Piecewise function can be explained as a set of subfunctions (each subfunction applied to an interval). Here we can observe a piecewise function that is composed of three sub functions.
-The first subfunction: between 0 and 3 (open interval in 3) there is a quadratic function, we can obtain it from the general form of the quadratic function when it has a vertex on (0,0) which is:
[tex]f(x)=ax^{2}[/tex]
[tex]a[/tex] will be a positive value if the parabola is opening upward and negative if it is opneing downward. This one opens upward so it is positive and it will be =1 because when we make [tex]x=1[/tex] on the graph corresponds to [tex]y=1[/tex] so:
[tex]f(x)=ax^{2}\\1=a(1^{2})\\a=1[/tex]
Our subfunction if [tex]0\leq x<3[/tex] is [tex]f(x)=x^{2}[/tex]
-the second and third subfunctions: both are straight lines defined each in an interval. The first between 3 and 6 (open interval in 3) and the second between 6 and 10. To build this two functions we need to understand the general form of the line function:
[tex]f(x)=mx+b[/tex]
where m is the slope of the line and b is the y intercept.
we calculate first m and b for both lines with the m and b equation (we must take two coordintates of each line ([tex]x_{1},y_{1},x_{2},y_{2}[/tex]):
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]b=y-mx[/tex]
First line (4,6),(5,5)
[tex]m=\frac{5-6}{5-4}}=-1[/tex]
[tex]b=(5)-(-1)(5)=10[/tex]
Second line (8,7),(10,10)
[tex]m=\frac{10-7}{10-8}}=\frac{3}{2}}=1.5[/tex]
[tex]b=(10)-(\frac{3}{2}})(10)=-5[/tex]
Now we can build both the equations:
First Lline ([tex]3<x\leq 6[/tex])
[tex]f(x)=mx+b\\f(x)=(-1)x+10\\f(x)=10-x[/tex]
Second Line([tex]6\leq x\leq 10[/tex]):
[tex]f(x)=mx+b\\f(x)=(\frac{3}{2}})x-5[/tex]
