My understanding about checking if the function value lies on the graph is putting the x value and checking if it lies on the graph. Now in this example I'm a bit confused, when I solve a b and c, a gives function which will cover values greater than 1, and c will do the reverse. I think b option will cover the values in both + and -ve x axis. Am I interpreting it correctly?
From inspection, it is clear the graph passes through $(0, 0)$, the origin. Neither $y = \vert 3x-3\vert$ nor $y = \vert3x-1\vert$ satisfy this.
Looking at option $(B)$, at $x = 0$, you have $x < 3$, so $y = -x$, meaning $y = 0$. Hence, $(B)$ satisfies this condition. Checking the other two points, you can see both $(3, -3)$ and $(6, 0)$ satisfy $y = x-6$ (because $x \geq 3$). Hence, option $(B)$ is correct.
This function is known known as a “piecewise function” since the function contains two “sub-functions” which apply for a certain part of the domain. If $x \geq 3$, you have a different function than if $x < 3$ (and they’re independent of each other). The absolute value function is a piecewise function because $\vert x\vert = x$ if $x \geq 0$ and $\vert x\vert = -x$ if $x < 0$, and option $(B)$’s function is actually the absolute value function $y = \vert x-3\vert -3$ but it’s written as two separate functions.
$$x\geq 3 \implies x-3 \geq 0 \implies y = x-3-3 \implies y = x-6$$
$$x < 3 \implies x-3 < 0 \implies y = -(x-3)-3 \implies y = -x+3-3 \implies y = -x$$