## College Algebra (6th Edition)

Reading the graphs, $\left[\begin{array}{lll} a_{1}=-4 & & b_{1}=4\\ a_{2}=-2 & & b_{2}=2\\ a_{3}=0 & & b_{3}=0\\ a_{4}=2 & & b_{4}=-2\\ a_{5}=4 & & b_{5}=-4 \end{array}\right]$ $\displaystyle \sum_{i=1}^{5}a_{i}^{2}=(-4)^{2}+(-2)^{2}+0^{2}+2^{2}+4^{2}$ $=16+4+0+4+16=40$ $\displaystyle \sum_{i=1}^{5}b_{i}^{2}=4^{2}+2^{2}+0^{2}+(-2)^{2}+(-4)^{2}=$ $=16+4+0+4+16=40$ $\displaystyle \sum_{i=1}^{5}a_{i}^{2}+\sum_{i=1}^{5}b_{i}^{2}=40+40=80$