sketch a graph of the function $f(x)=2cos(x - \frac{3pi}{4})+1$.

Explanation:

Step1: Identify the general form

The general form of a cosine - function is $y = A\cos(Bx - C)+D$. For the function $f(x)=2\cos(x-\frac{3\pi}{4}) + 1$, we have $A = 2$, $B = 1$, $C=\frac{3\pi}{4}$, and $D = 1$.

Step2: Find the amplitude

The amplitude $|A|$ gives the vertical distance from the mid - line to the maximum or minimum of the function. Here, $|A|=2$.

Step3: Find the period

The period of the cosine function $y = A\cos(Bx - C)+D$ is $T=\frac{2\pi}{|B|}$. Since $B = 1$, the period $T = 2\pi$.

Step4: Find the phase shift

The phase shift is given by $\frac{C}{B}$. Here, $\frac{C}{B}=\frac{3\pi}{4}$, so the graph of $y = 2\cos(x)$ is shifted to the right by $\frac{3\pi}{4}$ units.

Step5: Find the vertical shift

The vertical shift is $D = 1$. So the mid - line of the graph is $y = 1$.

Step6: Find key points

We know that for $y=\cos(x)$, the key points in one period $[0,2\pi]$ are $(0,1),(\frac{\pi}{2},0),(\pi,- 1),(\frac{3\pi}{2},0),(2\pi,1)$. For $y = 2\cos(x-\frac{3\pi}{4})+1$: When $x-\frac{3\pi}{4}=0$, i.e., $x=\frac{3\pi}{4}$, $y=2\times1 + 1=3$; When $x-\frac{3\pi}{4}=\frac{\pi}{2}$, i.e., $x=\frac{5\pi}{4}$, $y=2\times0 + 1=1$; When $x-\frac{3\pi}{4}=\pi$, i.e., $x=\frac{7\pi}{4}$, $y=2\times(-1)+1=-1$; When $x-\frac{3\pi}{4}=\frac{3\pi}{2}$, i.e., $x=\frac{9\pi}{4}$, $y=2\times0 + 1=1$; When $x-\frac{3\pi}{4}=2\pi$, i.e., $x=\frac{11\pi}{4}$, $y=2\times1 + 1=3$. Plot these key points and draw a smooth cosine - curve with mid - line $y = 1$, amplitude $2$, and period $2\pi$ shifted to the right by $\frac{3\pi}{4}$ units.

Answer:

The graph of $y = 2\cos(x-\frac{3\pi}{4})+1$ has a mid - line $y = 1$, amplitude $2$, period $2\pi$, and is shifted to the right by $\frac{3\pi}{4}$ units. Key points in one period are $(\frac{3\pi}{4},3),(\frac{5\pi}{4},1),(\frac{7\pi}{4},-1),(\frac{9\pi}{4},1),(\frac{11\pi}{4},3)$. Plot these points and draw a smooth cosine - curve.