Part 1)
Let's analyze each part of the function:
FROM x = 0 TO x = 5:
We can find the equation of this line by using The Slope-Intercept Form of the Equation of a Line, that states:
[tex]The \ graph \ of \ the \ equation: \\ \\ y=mx+b \\ \\ is \ a \ line \ whose \ slope \ is \ m \ and \ whose \ y-intercept \ is \ (0, b)[/tex]
So the y-intercept here is [tex](0,4)[/tex] and [tex]m=\frac{4}{5}[/tex], therefore:
[tex]\boxed{y=\frac{4}{5}x+4}[/tex]
FROM x = 0 TO x = 5:
From the previous line, we know that at [tex]x=5[/tex] the output is:
[tex]y=\frac{4}{5}x+4 \\ \\ y=\frac{4}{5}(5)+4 \\ \\ y=4+4=8[/tex]
So the point [tex]P_{1}(5,8)[/tex] lies on both lines.
For this new line, the slope is [tex]m=-\frac{3}{5}[/tex]
So, with the Point-Slope Form of the Equation of a Line we can find the equation of this other line:
[tex]The \ equation \ of \ the \ line \ with \ slope \ m \\ passing \ through \ the \ point \ (x_{1},y_{1}) \ is:\\ \\ y-y_{1}=m(x-x_{1})[/tex]
So:
[tex]y-8=-\frac{3}{5}(x-5) \\ \\ y=8-\frac{3}{5}x+3 \\ \\ \boxed{y=-\frac{3}{5}x+11}[/tex]
The graph is shown below.
Part 2)
The graph of the linear function [tex]f(x)=ax+b[/tex] is a line with slope [tex]m=a[/tex] and [tex]y-intercept[/tex] at [tex](0,b)[/tex]. From the items, we can assure that the following equations are linear functions:
[tex]\bullet 3y=2x-1.5 \ because \ y=\frac{2}{3}x-\frac{1}{2} \\ \\ \\ \bullet y=1 \ because \ y=m(0)+1 \\ \\ \\ \bullet 5(x+y)=-25 \ because \ 5x+5y=-25 \ \therefore 5y=-5x-25 \therefore y=-x-5[/tex]
In conclusion, the other functions are nonlinear and they are:
[tex]\bullet \ y=x^2+3 \\ \\ \bullet \ y=x^3 \\ \\ \bullet \ y=12-x^2[/tex]
