Y = mx + c
If y = 4, and x = 0, then:
4 = -1 (0) + c
c = 4
If c = 4:
y = -1 (2) + 4
y = -2 +4
y = 2
If you plug in the different values of x in each equation, the answer should be:
2,0,-2
Step one know that across from 128 is 128 because of vertical angles
step two a straight line =180 so 128 -180 = 52
step three know you know 3k+4=52 so subtract 4 from 52 and that gives you 48 then divide by 3 to find k and 48/3 is 16
so k +16
because 16x3 is 48 then add 4 which is 52 and 52 plus 128 is 180

![\sf \left[\begin{array}{cc}\sf 4&\sf 6\\ \sf 5 &\sf 8 \\ \sf 3 &\sf -2\end{array}\right]-\left[\begin{array}{cc}\sf 2&\sf 3\\ \sf 1 &\sf 4 \\ \sf -5&\sf3\end{array}\right]](https://tex.z-dn.net/?f=%5Csf%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%204%26%5Csf%206%5C%5C%20%5Csf%205%20%26%5Csf%208%20%5C%5C%20%5Csf%203%20%26%5Csf%20-2%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%202%26%5Csf%203%5C%5C%20%5Csf%201%20%26%5Csf%204%20%5C%5C%20%5Csf%20-5%26%5Csf3%5Cend%7Barray%7D%5Cright%5D)
Just substract corresponding terms
![\\ \sf\longmapsto \left[\begin{array}{cc}\sf 2 &\sf 3\\ \sf 4&\sf4\\ \sf 8&\sf -5\end{array}\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%202%20%26%5Csf%203%5C%5C%20%5Csf%204%26%5Csf4%5C%5C%20%5Csf%208%26%5Csf%20-5%5Cend%7Barray%7D%5Cright%5D)
Option B
Answer: The graph is attached.
Step-by-step explanation: The given functions whose graphs are to be compared are as follows:

In the attached figure, the graphs of both (A) and (B) are shown. We can easily see see from there, the shapes of both the graphs are same.
But, at x = 0, y = ∞ and at x = ∞, y = 0 in graph (A).
At x = 0, y = ∞ and at x = ∞, y = 6 in graph (B).
Thus, the comparison can be seen in the figure very clearly.