Need help with this function problem!! (not easy)?

Question

Biologists have noticed that the chirping rate of crickets of a certain species is related to temperature, and the relationship appears to be very nearly linear. A cricket produces 113 chirps per minute at 70°F and 173 chirps per minute at 80°F.
a. Find a linear equation that models the temperature T as a function of the number of chirps per minute N

b. what is the slope of this graph?

Answer 1

slope: (173-113) / (80-70) =

60 / 10 =

6

y=mx+b : m=6, pick any of the two points for x,y

113=(70)(6)+b

113=420+b

b= -307



y=6x-307 however it is not logical if the answer produces less than 0 chirps so....

0=6x-307

6x=307

x=51and1/6

now let x=T and y=N

N=6T-307 N>= 51and 1/6 degrees

Answer 2

We want an equation of the form



T = cN + f.



This is the standard form of a linear equation for T as a function of N (we are told that the relationship is very nearly linear). Here, you could use any symbols for c and f (I just choose c for coefficient and f for "freeze", when the cricket chirps zero per minute).



We have two solution points:



80 = 173c + f



and



70 = 113c + f.



Subtracting the second from the first, we have



10 = 60c or c = 1/6 chirps per minute/degree Fahrenheit.



This coefficient value is the slope of the graph (the answer to part b).



To give a closed form for the answer to a, we substitute the value of c in either of the two solution statements, for example,



80 = 173(1/6) + f



or



f = (480 - 173)/6 = 307/6 = 51 1/6 degrees Fahrenheit.



So, the general formula is



T = (N + 307)/6 or T = N/6 + 51 1/6.



You can check this: 113 + 307 is 420 and 420 divided by 6 is 70.



In fact, you know that this linear approximation formula is only good in a limited range. For example, negative chirps per minute at 40 degrees F is not possible. Similarly, a cricket will not chirp at about 658 chirps per minute (more than 10 chirps per second) in a pot of boiling water at 212 degrees F. The really interesting question (which the biologists should answer for us) is what are the lower and upper bounds on this linear relation.

Answer 3

First find the slope of the line given by (y2 - y1) / (x2 - x1) where y is the temperature and x is the chirping rate.



(x1,y1) = (113,70) and (x2,y2) = (173,80)



So (y2 - y1) / (x2 - x1) = (80 - 70) / (173 - 113) = 10 / 60 = 1/6 = slope



The equation of a straight line is y = mx + b where m is the slope and b is the y-intercept.



So far we have y = 1/6 x + b. To find b, plug in one of the two points. I'll use (x1,y1).



y = 1/6 x + b

70 = 1/6 (113) + b

70 = 18.8 + b

51.2 = b



So our final equation is

y = 1/6 x + 51.8