Question 9 0/1 pt 9 3 19 0 Details Write an equation for the polynomial graphed below u a A uit -4 -3 -2 -1 2 R 4 Q y(x) = Submit QuestionQuestion 10 0/1 pt 9 3 19 0 Details Determine an equation for the pictured graph. Write your answer in factored form and assume the leading coefficient is +1. y = Submit QuestionQuestion 11 0/1 pt 9 3 19 0 Details Pick the graph of the following function. f(2) = 203- 202+2 -5 -4 -3 -2 -1 123 4 5-5 -4 -3-2 1 2 3 4 5 -5 -4 -3 -2 1 12 3 4 5 in to -5 -4 -3 2 3 4 urt Submit QuestionMatch the function with the graph: - v - a a. -10 -9 -8 -7 6 - $ 9 10 b. - 10 -9 -8 -7 6- C. - 10 -9-8-d. 75 8 76 -5 4 -3 -2 - 10 Submit QuestionDetermine the roots of the function x = -2,0, 1 So the possible equation is it is squared y = f(x) - a (x+2) (x-0) (x-1 ) because it bounces back = ax (x+2) (x-1 ) + - x ( x + 2) ( x - 1)? Substitute any point of the graph ( not the x-intercept ) and solve for a. 2 Consider the point ( - 1 , 4 ) : 4 = a (-1) (- 7+2) (-1-1)2 4 = a (-4 ) a = -1 Polynomial Function: Factored Form Thus , x =x' order of roots 4 = - 1x (x+2) (x - 1 ) f(x) =ox(x+2)(x-3) sign of leading x-intercepts (roots) coefficient (-2.0). (0.0). (3.0)Determine the roots of the function X = -2, 1, 3 multi plicity of 2 because it bounces So the possible equation is back y = f (x ) = a (x+2) (x-1) (x- 3) Substitute any point of the graph ( not the x-intercept ) and solve for a. f ( x ) = 17 ( * + 2 ) ? ( x - 1 ) ? ( x - 3 ) : + Input ... Consider the point ( 0 , - 1 ) : - 1 = a (0+2 ) (0 -1 ) (0 -3 ) - 1 - a ( 12 ) 12 Thus , AS 12 X 12 ) ( x - 1 ) ( x- 3 )