Fit a quadratic function to these three points (-1, -11) (0,-3) and (3,-27)

ANSWER[tex]y = - 4 {x}^{2} + 4x - 3[/tex]EXPLANATIONLet the quadratic function be [tex]y = a {x}^{2} + bx + c[/tex]We substitute each point to find the constants, a,b, and c.Substitute: (x=0,y=-3)[tex] - 3 = a {(0)}^{2} + b(0) + c[/tex][tex] \implies \: c = - 3...(1)[/tex]Substitute: (x=-1,y=-11) and c=-3[tex] - 11 = a {( - 1)}^{2} + b( - 1) + - 3[/tex][tex] \implies \: - 11 = a - b - 3[/tex][tex] \implies \: a - b = - 8...(2)[/tex]Substitute: (x=3,y=-27) and c=-3[tex] -27= a {( 3)}^{2} + b( 3) + - 3[/tex][tex] \implies \: - 27 = 9a + 3b - 3[/tex][tex]\implies \: 3a + b = - 8...(3)[/tex]Add equations (3) and (2)[tex]3a + a = - 8 + - 8[/tex][tex]4a = - 16[/tex][tex]a = - 4[/tex]Put a=-4 in equation (2)[tex] - 4 - b = - 8[/tex][tex] - b = - 8 + 4 [/tex][tex] - b = - 4[/tex][tex]b = 4[/tex]The quadratic equation is [tex]y = - 4 {x}^{2} + 4x - 3[/tex]