Question

I’m unsure of what this is. help someone

Answer 1

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:

$The \ graph \ of \ the \ equation: \\ \\ y=mx+b \\ \\ is \ a \ line \ whose \ slope \ is \ m \ and \ whose \ y-intercept \ is \ (0, b)$

So the y-intercept here is $(0,4)$ and $m=\frac{4}{5}$, therefore:

$\boxed{y=\frac{4}{5}x+4}$

FROM x = 0 TO x = 5:

From the previous line, we know that at $x=5$ the output is:

$y=\frac{4}{5}x+4 \\ \\ y=\frac{4}{5}(5)+4 \\ \\ y=4+4=8$

So the point $P_{1}(5,8)$ lies on both lines.

For this new line, the slope is $m=-\frac{3}{5}$

So, with the Point-Slope Form of the Equation of a Line we can find the equation of this other line:

$The \ equation \ of \ the \ line \ with \ slope \ m \\ passing \ through \ the \ point \ (x_{1},y_{1}) \ is:\\ \\ y-y_{1}=m(x-x_{1})$

So:

$y-8=-\frac{3}{5}(x-5) \\ \\ y=8-\frac{3}{5}x+3 \\ \\ \boxed{y=-\frac{3}{5}x+11}$

The graph is shown below.

Part 2)

The graph of the linear function $f(x)=ax+b$ is a line with slope $m=a$ and $y-intercept$ at $(0,b)$. From the items, we can assure that the following equations are linear functions:

$\bullet 3y=2x-1.5 \ because \ y=\frac{2}{3}x-\frac{1}{2} \\ \\ \\ \bullet y=1 \ because \ y=m(0)+1 \\ \\ \\ \bullet 5(x+y)=-25 \ because \ 5x+5y=-25 \ \therefore 5y=-5x-25 \therefore y=-x-5$

In conclusion, the other functions are nonlinear and they are:

$\bullet \ y=x^2+3 \\ \\ \bullet \ y=x^3 \\ \\ \bullet \ y=12-x^2$