what is the equation in standard form of a parabola that models the values in the table x -2 0 4 f(x) 0 -6 78

Answer:  [tex]y=4x^2+5x-6[/tex] Step-by-step explanation: Given points are :   x     =    -2      0       4y=f(x) =    0       -6      78 We have to model the parabola with the help of these pointsSolution: We consider a standard equation of parabola [tex]y= ax^2+bx+c[/tex] now, we put the points in the equation we get,at (-2,0) is    4a-2b+c=0at (0,-6) is    c=-6at (4,78) is    78= 16a+4b+c  now, solving these equation we get, a= 4    , b= 5    , c= -6 so the equation formed with these points is   [tex]y=4x^2+5x-6[/tex] we can see this in the graph attached.