0.9273 + 2n\pi,0)$ or we can rewrite it in terms of a single - period shift as $(0.9273 + n\pi,0)$ considering the nature of the tangent function's repeating pattern.