Hence it is a simply rearrangement of the equation to start with, in order to make the subject:
This is the graph in 'slope-intercept' form. From here it is easy to see that gradient = and that y-intercept = 490.
The easiest way to draw a straight-line graph, such as this one, is to plot the y-intercept, in this case (0, 490), then plot another point either side of it at a fair distance (for example substitute = -5 and = 5 to procure two more sets of co-ordinates). These can be joined up with a straight line to form a section of the graph, which would otherwise extend infinitely either side - use the specified range in the question for x-values, and do not exceed it (clearly here the limit of -values is 0 ≤ x ≤ 735, since neither x nor y can be negative within the context of the question - the upper limit was found by substituting = 0).
In function notation, the graph is:
The graph of this function represents how the value of the function varies as the value of x varies. Looking back at the question context, this graph specifically represents how many wraps could have been sold at each number of sandwich sales, in order to maintain the same profit of $1470.
When the profit is higher, the gradient is not changed (this is defined by the relationship between the $2 and $3 prices, not the overall profit) - instead the -intercept is higher:
Therefore we have gleaned that the new y-intercept is.
Clearly I cannot see the third straight line. However the method for finding the equation of a straight line graph is fairly simple:
1. Select two points on the line and write down their coordinates
2. The gradient of the line =
3. Find the change in (Δ
4. Find the change in (Δ
5. Divide the result of stage 3 by the result of stage 4
6. This is your gradient
7. Take one of your sets of coordinates, and arrange them in the form , where your is the gradient you just calculated
8. There is only one variable left, which is (the y-intercept). Simply solve for this
9. Now generalise the equation, in the form , by inputting your gradient and y-intercept whilst leaving the coordinates as and
For example if the two points were (1, 9) and (4, 6):
Δ = 6 - 9 = -3
Δ = 4 - 1 = 3
= = -1
I choose the point (4, 6)
6 = (-1 * 4) + c
6 = c - 4
c = 10
Therefore, generally,
Within the context of the question, I imagine the prices of the two lunch specials will be the same in the third month and hence the gradient will still be - this means steps 1-6 can be omitted. Furthermore if the axes are clearly labelled, you may even be able to just read off the y-intercept and hence dispose with steps 1-8!
over the interval 
is given by:![Average\ rate\ of\ change= \frac{[-(8)^2-4(8)]-[-(6)^2-4(6)]}{8-6} \\ \\ = \frac{(-64-32)-(-36-24)}{2} = \frac{-96-(-60)}{2} = \frac{-96+60}{2} = \frac{-36}{2} \\ \\ -18](https://tex.z-dn.net/?f=Average%5C%20rate%5C%20of%5C%20change%3D%20%5Cfrac%7B%5B-%288%29%5E2-4%288%29%5D-%5B-%286%29%5E2-4%286%29%5D%7D%7B8-6%7D%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B%28-64-32%29-%28-36-24%29%7D%7B2%7D%20%3D%20%5Cfrac%7B-96-%28-60%29%7D%7B2%7D%20%3D%20%5Cfrac%7B-96%2B60%7D%7B2%7D%20%3D%20%5Cfrac%7B-36%7D%7B2%7D%20%20%5C%5C%20%20%5C%5C%20-18)
