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.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3