3.2 Linear Functions

Example 3.3

Part 1

We have two points, so we simply compute the slope using (rise/run). These points are of the form (yearly salary, income tax).

Pt1 = {15000,579};
Pt2 = {15500, 605};

N[(Pt2[[2]]-Pt1[[2]])/(Pt2[[1]]-Pt1[[1]])]

N[(Pt1[[2]]-Pt2[[2]])/(Pt1[[1]]-Pt2[[1]])]

Part 2

Now that we have the slope, we can simply determine how far up from 579 we go if we start with $15,000 and go to the right $350 to end up at $15,350 (the run).  Then we add this quantity to the rise to get:

350*0.052

%+579

So the extra tax was $18.20  going from $15,000 to $15,350, thus the total tax for earning $15,350 is $597.20.

Example 3.4

It costs $192.00 to rent a booth, but you sell necklaces at the booth for $32.00 each.

Part 1

Income is revenue minus costs.  The only cost is the $192.00, the income is $32.00 *n, where n is the number of necklaces sold.

Part 2

We have described the formula in words (mostly) from Part 1.

P[n_]:=32*n-192

Part 3

P(25) is the function notation for the net income of selling 25 necklaces, and as we see below, this is $608.00

P[25]

Parts 4,5,6

Vertical intercept is when n=0, i.e. P(0) = -192

P[0]

Horizontal intercept is when the output is 0, i.e. P(n) = 0.

Solve[P[n]==0,n]

The intercepts are interpreted as follows:  

Vertical intercept is how much money is earned if no necklaces are sold, in this case the result is a net loss of $192.00. 

Horizontal intercept is the break even point, i.e. how many necklaces need to be sold to end up with a net income of $0.00.  This was 6 necklaces.

The slope is m=32, and in practical terms is the increase in net income per necklace sold.

AxisPts = {{6,0},{0,P[0]}}

AxisPtsPlot = ListPlot[AxisPts,PlotStyle->{Red}, PlotMarkers->{"▽",16}];

LinePlot = Plot[P[n],{n,0,30},AxesLabel->{"necklaces","revenue"}];

Show[LinePlot,AxisPtsPlot]

Example 3.5

The great temperature conversion problem!  To convert degrees Celsius to degrees Fahrenheit, we need a function F = F(C).  We are given two points: (30,86), and (40,104).

Pt1 = {30,86};
Pt2 = {40,104};

N[(Pt2[[2]]-Pt1[[2]])/(Pt2[[1]]-Pt1[[1]])]

Once we have the slope m=1.8, we need to solve for the intercept.  Note that using either point results in the intercept value of b=32.

Solve[86==1.8*30+b,b]

Solve[104==1.8*40+b,b]

Fa[Ce_]:=1.8*Ce+32

Part 2

We simple need to solve the equation F(C) = 212 for C.  From the Solve command below, we see it is 100 degrees Celsius.

Solve[Fa[Ce] == 212, Ce]

Part 3

The slope is m = 1.8 and tells us that each degree Celsius is equal to 1.8 degrees Fahrenheit as the labels on m are ((degrees F)/(degrees C)).

Exercise 4

Percentage of Americans living below the poverty line can be modeled by the formula:

P(t) = 11.36 + 0.28t

Where t is the time in years since 2000.

P[t_]:=11.36+0.28*t

Part 1

This is simply P(10), which we evaluate to be 14.16% below.

P[10]

Parts 2 and 3

The slope is m=0.28 its labels are in percent/year, which means that the percentage of Americans living below the poverty line rises by 0.28 percent per year.

Exercise 15

The problem states a linear relationship between speed S of a driver and curvature D of the road on which they drive. Here, S = S(D), with S(0) = 46.26.  The speed decreases by 0.746 miles per hour per additional degree of curvature.

Part a.

Here we have a point (0,46.26) which happens to be the vertical intercept.  The slope is given by m=-0.746 with units mph/degree of curvature.  Thus, our equation is

SP[D_]:=-0.746*D+46.26

Part b.

Function notation is S(10) and from the calculation below, we have S(10) = 38.8 mph.

SP[10]