![c)y\ge(1)/(4)x-4](https://img.qammunity.org/2023/formulas/mathematics/college/vdpswulsyop14bb3rtzc6ry1jm5191k8ei.png)
Step-by-step explanation
Step 1
let's find the equation of the line
a)slope
the slope of a line is given by:
![\begin{gathered} \text{slope}=\frac{chang\text{e in y}}{\text{change in x}}=(y_2-y_(1|))/(x_2-x_1) \\ \text{where} \\ P1(x_1,y_1) \\ \text{and } \\ \text{P2(x}_2,y_2) \\ \text{are 2 points from the line} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5t5t9bjq4mj0ap28qc4x2cqxmn0hwjjsqw.png)
so
pick up 2 points from the line
let
P1=(0,-4)
P2=(4,-3)
now, replace in the equation to find the slope
![\begin{gathered} \text{slope}=(y_2-y_(1|))/(x_2-x_1) \\ \text{slope}=(-3-(-4))/(4-0)=(-3+4)/(4)=(1)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2zpy4911ij8ll6il1s7q1opwfmp8t8975x.png)
Step 2
b) the equation of the line, it can be found by using the poitn slope formula
![\begin{gathered} y-y_1=m(x-x_1) \\ \text{where m is the slope and } \\ P1(x_1,y_1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lci0viiakh5qrhwu31cv68ma7wpg9vfo8y.png)
then, let
P1=(0,-4)
slope=1/4
replace and isolate y
![\begin{gathered} y-y_1=m(x-x_1) \\ y-(-4)=(1)/(4)(x-0) \\ y+4=(1)/(4)x \\ \text{subtract 4 in both sides} \\ y+4-4=(1)/(4)x-4 \\ y=(1)/(4)x-4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/97d5bmxkcuqmi8zv4p5kydqrl34euub95m.png)
Step 3
finally, we need the shaded region, it means all the values over the line, in other words, the values greater than the function, so
![\begin{gathered} y=(1)/(4)x-4\Rightarrow shaded\text{ region }\Rightarrow y\ge(1)/(4)x-4 \\ \text{the line is continous so}\Rightarrow\ge \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jfvpxdx3gp0hrw3r2oprsaxc3awi0ffitp.png)
therfore, the answer is
![c)y\ge(1)/(4)x-4](https://img.qammunity.org/2023/formulas/mathematics/college/vdpswulsyop14bb3rtzc6ry1jm5191k8ei.png)
I hope this helps you