Answer:
[tex]f(x)\left \{ {{3x+1 if x\leq 0} \atop {-3x+1 if x>0}} \right.[/tex]
Step-by-step explanation:
So if we first graph the given equation, we'll see the graph I've attached below.
Remember that piecewise functions are functions that change based on the circumstances. I know that sounds super confusing, but it's actually really simple!
In this case, for example, we see the line increasing from -∞ to [tex]0[/tex], and then suddenly going downwards and decreasing. That's a good spot for us to notice because that indicates a change. We notice that the function looks different when [tex]x<0[/tex] or [tex]x>0[/tex]. If you break the function into those two parts, you see that they are just linear equations, but they're only visible when x is either greater than or less than 0.
Now that we notice this pattern, we can find the equation of the lines for both lines.
The points (-3,-8) and (-1,-2) are points on the first line, the one that increases (on the left). We can use those points to find the slope of the first line. Remember the slope equation:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
Plug in your points:
[tex]m=\frac{-2-(-8)}{-1-(-3)}[/tex]
[tex]m=\frac{6}{2}[/tex]
[tex]m=3[/tex]
So, the slope of the first line is 3. The y-intercept, looking at the graph, is 1. The equation of the first line is [tex]y=3x+1[/tex]. We'll need this later.
Let's do the same thing for the second line. Just looking at the graph, we can see that this is the same exact line, just with a negative slope. So, the equation for the second line is [tex]y=-3x+1[/tex].
So now we can set up a piecewise function.
[tex]f(x)\left \{ {{3x+1} \atop {-3x+1}} \right.[/tex]
The two functions in the bracket are the two different functions used in this graph. Now we need to figure out where each function is effective. Well, they share a y-intercept. Remember that a true function cannot have two points with the same x value. So the first function is effective to the left of x=0, while the second is effective to the right of x=0. In other words, when [tex]x\leq 0[/tex], [tex]f(x)=3x+1[/tex]. But, when [tex]x>0[/tex], [tex]f(x)=-3x+1[/tex]. Now our piecewise function looks like this:
[tex]f(x)\left \{ {{3x+1 if x\leq 0} \atop {-3x+1 if x>0}} \right.[/tex]
And that is our piecewise function for the original function.
I know this is confusing, so please let me know if you have any questions! I hope this helps!
