X= Intercepts HELP!!!!!!?

Question

Find the x-intercepts

Y=x^2 - 3x -4

Y= 2x+4

Two different questions I have tried to work them out I keep getting stuck!!!

Answer 1

to find x intercept, plug 0 for y
0 = x^2 - 3x - 4

factor
0 = (x - 4) (x+1)
x = 4 or -1

0 = 2x + 4
-4 = 2x
x = -2

Answer 2

ok.
in order to get the "x intercepts", you want to find out where the function crosses the x axis. or, where it has the points (x,0)
in order to do that, set each equation, y =0 and solve for x

for the first one
0 = x^2 -3x - 4
x^2 -3x -4 can be factored to (x -4)(x +1)
thus,
x-4 = 0
or
x+1 =0
add 4 to the first one, you get x=4
subtract 1 from the second one, you get x=-1
so the two intercepts are (-1,0) and (4,0)

do the same sort of thing for the second one
let y=0 again and solve for x
0 = 2x + 4
subtract 4 from each side
-4=2x
divide by 2
x = -2
so the x intercept is (-2,0)

Answer 3

You find the x-intercepts by setting y=0 and solving. Doing this for the first equation you get (x-4)(x+1)=0 which means the x-intercepts are 4 and -1. It is even easier for the second equation: x=-2.

Answer 4

The x intercepts are defined when you set y to equal zero these are what x will equal when Y is equal to 0.
Y=X^2-3X-4
what we are going to do is factor this out.
(X+1)*(X-4)
now we will set these to equal zero
X+1=0 X-4=0
X+1-1=0-1 X-4+4=0+4
X=(-1) X=4
so your x intercepts are (-1) and 4

Y=2x+4
we will set Y to equal 0
0=2X+4
0-4=2X+4-4
-4=2X
-4/2=2X/2
-2=X
x-intercept is -2

Answer 5

you have x intercepts when y = 0

y = x^2 - 3x - 4 = 0

x^2 - 3x - 4=(x-4)(x+1)=0

x = 4 or x = -1

intercepts are (4,0) an (-1,0)

2)
Y= 2x+4 = 0
x + 2 = 0
x = -2

intercept is (-2,0)

Answer 6

x intercept is when y = 0

x^2 - 3x - 4 = 0
x^2 - 3x + (3/2)^2 - (3/2)^2 - 4= 0
(x - 3/2)^2 - (9 + 16)/4 = 0
(x - 3/2)^2 - 25/4 = 0
(x - 3/2 - 5/2)(x - 3/2 + 5/2) = 0
(x - 4)(x + 1) = 0
x = 4
x = -1

2x + 4 = 0
x = -2
y = 4