# Finding function through a graph 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?

• Note that $f(0) = 0$, which of the options satisfy this constraint? – caverac Nov 30 '18 at 18:17
• @caverac none of these first 3 options are satisfying this condition!? then? – shawn k Nov 30 '18 at 18:20
• $(B)$ clearly satisfies that. – KM101 Nov 30 '18 at 18:21
• Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$ – caverac Nov 30 '18 at 18:21
• Option $(B)$ is what is known as a piecewise function (really just an absolute value function in this case). At $x = 0$, $x < 3$ so you use $y = -x$. – KM101 Nov 30 '18 at 18:22

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.
$$x\geq 3 \implies x-3 \geq 0 \implies y = x-3-3 \implies y = x-6$$
$$x < 3 \implies x-3 < 0 \implies y = -(x-3)-3 \implies y = -x+3-3 \implies y = -x$$