Question 447691
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Just so that no one who reads this later gets lost, let's take it from the top:


You are looking for a linear function, *[tex \Large f(x)] such that if *[tex \Large x\ =\ 25] then *[tex \Large y\ =\ 16] -- that gives you the point (26,16).  Also if *[tex \Large x\ =\ 40] then *[tex \Large y\ =\ 26] -- the point (40, 26).


You did the slope calculation correctly:  *[tex \Large \frac{y_2\ -\ y_1}{x_2\ -\ x_1}\ =\ \frac{26\ -\ 16}{40\ -\ 25}\ =\ \frac{10}{15}\ =\ \frac{2}{3}]


And then, using the point-slope form of the equation of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of a point on the line and *[tex \Large m] is the slope:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 16\ =\ \frac{2}{3}(x\ -\ 25)]


from which you derived:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{2}{3}x\ -\ \frac{2}{3}]


And correctly discovered that the *[tex \Large y]-intercept is *[tex \Large -\frac{2}{3}]


But that IS your function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{2}{3}x\ -\ \frac{2}{3}]


Pick any value you like for *[tex \Large x], insert that value in place of *[tex \Large x], and do the arithmetic.


Let's check it out:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}(25)\ -\ \frac{2}{3}\ =\ \frac{48}{3}\ =\ 16]


That one checks, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}(40)\ -\ \frac{2}{3}\ =\ \frac{78}{3}\ =\ 26]


and this checks, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}(50)\ -\ \frac{2}{3}\ =\ \frac{98}{3}\ =\ 32\frac{2}{3}\ \neq\ 33]


Well, almost, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}(100)\ -\ \frac{2}{3}\ =\ \frac{??}{3}\ =\ ??]


I'll let you do the arithmetic for the last one.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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