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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 $y = vert 3x-3vert$ nor $y = vert3x-1vert$ 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 $vert xvert = x$ if $x geq 0$ and $vert xvert = -x$ if $x < 0$, and option $(B)$’s function is actually the absolute value function $y = vert x-3vert -3$ but it’s written as two separate functions.

$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$

$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = x$$