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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?

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    $\begingroup$ Note that $f(0) = 0$, which of the options satisfy this constraint? $\endgroup$ – caverac Nov 30 '18 at 18:17
  • $\begingroup$ @caverac none of these first 3 options are satisfying this condition!? then? $\endgroup$ – shawn k Nov 30 '18 at 18:20
  • $\begingroup$ $(B)$ clearly satisfies that. $\endgroup$ – KM101 Nov 30 '18 at 18:21
  • $\begingroup$ Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$ $\endgroup$ – caverac Nov 30 '18 at 18:21
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    $\begingroup$ 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 $y = -x$. $\endgroup$ – 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.

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

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

  • $\begingroup$ +1 for awesomeness! $\endgroup$ – shawn k Nov 30 '18 at 18:35

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