Problem 6 on 3.3

Data6 = {{2012, 27870},{2013,29004},{2014,30138}, {2015,31272},{2016,32406}}

Data6PTSPlot = ListPlot[Data6, PlotMarkers->"X"]

Data6[[1]]

Data6[[2]]

Data6[[2]]-Data6[[1]]

Data6[[3]]-Data6[[2]]

Data6[[4]]-Data6[[3]]

Data6[[4]]-Data6[[2]]

(Data6[[4]][[2]]-Data6[[2]][[2]])/(Data6[[4]][[1]]-Data6[[2]][[1]])

(Data6[[5]][[2]]-Data6[[1]][[2]])/(Data6[[5]][[1]]-Data6[[1]][[1]])

so m is 1134, the change in y over the change in x, so to find the equation of the line y=mx+b, we just need to find b.

Data6

Data6[[3]][[2]]==1134*Data6[[3]][[1]]+b

Solve[%,b]

Data6[[5]][[2]]==1134*Data6[[5]][[1]]+b

Solve[%,b]

Remember: Alt+7 gives you a text cell...

The equation for our data appears to be y = 1134*x-2253738, here x is a year, and y is the tuition cost for that year

Data6

Data6LinePlot = Plot[1134*x-2253738,{x,2011,2017},PlotStyle->{Green, Thickness[0.005], Dashed}]

Show[Data6PTSPlot,Data6LinePlot]

F[x_] = 1134*x-2253738

Data6

{F[2012],F[2013],F[2014],F[2015],F[2016]}

F[2021] represents cost (tuition) in year 2021, which comes out to $38,076

F[2021]