graph the trigonometric function.\n\n$y = 2cos3x + 1$\n\nplot all points corresponding to minima and maxima…

Answer:

The key points for one cycle of the function $y = 2 \cos(3x) + 1$ are: Maximum: $(0, 3)$ Midline: $(\frac{\pi}{6}, 1)$ Minimum: $(\frac{\pi}{3}, -1)$ Midline: $(\frac{\pi}{2}, 1)$ Maximum: $(\frac{2\pi}{3}, 3)$

(The user would then plot these points on the provided grid and use the graphing tool to draw the cosine curve through them.)

Explanation:

Step1: Identify function parameters.

The function is $y = A \cos(Bx) + D$ with $A=2$, $B=3$, $D=1$. Amplitude: $|A| = 2$. Period: $P = \frac{2\pi}{|B|} = \frac{2\pi}{3}$. Midline: $y = D = 1$. Maximum value: $D + |A| = 1 + 2 = 3$. Minimum value: $D - |A| = 1 - 2 = -1$.

Step2: Determine x-values for key points.

One cycle starts at $3x=0$ and ends at $3x=2\pi$. Key points occur at $3x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$. $x = 0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}$.

Step3: Calculate y-values for key points.

For $x=0$: $y = 2 \cos(3 \cdot 0) + 1 = 2 \cos(0) + 1 = 2(1) + 1 = 3$. Point: $(0, 3)$ (Maximum). For $x=\frac{\pi}{6}$: $y = 2 \cos(3 \cdot \frac{\pi}{6}) + 1 = 2 \cos(\frac{\pi}{2}) + 1 = 2(0) + 1 = 1$. Point: $(\frac{\pi}{6}, 1)$ (Midline). For $x=\frac{\pi}{3}$: $y = 2 \cos(3 \cdot \frac{\pi}{3}) + 1 = 2 \cos(\pi) + 1 = 2(-1) + 1 = -1$. Point: $(\frac{\pi}{3}, -1)$ (Minimum). For $x=\frac{\pi}{2}$: $y = 2 \cos(3 \cdot \frac{\pi}{2}) + 1 = 2 \cos(\frac{3\pi}{2}) + 1 = 2(0) + 1 = 1$. Point: $(\frac{\pi}{2}, 1)$ (Midline). For $x=\frac{2\pi}{3}$: $y = 2 \cos(3 \cdot \frac{2\pi}{3}) + 1 = 2 \cos(2\pi) + 1 = 2(1) + 1 = 3$. Point: $(\frac{2\pi}{3}, 3)$ (Maximum).