Hello!
The answer is:
Why?
To graph a piecewise function, we need to graph each of the function that compounds the principal function (piecewise function) for the given values of the domain.
So,
For the first function, we have:
[tex]f(x)=y=3x-5\\\\y=3x-5[/tex]
We have that it's a positive slope line with y-intercept at -5, so, calculating the x-intercept we have by making y equal to 0, we have:
[tex]y=3x-5[/tex]
[tex]0=3x-5[/tex]
[tex]x=\frac{5}{3}=1.67[/tex]
Also, we have the domain for the function:
[tex]y=3x-5\leq -1[/tex]
Which domain is coming from the negative infinite to -1:
Domain: (-∞, -1]
Hence, we have that the function has a positive slope, intercepts the y-axis at (0,-5) and the x-axis at (1.67,0), also, it exists from -∞ to -1.
- For the second function, we have:
[tex]f(x)=y=-2x+3\\\\y=-2x+3[/tex]
We have that it's a negative slope line with y-intercept at 3, so, calculating the x-intercept we have by making y equal to 0, we have:
[tex]y=-2x+3[/tex]
[tex]0=-2x+3[/tex]
[tex]x=\frac{3}{2}[/tex]
Also, we have the domain for the function:
[tex]-2x+3, -1<x<4[/tex]
Which domain is coming from the negative infinite to -1:
Domain: (-1,4)
Hence, we have that the function has a negative slope, intercepts the y-axis at (0,3) and the x-axis at (1.5,0), also, it exists from (-1 to 4)
- For the third function, we have:
[tex]y=2[/tex]
We have that it's a horizontal line, existing from 4.
The domain for the function will be:
[tex]2,x\geq 4[/tex]
or:
Domain: [4,∞)
- Graphing:
First function: Green line
Second function: Black line
Third function: Red line
Note: I have attached two pictures for better understanding, in the first picture we can see the functions that compound the piecewise function without the domain conditions, in the second picture we can see the functions with the domain conditions given by the piecewise function.
Have a nice day!
