The only thing we need to here to find who is correct is evaluate the function

at

. To do that we are going to replace

with

in the function:





We can conclude that Lynn is correct.

is indeed 14.
Check the picture below.
so let's find the lengths of those two sides in red, since are the length and width of the rectangle.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[-3-(-6)]^2+[6-3]^2}\implies d=\sqrt{(-3+6)^2+(6-3)^2} \\\\\\ d=\sqrt{9+9}\implies \boxed{d=\sqrt{18}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-3%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B%5B-3-%28-6%29%5D%5E2%2B%5B6-3%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-3%2B6%29%5E2%2B%286-3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B9%2B9%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B18%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})~\hfill d=\sqrt{[-2-(-6)]^2+[-1-3]^2} \\\\\\ d=\sqrt{(-2+6)^2+(-1-3)^2}\implies d=\sqrt{16+16}\implies \boxed{d=\sqrt{32}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{(\sqrt{18})(\sqrt{32})}\implies \sqrt{18\cdot 32}\implies \sqrt{576}\implies 24](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29~%5Chfill%20d%3D%5Csqrt%7B%5B-2-%28-6%29%5D%5E2%2B%5B-1-3%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B%28-2%2B6%29%5E2%2B%28-1-3%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B16%2B16%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B32%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20rectangle%7D%7D%7B%28%5Csqrt%7B18%7D%29%28%5Csqrt%7B32%7D%29%7D%5Cimplies%20%5Csqrt%7B18%5Ccdot%2032%7D%5Cimplies%20%5Csqrt%7B576%7D%5Cimplies%2024)
A Non-example is like there we not enough to grapha
You didn't give the fourth zero, but the answer is still false. If you have a root or an imaginary number as a zero, then its conjugate is also a zero. So if 8i is a zero, then -8i must also be a zero, and if 4i is a zero, then -4i must be a zero, with those zeros and -4, the number of zeroes exceeds the number of zeroes that a fourth degree polynomial can have.
Answer: Bears: 1.80 Saints: 1.40
Step-by-step explanation: Hope this helps!!