Sal's sandwich shop sells wraps and sandwiches as part of its lunch specials. the profit on every sandwich is $2 and the profit on every wrap is $3. sal made a profit of $1,470 from lunch specials last month. the equation 2x + 3y = 1,470 represents sal's profits last month, where x is the number of sandwich lunch specials sold and y is the number of wrap lunch specials sold.
change the equation to slope-intercept form. identify the slope and y-intercept of the equation. be sure to show all your work.
describe how you would graph this line using the slope-intercept method. be sure to write using complete sentences.
write the equation in function notation. explain what the graph of the function represents. be sure to use complete sentences.
graph the function. on the graph, make sure to label the intercepts. you may graph your equation by hand on a piece of paper and scan your work or you may use graphing technology.
suppose sal's total profit on lunch specials for the next month is $1,593. the profit amounts are the same: $2 for each sandwich and $3 for each wrap. in a paragraph of at least three complete sentences, explain how the graphs of the functions for the two months are similar and how they are different.
below is a graph that represents the total profits for a third month. write the equation of the line that represents this graph. show your work or explain how you determined the equations.
graph of line going through ordered pairs (0, 300) and (450, 0)
03.03 linear functions—option 1 rubric
requirements possible points student points
student changes equation to slope-intercept form. student shows all work and identifies the slope and y-intercept of the equation. 4
student writes a description, which is clear, precise, and correct, of how to graph the line using the slope-intercept method. 4
student changes equation to function notation. student explains clearly what the graph of the equation represents. 2
student graphs the equation and labels the intercepts correctly. 4
student writes at least three sentences explaining how the graphs of the two equations are the same and how they are different. 4
student writes the equation of the graph in either form correctly and shows work or explains the process. 2

Step-by-step explanation: For part 1, it makes sense that the form is y = m + c. So, 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 the 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 easy terms, 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 531. 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!

may i see the picture? it would a lot, xox