91Ó°ÊÓ

First, we test for symmetry with respect to the line \(\theta= \pi/2\), the pole, and the polar axis.\n\nWe replace \(\theta\) with \(-\theta\). If the equation remains unchanged, then it is symmetric about the polar axis. In our case, replacing \(\theta\) with \(-\theta\) in the equation \(r=4 \cos \theta + 4 \sin \theta\), we get \(r = 4 \cos(-\theta) + 4 \sin(-\theta)\), which simplifies to \(r = 4 \cos \theta - 4 \sin \theta\), which is not symmetric about the polar axis because it does not match with the original equation.\n\nNext, we replace \(r\) with \(-r\). If the equation remains unchanged, then it is symmetric about the pole. In our case, replacing \(r\) with \(-r\) we get \(-r = 4 \cos \theta - 4 \sin \theta\), which is not the same as the original equation, thus not symmetric about the pole.\n\nFinally, we replace \(\theta\) with \(\pi - \theta\). If the equation remains unchanged, then it is symmetric about the line \(\theta = \pi/2\). In our equation, replacing \(\theta\) with \(\pi - \theta\), we get \(r = 4 \cos(\pi - \theta) + 4 \sin(\pi - \theta)\), which simplifies to \(r = -4 \cos \theta - 4 \sin \theta\), which is not symmetric about the line \(\theta = \pi/2\).\n\nTherefore, the equation does not exhibit any form of symmetry.

In order to plot the graph, it may be easier to convert to Cartesian coordinates, remembering that \(r^2 = x^2 + y^2\), \(x = r \cos \theta\) and \(y = r \sin \theta\). \n\nSubstituting \(x\) and \(y\) into our equation, we have \(r^2 = 4r \cos \theta + 4r \sin \theta\). Replacing \(r \cos \theta\) with \(x\) and \(r \sin \theta\) with \(y\), we get that \(r^2 = 4x + 4y\). Finally, replacing \(r^2\) with \(x^2 + y^2\), we get \(x^2 + y^2 = 4x + 4y\). Rearranging this equation to a standard circle form, we get \((x-2)^2 + (y-2)^2 = 8\).

Finally, we plot the graph using the Cartesian coordinates. The graph is a circle with center (2,2) and radius \(\sqrt{8}\). This way the polar equation is graphed in x and y coordinates.