Problem:A non-linear system consists of two functions: f(x)=x²+2x+1 and g(x)=3-x-x². Solve this system in two different ways. Your choices are: Table, Graph, or Algebraically.A. Make a table of values for the functions. The table may be horizontal or vertical but it must have a minimum of five x-values and the corresponding function values showing each solution, one value lower, one value higher, and one between the two solutions. Indicate the solutions by marking the x-values and the corresponding function values that are equal.B. Solve the system algebraically. (Hint: set the two functions equal to each other and solve the resulting function.) You should obtain a quadratic equation. Solve it either by factoring or using the quadratic formula. Give the x-values of the solution set, then evaluate the original function to find the corresponding y-values. Give the results as ordered pairs of exact values.C. Plot a graph of the functions over an interval sufficient to show the solutions. You may carefully sketch or plot your graph manually or use Desmos or other technology. Clearly indicate and label on the graph the x and y values of the solution(s).

Answer:Part B. see the procedurePart C. see the procedureStep-by-step explanation:we have[tex]f(x)=x^{2}+2x+1[/tex] -----> equation A[tex]g(x)=3-x-x^{2}[/tex] -----> equation B  Part B. Solve the system algebraicallyequate the equation A and the equation B[tex]x^{2}+2x+1=3-x-x^{2}[/tex][tex]2x^{2}+3x-2=0[/tex]The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to [tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex] in this problem we have [tex]2x^{2}+3x-2=0[/tex]so [tex]a=2\\b=3\\c=-2[/tex] substitute in the formula [tex]x=\frac{-3(+/-)\sqrt{3^{2}-4a(2)(-2)}} {2(2)}[/tex] [tex]x=\frac{-3(+/-)\sqrt{25}} {4}[/tex] [tex]x=\frac{-3(+/-)5} {4}[/tex] [tex]x1=\frac{-3(+)5} {4}=0.5[/tex] [tex]x2=\frac{-3(-)5} {4}=-2[/tex] Find the values of yFor x=0.5[tex]f(0.5)=0.5^{2}+2(0.5)+1=2.25[/tex]For x=-2[tex]f(-2)=(-2)^{2}+2(-2)+1=1[/tex]the solutions are the points(0.5,2.25) and (-2,1)Part C. Solve the system by graphusing a graphing toolwe know that The solution of the non linear system is the intersection point both graphsThe intersection points are (0.5,2.25) and (-2,1)thereforeThe solutions are the points (0.5,2.25) and (-2,1)see the attached figure