Which explains how to show that x= -8/3 Is a root of 9x^2 +48x+64=0? a)Graph 9x^2+48x+64=0 to show that x= -8/3 is where the parabola crosses the y- axis.

b)Graph 9x^2 +48x+64=0 to show that the parabola's vertex is just above the value x=-8/3

c)Substitute x= —8/3 back into 9^2+48x+64=0 and simplify to show the values of a and c remain perfect squares.

d)Substitute x= -8/3 back into 9x^2 +48x+64=0 and simplify to show that a true statement results.