Question

g(x) = x^2 - 4x- 12

What are the x intercepts?

Answer 1

(x-2)^2)-16

(x-6)(x+2)

Answer 2

at first i added -4 and +4 for this question
x^2 - 4x- 12 +4 -4
(x^2 - 4x+4)-16
(x-2)^2)-16
(x-6)(x+2)

Answer 3

the xints occur when g(x) = 0

Answer 4

x=-2
x=6
proof
x^2-4x-12=0
x-intercepts means y=0
comparing with general equation
ax^2+bx+c=0
12*1=12
factors of 12 used = 2 , 6
hence,
x^2-6x+2x-12=0
x(x-6)+2(x-6)=0
(x-6)(x+2)=0
therefore x=-2
x=6

Answer 5

y=2/3x-3.............eqn 1
2x-3y=9............eqn 2
put eqn 1 into eqn 2
2x-3(2/3x-3)=9
2x-2x+9=9
2x-2x+9-9=0
hence x and y has no values