An x-intercept is namely a "solution" or "zero" or "root" often called, and when that happens, y = 0, just like with any other x-intercept.

so what is its x-intercept anyway?
y=6x-3
put a point on (0,-3) this is your y intercept
go up 6 and over 1 this should land you at (1,0) do this again and your at (2,6)
go back to the y intercept now go down 6 and over 1, this should land you (-1,-6)
continue to do this until you graph is filled
You are subtracting -16-4
So I would assume your answer would be C
I hope this answer helps!
Answer:
-3
Step-by-step explanation:
Let x = the number
Then 3x = its triple, and the condition is:
x - 8 < 3x
Subtract x from each side
-8 < 2x
Divide each side by 2
-4 < x
Reverse the inequality
x > -4
The smallest integer greater than -4 is -3.
Check:
-3 - 8 < 3(-3)
-11 < -9
It checks.