Question

Determine the amplitude and the period for the function. Sketch the graph of the function over one period. $$ y=\cos \left(2 x+\frac{\pi}{4}\right) $$

Short answer

The amplitude of the function \(y=\cos \left(2 x+\frac{\pi}{4}\right)\) is 1, and its period is \(\pi\). The graph of the function can be sketched over one period using key points at \(x=0\), \(x=\frac{\pi}{4}\), \(x=\frac{\pi}{2}\), \(x=\frac{3\pi}{4}\), and \(x=\pi\).

Step-by-step solution

Determine the Amplitude

The amplitude of the function is the absolute value of the coefficient of the cosine term. In our case, the coefficient of the cosine term is 1, so the amplitude is \(|\,1\,|\), which is 1.

Determine the Period

To find the period of the function, we can use the formula: \[\text{Period} = \frac{2\pi}{\left|\text{coefficient of the angle}\right|}\] In our case, the coefficient of the angle, which is the value multiplying x inside the cosine, is 2. So, the period of the function is \(\frac{2\pi}{\left|\,2\,\right|}=\pi\).

Sketch the Graph

To sketch the graph over one period, we need to plot the values of the function in the interval defined by the starting point of the period, which is 0, and the endpoint of the period, which is one period ahead from the starting point: \((0, \pi)\). Let's consider some key points in the sketch: 1. For \(x=0\), the function value is \(\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\). 2. For \(x=\frac{\pi}{4}\), the function value is \(\cos\left(\frac{3\pi}{4}\right)=-\frac{\sqrt2}{2}\). 3. For \(x=\frac{\pi}{2}\), the function value is \(\cos\left(\frac{5\pi}{4}\right)=-\frac{\sqrt2}{2}\). 4. For \(x=\frac{3\pi}{4}\), the function value is \(\cos\left(\frac{7\pi}{4}\right)=\frac{\sqrt2}{2}\). 5. At the end of one period, \(x=\pi\), the function value is \(\cos\left(\frac{9\pi}{4}\right)=\frac{\sqrt2}{2}\). With these points, we can sketch the graph of the function \(y=\cos \left(2 x+\frac{\pi}{4}\right)\) over one period. In conclusion, the function \(y=\cos \left(2 x+\frac{\pi}{4}\right)\) has an amplitude of 1 and a period of \(\pi\).