Question

graph the function
$y = \frac{6}{5} \cot x$

options: a, b, c, d (each with a graph)

Explanation:

Step1: Analyze the general form of cotangent function

The general form of the cotangent function is \(y = A\cot(Bx - C)+D\). For the function \(y=\frac{6}{5}\cot x\), we have \(A = \frac{6}{5}\), \(B = 1\), \(C=0\), \(D = 0\). The period of the cotangent function \(y = A\cot(Bx)\) is \(\frac{\pi}{|B|}\), so the period of \(y=\frac{6}{5}\cot x\) is \(\pi\) (since \(B = 1\)). The vertical asymptotes of the basic cotangent function \(y=\cot x\) are at \(x = n\pi\), \(n\in\mathbb{Z}\). For \(y=\frac{6}{5}\cot x\), the vertical asymptotes are also at \(x=n\pi\), \(n\in\mathbb{Z}\). When \(x=\frac{\pi}{2}\), \(y=\frac{6}{5}\cot(\frac{\pi}{2})=0\). The amplitude - like factor \(|A|\) affects the "steepness" of the graph. Since \(A=\frac{6}{5}\), the graph of \(y = \frac{6}{5}\cot x\) is a vertical stretch of the graph of \(y=\cot x\) by a factor of \(\frac{6}{5}\).

Step2: Eliminate wrong options

• Option A: The basic cotangent function \(y=\cot x\) has a graph that decreases in each period. If we consider the sign, when \(x\in(0,\pi)\), \(\cot x>0\) for \(x\in(0,\frac{\pi}{2})\) and \(\cot x < 0\) for \(x\in(\frac{\pi}{2},\pi)\). Option A has positive values on both sides of the \(y\) - axis in the interval \((-\pi,\pi)\) which is wrong for \(y=\frac{6}{5}\cot x\).
• Option B: The basic cotangent function \(y = \cot x\) is positive in \((0,\frac{\pi}{2})\) and negative in \((\frac{\pi}{2},\pi)\). Option B has the wrong sign pattern.
• Option D: When \(x=\frac{\pi}{2}\), \(y = 0\). Option D does not pass through the point \((\frac{\pi}{2},0)\)

Answer:

C.